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Free Oracle Tuning Book

with 2 comments

Quick Start Guide to Oracle Query TuningWho can resist a free Rich Nimeiec book on SQL Tuning? O.K., those who know everything can resist. If you’re like me, this is an opportunity to learn from Rich. Click on the book image or this link to get a free copy, or if you want to pay $10 for a copy click here to buy Quick Start Guide to Oracle Query Tuning: Tips for DBAs and Developers from Amazon.com.

The book is four chapters long, is a 129 pages in length, and is in a PDF format. The outline is:

  1. Query Tuning: Developer and Beginning DBA
  2. Query Tuning: Basics for DBAs and Developers
  3. Advanced Performance Tuning
  4. Tips for Tuning When You Have Everything Tuned

Enjoy reading it. His more comprehensive book is Oracle Database 11g Release 2 Performance Tuning Tips & Techniques (Oracle Press) and it’s $30, but it’s written for an advanced audience (more or less OCA or higher).

Written by maclochlainn

August 31st, 2015 at 11:24 am

Use an object in a query?

without comments

Using an Oracle object type’s instance in a query is a powerful capability. Unfortunately, Oracle’s SQL syntax doesn’t make it immediately obvious how to do it. Most get far enough to put it in a runtime view (a subquery in the FROM clause), but then they get errors like this:

SELECT	 instance.get_type()
         *
ERROR AT line 4:
ORA-00904: "INSTANCE"."GET_TYPE": invalid identifier

The problem is how Oracle treats runtime views, which appears to me as a casting error. Somewhat like the ORDER BY clause irregularity that I noted in July, the trick is complete versus incomplete syntax. The following query fails and generates the foregoing error:

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SELECT instance.get_type() AS object_type
,      instance.to_string() AS object_content
FROM  (SELECT dependent()AS instance
       FROM   dual);

If you add a table alias, or name, to the runtime view on line 4, it works fine:

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SELECT cte.instance.get_type() AS object_type
,      cte.instance.to_string() AS object_content
FROM  (SELECT dependent() AS instance
       FROM   dual) cte;

That is the trick. You use an alias for the query, which assigns the alias like a table reference. The reference lets you access instance methods in the scope of a query. Different columns in the query’s SELECT-list may return different results from different methods from the same instance of the object type.

You can also raise an exception if you forget the open and close parentheses for a method call to a UDT, which differs from how Oracle treats no argument functions and procedures. That type of error would look like this:

SELECT cte.instance.get_type AS object_type
       *
ERROR AT line 1:
ORA-00904: : invalid identifier

It is an invalid identifier because there’s no public variable get_type, and a method is only found by using the parenthesis and a list of parameters where they’re required.

The object source code is visible by clicking on the expandable label below.

As always, I hope this helps those solving problems.

Written by maclochlainn

August 22nd, 2015 at 5:23 pm

ORDER BY CASE

with 8 comments

Sometimes I give parts of a solution to increase the play time to solve a problem. I didn’t anticipate a problem when showing how to perform a sort operation with a CASE statement. It’s a sweet solution when you need to sort something differently than a traditional ascending or descending sort.

I gave my students this ORDER BY clause as an example:

  CASE
    WHEN filter = 'Debit' THEN 1
    WHEN filter = 'Credit' THEN 2
    WHEN filter = 'Total' THEN 3
  END;

It raises the following error in MySQL for students:

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ORDER BY
  CASE
    WHEN filter = 'Debit' THEN 1
    WHEN filter = 'Credit' THEN' at line 6

It raises the following error in Oracle for some students:

  CASE
  *
ERROR AT line 7:
ORA-01785: ORDER BY item must be the NUMBER OF a SELECT-list expression

So, I built a little test case to replicate the problem and error message they encountered:

SQL> SELECT 'Debit' AS filter FROM dual
  2  UNION ALL
  3  SELECT 'Credit' AS filter FROM dual
  4  UNION ALL
  5  SELECT 'Total' AS filter FROM dual
  6  ORDER BY
  7    CASE
  8      WHEN filter = 'Debit' THEN 1
  9      WHEN filter = 'Credit' THEN 2
 10      WHEN filter = 'Total' THEN 3
 11    END;

They said, great but how can you fix it? That’s simple, with a Common Table Expression (CTE) in Oracle or with an inline view in MySQL. The Oracle CTE solution is:

  1  WITH results AS
  2  (SELECT 'Debit' AS filter FROM dual
  3   UNION ALL
  4   SELECT 'Credit' AS filter FROM dual
  5   UNION ALL
  6   SELECT 'Total' AS filter FROM dual)
  7  SELECT filter
  8  FROM   results
  9  ORDER BY
 10    CASE
 11	 WHEN filter = 'Debit'  THEN 1
 12	 WHEN filter = 'Credit' THEN 2
 13	 WHEN filter = 'Total'  THEN 3
 14    END;

There are two MySQL solutions. One simply removes the FROM dual clauses from the query components and the other uses an inline view in the FROM clause. This is the inline view:

SELECT filter
FROM  (SELECT 'Debit' AS filter FROM dual
       UNION ALL
       SELECT 'Credit' AS filter FROM dual
       UNION ALL
       SELECT 'Total' AS filter FROM dual) resultset
ORDER BY
  CASE
    WHEN filter = 'Debit' THEN 1
    WHEN filter = 'Credit' THEN 2
    WHEN filter = 'Total' THEN 3
  END;

This is the solution without the FROM dual clauses:

SELECT 'Debit' AS filter
UNION ALL
SELECT 'Credit' AS filter
UNION ALL
SELECT 'Total' AS filter
ORDER BY
  CASE
    WHEN filter = 'Debit' THEN 1
    WHEN filter = 'Credit' THEN 2
    WHEN filter = 'Total' THEN 3
  END;

Both MySQL solutions yield the following:

+--------+
| filter |
+--------+
| Debit  |
| Credit |
| Total  |
+--------+
3 rows in set (0.00 sec)

It puts the fabricating query inside a result set, and then lets you use the column alias to filter the set. If you have a better approach, please share it.

Written by maclochlainn

July 8th, 2015 at 10:06 pm

Mac SQL Developer Install

without comments

This how you install SQL Developer on Mac OS Yosemite. The first thing you need to do is download and install Java 8, not Java 7 on your Mac OS Yosemite as suggested on some web sites. You can determine whether or not Java is installed by running the following command:

Mac-Pro-3:~ username$ java -version
No Java runtime present, requesting install.

You must accept the Java license to install Java 8 on the Mac OS X operating system:

YosemiteInstallJava_01

You have the option of installing the Java SDK or JDK. I’ve opted to install Netbeans 8 with JDK 8u45, as you can tell from the screen capture after you launched the file:

YosemiteInstallJava_02

It is a standard Mac OS installation, which is why I didn’t bother showing any dialog messages. After installing the Java JDK or SDK, you should download SQL Developer 4.1 from Oracle’s web site. Below is a screen shot of the Oracle download web page where I’ve accepted the license agreement:

SQLDeveloperDownload

If you attempt to launch the installation and you’ve set your Mac Security to the “Mac App Store and identified developers” setting, you should raise the following exception:

SQLDeveloperInstall_01

If you reset the Mac Security to an “Anywhere” setting, you can install Oracle SQL Developer on Yosemite. Just make sure you reset it to the “Mac App Store and identified developers” setting after you install SQL Developer.

If you launch SQL Developer with the Security “Anywhere” setting, it displays the following dialog:

SQLDeveloperInstall_02

After you launch the program, you will see the following progress dialog:

SQLDeveloperInstall_03

The last step of the installation launches SQL Developer, as shown below:

SQLDeveloperInstall_04

Click the Connections icon to create an initial connection, like the following:

SQLDeveloperInstall_05

After connecting to the database, you can write and execute a query as shown in the next screen capture:

SQLDeveloperInstall_06

As always, I hope that this helps those who require an example to install SQL Server on a Mac OS.

Written by maclochlainn

June 12th, 2015 at 3:08 am

Bash Arrays & Oracle

with 2 comments

Last week, I wrote about how to use bash arrays and the MySQL database to create unit and integration test scripts. While the MySQL example was nice for some users, there were some others who wanted me to show how to write bash shell scripts for Oracle unit and integration testing. That’s what this blog post does.

If you don’t know much about bash shell, you should start with the prior post to learn about bash arrays, if-statements, and for-loops. In this blog post I only cover how to implement a bash shell script that runs SQL scripts in silent mode and then queries the database in silent mode and writes the output to an external file.

I’ve copied the basic ERD for the example because of a request from a reader. In their opinion it makes cross referencing the two posts unnecessary.

LittleERDModel

To run the bash shell script, you’ll need the following SQL files, which you can see by clicking not he title below. There are several differences. For example, Oracle doesn’t support a DROP IF EXISTS syntax and requires you to write anonymous blocks in their PL/SQL language; and you must explicitly issue a QUIT; statement even when running in silent mode unlike MySQL, which implicitly issues an exit.

If you don’t have a sample test schema to use to test this script, you can create a sample schema with the following create_user.sql file. The file depends on the existence of a users and temp tablespace.

Click the link below to see the source code for a script that let’s you create a sample user account as the system user:

The following list_oracle.sh shell script expects to receive the username, password, and fully qualified path in that specific order. The script names are entered manually in the array because this should be a unit test script.

This is an insecure version of the list_oracle.sh script because you provide the password on the command line. It’s better to provide the password as you run the script.

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#!/usr/bin/bash
 
# Assign user and password
username="${1}"
password="${2}"
directory="${3}"
 
echo "User name:" ${username}
echo "Password: " ${password}
echo "Directory:" ${directory}
 
# Define an array.
declare -a cmd
 
# Assign elements to an array.
cmd[0]="actor.sql"
cmd[1]="film.sql"
cmd[2]="movie.sql"
 
# Call the array elements.
for i in ${cmd[*]}; do
  sqlplus -s ${username}/${password} @${directory}/${i} > /dev/null
done
 
# Connect and pipe the query result minus errors and warnings to the while loop.
sqlplus -s ${username}/${password} @${directory}/tables.sql 2>/dev/null |
 
# Read through the piped result until it's empty.
while IFS='\n' read actor_name; do
  echo $actor_name
done
 
# Connect and pipe the query result minus errors and warnings to the while loop.
sqlplus -s ${username}/${password} @${directory}/result.sql 2>/dev/null |
 
# Read through the piped result until it's empty.
while IFS='\n' read actor_name; do
  echo $actor_name
done

The IFS (Internal Field Separator) works with whitespace by default. The IFS on lines 29 and 37 sets the IFS to a line return ('\n'). That’s the trick to display the data, and you can read more about the IFS in this question and answer post.

You can run the shell script with the following syntax:

./list_oracle.sh sample sample /home/student/Code/bash/oracle > output.txt

You can then display the results from the output.txt file with the following command:

cat output.txt command:

It will display the following output:

User name: sample
Password:  sample
Directory: /home/student/Code/bash/oracle
 
Table Name
------------------------------
MOVIE
FILM
ACTOR
 
Actors in Films
----------------------------------------
Chris Hemsworth, Thor
Chris Hemsworth, Thor: The Dark World
Chris Pine, Star Trek
Chris Pine, Star Trek into Darkness
Chris Pratt, Guardians of the Galaxy

As always, I hope this helps those looking for a solution.

Written by maclochlainn

May 21st, 2015 at 1:16 am

Bash Arrays & MySQL

with 2 comments

Student questions are always interesting! They get me to think and to write. The question this time is: “How do I write a Bash Shell script to process multiple MySQL script files?” This post builds the following model (courtesy of MySQL Workbench) by using a bash shell script and MySQL script files, but there’s a disclaimer on this post. It shows both insecure and secure approaches and you should avoid the insecure ones.

LittleERDModel

It seems a quick refresher on how to use arrays in bash shell may be helpful. While it’s essential in a Linux environment, it’s seems not everyone masters the bash shell.

Especially, since I checked my Learning the Bash Shell (2nd Edition) and found a typo on how you handle arrays in the bash shell, and it’s a mistake that could hang newbies up (on page 161). Perhaps I should update my copy because I bought it in 1998. 😉 It was good then, and the new edition is probably better. The error is probably corrected in the current Learning the Bash Shell, but if not, the following examples show you how to use arrays in loops.

Naturally, these do presume some knowledge of working with bash shell, like the first line always is the same in any bash shell script. That you open an if-statement with an if and close it with a fi, and that you else-if is elif; and that a semicolon between a for-statement and the do statement is required when they’re on the same line because they’re two statements.

If you’re new to bash shell arrays, click on the link below to expand a brief tutorial. It takes you through three progressive examples of working with bash arrays.

Only one more trick needs to be qualified before our main MySQL examples. That trick is how you pass parameters to a bash shell script. For reference, this is the part that’s insecure because user command histories are available inside the Linux OS.

Here’s a hello_whom.sh script to demonstrates the concept of parameter passing:

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#!/usr/bin/bash
 
# This says hello to the argument while managing no argument.
if [[ ${#} = 1 ]]; then
  echo 'The '${0}' program says: "Hello '${1}'!"'
elif [[ ${#} > 1 ]]; then
  echo 'The '${0}' program wants to know if you have more than one name?'
else
  echo 'The '${0}' program wants to know if you have a name?'
fi

If you need more on how parameters are passed and managed, you can check a prior blob post on Handling bash Parameters, or check the bash help pages. The following leverages bash arrays to run scripts and query the MySQL database from the command line.

You will need the three batch SQL files first, so here they are:

The following list_mysql.sh shell script expects to receive the username, password, database and fully qualified path in that specific order. The script names are entered manually because this should be a unit test script. Naturally, you can extend the script to manage those parameters but as mentioned I see this type of solution as a developer machine only script to simplify unit testing. Anything beyond that is risky!

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#!/usr/bin/bash
 
# Assign user and password
username="${1}"
password="${2}"
database="${3}"
directory="${4}"
 
# List the parameter values passed.
echo "Username:  " ${username}
echo "Password:  " ${password}
echo "Database:  " ${database}
echo "Directory: " ${directory}
echo ""
 
# Define an array.
declare -a cmd
 
# Assign elements to an array.
cmd[0]="actor.sql"
cmd[1]="film.sql"
cmd[2]="movie.sql"
 
# Call the array elements.
for i in ${cmd[*]}; do
  mysql -s -u${username} -p${password} -D${database} < ${directory}/${i} > /dev/null 2>/dev/null
done
 
# Connect and pipe the query result minus errors and warnings to the while loop.
mysql -u${username} -p${password} -D${database} <<<'show tables' 2>/dev/null |
 
# Read through the piped result until it's empty but format the title.
while IFS='\n' read list; do
  if [[ ${list} = "Tables_in_sampledb" ]]; then
    echo $list
    echo "----------------------------------------"
  else
    echo $list
  fi
done
echo ""
 
# Connect and pipe the query result minus errors and warnings to the while loop.
mysql -u${username} -p${password} -D${database} <<<'SELECT CONCAT(a.actor_name," in ",f.film_name) AS "Actors in Films" FROM actor a INNER JOIN movie m ON a.actor_id = m.actor_id INNER JOIN film f ON m.film_id = f.film_id' 2>/dev/null |
 
# Read through the piped result until it's empty but format the title.
while IFS='\n' read actor_name; do
  if [[ ${actor_name} = "Actors in Films" ]]; then
    echo $actor_name
    echo "----------------------------------------"
  else
    echo $actor_name
  fi
done

The IFS (Internal Field Separator) works with whitespace by default. The IFS on lines 33 and 47 sets the IFS to a line return ('\n'). That’s the trick to display the data, and you can read more about the IFS in this question and answer post.

You can run this script with the following input parameters from the local directory where you deploy it. The a parameters are: (1) username, (2) password, (3) database, and (4) a fully qualified path to the SQL setup files.

./list_mysql.sh student student sampledb "/home/student/Code/bash/mysql"

With valid input values, the list_mysql.sh bash script generates the following output, which confirms inputs and verifies actions taken by the scripts with queries:

Username:   student
Password:   student
Database:   sampledb
Directory:  /home/student/Code/bash/mysql
 
Tables_in_sampledb
----------------------------------------
actor
film
movie
 
Actors in Films
----------------------------------------
Chris Hemsworth in Thor
Chris Hemsworth in Thor: The Dark World
Chris Pine in Star Trek
Chris Pine in Star Trek into Darkness
Chris Pine in Guardians of the Galaxy

If you forgot to provide the required inputs to the list_mysql.sh bash script, it alternatively returns the following output:

Username:  
Password:  
Database:  
Directory: 
 
./list_mysql.sh: line 25: /actor.sql: No such file or directory
./list_mysql.sh: line 25: /film.sql: No such file or directory
./list_mysql.sh: line 25: /movie.sql: No such file or directory

The secure way removes the password at a minimum! The refactored program will require you to manually enter the password for all elements of the array (three in this sample), and twice for the two queries. Here’s the refactored code:

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#!/usr/bin/bash
 
# Assign user and password
username="${1}"
database="${2}"
directory="${3}"
 
# List the parameter values passed.
echo "Username:  " ${username}
echo "Database:  " ${database}
echo "Directory: " ${directory}
echo ""
 
# Define an array.
declare -a cmd
 
# Assign elements to an array.
cmd[0]="actor.sql"
cmd[1]="film.sql"
cmd[2]="movie.sql"
 
# Call the array elements.
for i in ${cmd[*]}; do
  mysql -s -u${username} -p -D${database} < ${directory}/${i} > /dev/null 2>/dev/null
done
 
# Connect and pipe the query result minus errors and warnings to the while loop.
mysql -u${username} -p -D${database} <<<'show tables' 2>/dev/null |
 
# Read through the piped result until it's empty.
while IFS='\n' read list; do
  if [[ ${list} = "Tables_in_sampledb" ]]; then
    echo $list
    echo "----------------------------------------"
  else
    echo $list
  fi
done
echo ""
 
# Connect and pipe the query result minus errors and warnings to the while loop.
mysql -u${username} -p -D${database} <<<'SELECT CONCAT(a.actor_name," in ",f.film_name) AS "Actors in Films" FROM actor a INNER JOIN movie m ON a.actor_id = m.actor_id INNER JOIN film f ON m.film_id = f.film_id' 2>/dev/null |
 
# Read through the piped result until it's empty.
while IFS='\n' read actor_name; do
  if [[ ${actor_name} = "Actors in Films" ]]; then
    echo $actor_name
    echo "----------------------------------------"
  else
    echo $actor_name
  fi
done

Please let me know if you think there should be any more scaffolding for newbies in this post. As always, I hope this helps those looking for this type of solution.

Written by maclochlainn

May 17th, 2015 at 12:01 pm

Leaf node queries

without comments

A reader posted A dynamic level limiting hierarchical query about Oracle’s hierarchical queries. They wanted to know how to capture only the hierarchy above the level where the first leaf node occurs. They gave me the following hierarchy map as an example:

               1                                    2
        +-------------+                       +-----------+
        |             |                       |           |      
        3             5                       4           6
    +---------+    +-----------+           +-----+    +------+
    |         |    |           |           |     |    |      |
    7         9    11          13          8     10   12     14
+-----+   +-----+  +--+    +-------+                       +-----+ 
|     |   |     |     |    |       |                       |     |
15    17  19    21    23   27      29                     16     18
                                                                 +---+
                                                                     |
                                                                     20

You can find the node values and hierarchical level with the following query:

SELECT   tt.child_id
,        LEVEL
FROM     test_temp tt
WHERE    CONNECT_BY_ISLEAF  = 1
START
WITH     tt.parent_id IS NULL 
CONNECT
BY PRIOR tt.child_id = tt.parent_id
ORDER BY 2;

We really don’t need the node values to solve the problem. We only need the lowest LEVEL value returned by the query, which is 3. The combination of the MIN and CONNECT_BY_ISLEAF functions let us solve this problem without writing a PL/SQL solution. The subquery returns the lowest level value, which is the first level where a leaf node occurs.

SELECT   LPAD(' ', 2*(LEVEL - 1)) || tt.child_id AS child_id
FROM     test_temp tt
WHERE    LEVEL <= (SELECT   MIN(LEVEL)
                   FROM     test_temp tt
                   WHERE    CONNECT_BY_ISLEAF  = 1
                   START
                   WITH     tt.parent_id IS NULL 
                   CONNECT
                   BY PRIOR tt.child_id = tt.parent_id)
START
WITH     tt.parent_id IS NULL
CONNECT
BY PRIOR tt.child_id = tt.parent_id;

It returns:

               1                                    2
        +-------------+                       +-----------+
        |             |                       |           |      
        3             5                       4           6
    +---------+    +-----------+           +-----+    +------+
    |         |    |           |           |     |    |      |
    7         9    11          13          8     10   12     14

While I answered the question in a comment originally, it seemed an important trick that should be shared in its own post.

Written by maclochlainn

April 30th, 2015 at 4:31 pm

MySQL OCP Exams

with 6 comments

Planning out my year, I decided to take the Oracle OCP and MySQL OCP exams. I checked for review books and was pleasantly surprised to find the soon to be released OCP MySQL Database Administrator Exam Guide (Exam 1Z0-883). However, I noticed that the book was actually prepared for the obsolete and discountinued Exams 1Z0-870, 1Z0-873, and 1Z0-874. As it turns out, Steve O’Hearn has informed me that there isn’t a book and that the posting in Amazon.com is in error.

There isn’t an alternative review book for the OCP MySQL 5.6 Developer or Database Administrator Exams. The question that I have is simple: “How relevant is this book because it was prepared for the older exams?” There isn’t a table of content published on the Amazon.com site. If there was a table of contents it could help me determine how close the book’s content is to the new exam.

As a preparation to figure out the value of the book as a study guide, I’ve reviewed the current Oracle MySQL Training Objectives (listed below). The new MySQL OCP Developer and Administrator exams have the following descriptions and objectives:

  • MySQL 5.6 Developer 1Z0-882. Oracle provides the following outline for their MySQL for Developer (Ed 3) training course:

    Course Objectives

    • Describe the MySQL client/server architecture
    • Use MySQL client programs and common options
    • Program MySQL applications with Java and PHP connectors
    • Use a “NoSQL” approach to store and retrieve data
    • Design efficient tables
    • Create and delete database objects
    • Use expressions in SQL statements
    • Examine database metadata
    • Use SQL statements to modify table data
    • Maintain database integrity with transactions
    • Write multiple table queries
    • Create “virtual tables” containing specific data
    • Create user-defined variables, prepared statements, and stored routines
    • Create and manage triggers
    • Identify and deal with errors and exceptions in client programs
    • Write queries that are fast and effective, even under heavy loads
  • MySQL 5.6 Database Administrator 1Z0-883. Oracle provides the following outline for their MySQL for Database Administrators (Ed 3.1) training course:

    Course Objectives

    • Describe the MySQL Architecture
    • Install and Upgrade MySQL
    • Use the INFORMATION_SCHEMA database to access metadata
    • Perform the MySQL start and shutdown operations
    • Configure MySQL server options at runtime
    • Use available tools including MySQL Workbench
    • Evaluate data types and character sets for performance issues
    • Understand data locking in MySQL
    • Understand the use of the InnoDB storage engine with MySQL
    • Maintain integrity of a MySQL installation
    • Use triggers for administration tasks
    • Use Enterprise Audit and Pluggable Authentication
    • Configure advanced replication topologies for high availability
    • Describe introductory performance tuning techniques
    • Perform backup and restore operations
    • Automate administrative tasks with scheduled events

    As always, I hope this helps those who read it; and, in this case I hope it helps you make an effective decision on preparation resources for the MySQL 5.6 OCP exams.

Written by maclochlainn

April 24th, 2015 at 12:39 am

Oracle Cleanup a Schema

with one comment

Back in January 2014, I wrote a script to cleanup an Oracle student schema. It worked well until I started using APEX 4 in my student schema. You create the following 75 objects when you create an APEX 4 schema.

OBJECT TYPE    TOTAL
------------ -------
TABLE		  17
INDEX		  28
SEQUENCE	   5
TRIGGER 	  14
LOB		   9
FUNCTION	   2

Here’s the modified script that ignores the objects created automatically by Oracle APEX when you create a student workspace:

BEGIN
  FOR i IN (SELECT    object_name
            ,         object_type
            ,         last_ddl_time
            FROM      user_objects
            WHERE     object_name NOT IN
                       ('APEX$_WS_WEBPG_SECTION_HISTORY','APEX$_WS_WEBPG_SECTIONS_T1'
                       ,'APEX$_WS_WEBPG_SECTIONS_PK','APEX$_WS_WEBPG_SECTIONS'
                       ,'APEX$_WS_WEBPG_SECHIST_IDX1','APEX$_WS_TAGS_T1'
                       ,'APEX$_WS_TAGS_PK','APEX$_WS_TAGS_IDX2','APEX$_WS_TAGS_IDX1'
                       ,'APEX$_WS_TAGS','APEX$_WS_ROWS_T1','APEX$_WS_ROWS_PK'
                       ,'APEX$_WS_ROWS_IDX','APEX$_WS_ROWS','APEX$_WS_NOTES_T1'
                       ,'APEX$_WS_NOTES_PK','APEX$_WS_NOTES_IDX2','APEX$_WS_NOTES_IDX1'
                       ,'APEX$_WS_NOTES','APEX$_WS_LINKS_T1','APEX$_WS_LINKS_PK'
                       ,'APEX$_WS_LINKS_IDX2','APEX$_WS_LINKS_IDX1','APEX$_WS_LINKS'
                       ,'APEX$_WS_HISTORY_IDX','APEX$_WS_HISTORY','APEX$_WS_FILES_T1'
                       ,'APEX$_WS_FILES_PK','APEX$_WS_FILES_IDX2','APEX$_WS_FILES_IDX1'
                       ,'APEX$_WS_FILES','APEX$_ACL_T1','APEX$_ACL_PK','APEX$_ACL_IDX1'
                       ,'APEX$_ACL','CUSTOM_AUTH','CUSTOM_HASH','DEPT','EMP'
                       ,'UPDATE_ORDER_TOTAL')
            AND NOT ((object_name LIKE 'DEMO%' OR
                      object_name LIKE 'INSERT_DEMO%' OR
                      object_name LIKE 'BI_DEMO%') AND
                      object_type IN ('TABLE','INDEX','SEQUENCE','TRIGGER'))
            AND NOT (object_name LIKE 'SYS_LOB%' AND object_type = 'LOB')
            AND NOT (object_name LIKE 'SYS_C%' AND object_type = 'INDEX')
            ORDER BY object_type DESC) LOOP
 
    /* Drop types in descending order. */
    IF i.object_type = 'TYPE' THEN
 
      /* Drop type and force operation because dependencies may exist. Oracle 12c
         also fails to remove object types with dependents in pluggable databases
         (at least in release 12.1). Type evolution works in container database
         schemas. */
      EXECUTE IMMEDIATE 'DROP '||i.object_type||' '||i.object_name||' FORCE';
 
    /* Drop table tables in descending order. */
    ELSIF i.object_type = 'TABLE' THEN
 
      /* Drop table with cascading constraints to ensure foreign key constraints
         don't prevent the action. */
      EXECUTE IMMEDIATE 'DROP '||i.object_type||' '||i.object_name||' CASCADE CONSTRAINTS PURGE';
 
      /* Oracle 12c ONLY: Purge the recyclebin to dispose of system-generated
         sequence values because dropping the table doesn't automatically 
         remove them from the active session.
         CRITICAL: Remark out the following when working in Oracle Database 11g. */
      EXECUTE IMMEDIATE 'PURGE RECYCLEBIN';
 
    ELSIF i.object_type = 'LOB' OR i.object_type = 'INDEX' THEN
 
      /* A system generated LOB column or INDEX will cause a failure in a
         generic drop of a table because it is listed in the cursor but removed
         by the drop of its table. This NULL block ensures there is no attempt
         to drop an implicit LOB data type or index because the dropping the
         table takes care of it. */
      NULL;
 
    ELSE
 
      dbms_output.put_line('DROP '||i.object_type||' '||i.object_name||';');
      /* Drop any other objects, like sequences, functions, procedures, and packages. */
      EXECUTE IMMEDIATE 'DROP '||i.object_type||' '||i.object_name;
 
    END IF;
  END LOOP;
END;
/

As always, I hope this helps others.

Written by maclochlainn

April 19th, 2015 at 7:13 pm

MySQLdb Manage Columns

without comments

Sometimes trying to keep a post short and to the point raises other questions. Clearly, my Python-MySQL Program post over the weekend did raise a question. They were extending the query example and encountered this error:

TypeError: range() integer end argument expected, got tuple.

That should be a straight forward error message because of two things. First, the Python built-in range() function manages a range of numbers. Second, the row returned from a cursor is actually a tuple (from relational algebra), and it may contain non-numeric data like strings and dates.

The reader was trying to dynamically navigate the number of columns in a row by using the range() function like this (where row was a row from the cursor or result set):

    for j in range(row):

Naturally, it threw the type mismatch error noted above. As promised, the following Python program fixes that problem. It also builds on the prior example by navigatung an unknown list of columns. Lines 16 through 31 contain the verbose comments and programming logic to dynamically navigate the columns of a row.

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#!/usr/bin/python
 
# Import sys library.
import MySQLdb
import sys
 
try:
  # Create new database connection.
  db = MySQLdb.connect('localhost','student','student','studentdb')
  # Create a result set cursor.
  rs = db.cursor()
  rs.execute("SELECT item_title, item_subtitle, item_rating FROM item")
  # Assign the query results to a local variable.
  for i in range(rs.rowcount):
    row = rs.fetchone()
    # Initialize variable for printing row as a string.
    data = ""
    # Address an indefinite number of columns.
    count = 0
    for j in range(len(row)):
      # Initialize column value as an empty string.
      datum = ""
      # Replace column values when they exist.
      if str(row[count]) != 'None':
        datum = str(row[count])
      # Append a comma when another column follows.
      if count == len(row) - 1:
        data += datum
      else:
        data += datum + ", "
      count += 1
    # Print the formatted row as a string.
    print data
except MySQLdb.Error, e:
  # Print the error.
  print "ERROR %d: %s" % (e.args[0], e.args[1])
  sys.exit(1)
finally:
  # Close the connection when it is open.
  if db:
    db.close()

There are a couple Python programming techniques that could be perceived as tricks. Line 24 checks for a not null value by explicitly casting the column’s value to a string and then comparing its value against the string equivalent for a null. The MySQLdb returns a 'None' string for null values by default. The if-block on lines 27 through 30 ensure commas aren’t appended at the end of a row.

While the for-loop with a range works, I’d recommend you write it as a while-loop because its easier to read for most new Python programmers. You only need to replace line 20 with the following to make the change:

20
    while (count < len(row)):

Either approach generates output like:

The Hunt for Red October, Special Collectornulls Edition, PG
Star Wars I, Phantom Menace, PG
Star Wars II, Attack of the Clones, PG
Star Wars II, Attack of the Clones, PG
Star Wars III, Revenge of the Sith, PG-13
The Chronicles of Narnia, The Lion, the Witch and the Wardrobe, PG
RoboCop, , Mature
Pirates of the Caribbean, , Teen
The Chronicles of Narnia, The Lion, the Witch and the Wardrobe, Everyone
MarioKart, Double Dash, Everyone
Splinter Cell, Chaos Theory, Teen
Need for Speed, Most Wanted, Everyone
The DaVinci Code, , Teen
Cars, , Everyone
Beau Geste, , PG
I Remember Mama, , NR
Tora! Tora! Tora!, The Attack on Pearl Harbor, G
A Man for All Seasons, , G
Hook, , PG
Around the World in 80 Days, , G
Harry Potter and the Sorcerer's Stone, , PG
Camelot, , G

As always, I hope this helps those looking for clarity.

Written by maclochlainn

April 13th, 2015 at 10:05 pm