Archive for October, 2021
MySQL and JavaScript
Sometimes students truly amaze me with their curiosity. The question was simple from their perspective while we were discussing MySQL’s builtin string functions. How would you do something like this JavaScript logic without using literals or session variables?
// Declare a string and substring. var myStr = 'Get me from the string.' var mySubstr = 'me' // Assign the substring to variable by rescuing it from the larger string. var rescued = myStr.substring(myStr.indexOf(mySubstr),myStr.indexOf(mySubstr) + mySubstr.length) // Print the result. print(rescued) |
tested with MongoDB, like
mongo --nodb --norc < parsing.js |
returning:
MongoDB shell version v4.0.20 me bye |
They thought the question would show SQL’s limits as a problem solving and programming language because they didn’t see how MySQL could assign a variable for evaluation in the builtin functions.
They were surprised to see how I showed them that they could do it. Since they disallowed session variables, I built a sample table and inserted the string value in a text column before writing a query with a Common Table Expression (CTE).
The MySQL steps are:
-- Stable testing scenario with table values requires a test table. DROP TABLE IF EXISTS sample; CREATE TABLE sample ( text VARCHAR(100) ); -- Insert the literal string into the testing table. INSERT INTO sample ( text ) VALUES ('Get me from the string.'); -- Test using a WITH clause to place a variable in context for use -- in the query, relying on the fact that a Cartesian set of one -- column and row becomes a new column in all rows of the other -- table's set. WITH struct AS (SELECT 'me' AS result) SELECT SUBSTR(text,INSTR(text,struct.result),LENGTH(struct.result)) AS rescued FROM sample CROSS JOIN struct; |
It returns the following:
+---------+ | rescued | +---------+ | me | +---------+ 1 row in set (0.00 sec) |
Wow, SQL works like a programming language was the response of the class. It’s like anything else in technology, new stuff isn’t as cool as old stuff until you learn how to use it.
Oracle’s Sparse Lists
Oracle’s PL/SQL Programming Language is really quite nice. I’ve written 8 books on it and still have fun coding in it. One nasty little detail about Oracle’s lists, introduced in Oracle 8 as PL/SQL Tables according their documentation, is they rely on sequential numeric indexes. Unfortunately, Oracle lists support a DELETE method, which can create gaps in the sequential indexes.
Oracle calls a sequence without gaps densely populated and a sequence with gaps sparsely populated. This can cause problems when PL/SQL code inadvertently removes elements at the beginning, end, or somewhere in the middle of the list. That’s because a program can then pass the sparsely populated list as a parameter to another stored function or procedure where the developer may traverse the list in a for-loop. That traversal may raise an exception in a for-loop, like this when it has gaps in the index sequence:
DECLARE * ERROR AT line 1: ORA-01403: no data found ORA-06512: AT line 20 |
Oracle’s myriad built-in libraries don’t offer a function to compact a sparsely populated list into a densely populated list. This post provides a compact stored procedure that converts a sparsely populated list to a densely populated list.
The first step to using the compact stored procedure requires that you create an object type in SQL, like this list of 20-character strings:
DROP TYPE list; CREATE OR REPLACE TYPE list IS TABLE OF VARCHAR2(20); / |
Now, you can implement the compact stored procedure by passing the User-Defined Type as it’s sole parameter.
CREATE OR REPLACE PROCEDURE compact ( sparse IN OUT LIST ) IS /* Declare local variables. */ iterator NUMBER; -- Leave iterator as null. /* Declare new list. */ dense LIST := list(); BEGIN /* Initialize the iterator with the starting value, which is necessary because the first element of the original list could have been deleted in earlier operations. Setting the initial iterator value to the first numeric index value ensures you start at the lowest available index value. */ iterator := sparse.FIRST; /* Convert sparsely populated list to densely populated. */ WHILE (iterator <= sparse.LAST) LOOP dense.EXTEND; dense(dense.COUNT) := sparse(iterator); iterator := sparse.NEXT(iterator); END LOOP; /* Replace the input parameter with the compacted list. */ sparse := dense; END; / |
Before we test the compact stored procedure, let’s create deleteElement stored procedure for our testing:
CREATE OR REPLACE PROCEDURE deleteElement ( sparse IN OUT LIST , element IN NUMBER ) IS BEGIN /* Delete a value. */ sparse.DELETE(element); END; / |
Now, let’s use an anonymous block to test compacting a sparsely populated list into a densely populated list. The test program will remove the first, last, and one element in the middle before printing the sparsely populated list’s index and string values. This test will show you gaps in the remaining non-sequential index values.
After you see the gaps, the test program compacts the remaining list values into a new densely populated list. It then prints the new index values with the data values.
DECLARE /* Declare a four item list. */ lv_strings LIST := list('one','two','three','four','five','six','seven'); BEGIN /* Check size of list. */ dbms_output.put_line('Print initial list size: ['||lv_strings.COUNT||']'); dbms_output.put_line('==================================='); /* Delete a value. */ deleteElement(lv_strings,lv_strings.FIRST); deleteElement(lv_strings,3); deleteElement(lv_strings,lv_strings.LAST); /* Check size of list. */ dbms_output.put_line('Print modified list size: ['||lv_strings.COUNT||']'); dbms_output.put_line('Print max index and size: ['||lv_strings.LAST||']['||lv_strings.COUNT||']'); dbms_output.put_line('==================================='); FOR i IN 1..lv_strings.LAST LOOP IF lv_strings.EXISTS(i) THEN dbms_output.put_line('List list index and item: ['||i||']['||lv_strings(i)||']'); END IF; END LOOP; /* Call a procedure by passing current sparse collection and the procedure returns dense collection. */ dbms_output.put_line('==================================='); dbms_output.put_line('Compacting list.'); compact(lv_strings); dbms_output.put_line('==================================='); /* Print the new maximum index value and list size. */ dbms_output.put_line('Print new index and size: ['||lv_strings.LAST||']['||lv_strings.COUNT||']'); dbms_output.put_line('==================================='); FOR i IN 1..lv_strings.COUNT LOOP dbms_output.put_line('List list index and item: ['||i||']['||lv_strings(i)||']'); END LOOP; dbms_output.put_line('==================================='); END; / |
It produces output, like:
Print initial list size: [7] =================================== Print modified list size: [4] Print max index and size: [6][4] =================================== List list index and item: [2][two] List list index and item: [4][four] List list index and item: [5][five] List list index and item: [6][six] =================================== Compacting list. =================================== Print new index and size: [4][4] =================================== List list index and item: [1][two] List list index and item: [2][four] List list index and item: [3][five] List list index and item: [4][six] =================================== |
You can extend this concept by creating User-Defined Types with multiple attributes, which are essentially lists of tuples (to draw on Pythonic lingo).
MySQL SQL Filters
An interesting outcome of teaching SQL is discovering what skills new users require. One that I continuously rediscover is how to build a test case for various elements of SQL. This is a small article on querying with filters in the WHERE clause.
There are several of the exercises in Alan Beaulieu’s Learning SQL, 3rd Edition that would benefit from example setup. For example, Chapter 4 provides a snapshot of the payment table but doesn’t provide any instructions.
You can create an exercise_4_2 table with the following SQL statement if you plan to change the data:
CREATE TABLE exercise_4_2 AS SELECT payment_id , customer_id , amount , payment_date FROM payment WHERE payment_id BETWEEN 101 AND 120; |
Alternatively, you can create an exercise_4_2 view with the following SQL statement if you plan to only query the data:
CREATE VIEW exercise_4_2 AS SELECT payment_id , customer_id , amount , payment_date FROM payment WHERE payment_id BETWEEN 101 AND 120; |
After creating the new exercise_4_2 table or view from the payment table it will hold a subset of data. You can query all the rows from the new exercise_4_2 table or view with this statement:
SELECT * FROM exercise_4_2; |
It returns the following data set:
+------------+-------------+--------+---------------------+ | payment_id | customer_id | amount | payment_date | +------------+-------------+--------+---------------------+ | 101 | 4 | 8.99 | 2005-08-18 05:14:44 | | 102 | 4 | 1.99 | 2005-08-19 02:19:13 | | 103 | 4 | 2.99 | 2005-08-20 09:32:04 | | 104 | 4 | 6.99 | 2005-08-20 12:55:40 | | 105 | 4 | 4.99 | 2005-08-21 04:53:37 | | 106 | 4 | 2.99 | 2005-08-22 13:58:23 | | 107 | 4 | 1.99 | 2005-08-23 07:43:00 | | 108 | 5 | 0.99 | 2005-05-29 07:25:16 | | 109 | 5 | 6.99 | 2005-05-31 11:15:43 | | 110 | 5 | 1.99 | 2005-05-31 19:46:38 | | 111 | 5 | 3.99 | 2005-06-15 22:03:14 | | 112 | 5 | 2.99 | 2005-06-16 08:01:02 | | 113 | 5 | 4.99 | 2005-06-17 15:56:53 | | 114 | 5 | 2.99 | 2005-06-19 04:20:13 | | 115 | 5 | 4.99 | 2005-06-20 18:38:22 | | 116 | 5 | 4.99 | 2005-07-06 09:11:58 | | 117 | 5 | 2.99 | 2005-07-08 20:04:43 | | 118 | 5 | 4.99 | 2005-07-09 01:57:57 | | 119 | 5 | 5.99 | 2005-07-09 07:13:52 | | 120 | 5 | 1.99 | 2005-07-09 08:51:42 | +------------+-------------+--------+---------------------+ 20 rows in set (0.26 sec) |
With the exercise_4_2 table, you can test the exercises 4-1 and 4-2. Here are the two problems:
- The first exercise checks for rows where the customer_id is not equal to 5 and whether the amount is greater than 8 or payment_date is equal to ‘2005-08-23’. You can structure that question as the following query:
mysql> SELECT * -> FROM exercise_4_2 -> WHERE customer_id <> 5 -> AND (amount > 8 OR DATE(payment_date) = '2005-08-23');It would return the following two rows from the exercise_4_2 table:
+------------+-------------+--------+---------------------+ | payment_id | customer_id | amount | payment_date | +------------+-------------+--------+---------------------+ | 101 | 4 | 8.99 | 2005-08-18 05:14:44 | | 107 | 4 | 1.99 | 2005-08-23 07:43:00 | +------------+-------------+--------+---------------------+ 2 rows in set (0.14 sec)
- The second exercise checks for rows where the customer_id is equal to 5 and whether the amount is not greater than 6 or payment_date is not equal to ‘2005-06-19’. You can structure that question as the following query:
mysql> SELECT * -> FROM exercise_4_2 -> WHERE customer_id = 5 -> AND NOT (amount > 6 OR DATE(payment_date) = '2005-06-19');It would return the following eleven rows from the exercise_4_2 table:
+------------+-------------+--------+---------------------+ | payment_id | customer_id | amount | payment_date | +------------+-------------+--------+---------------------+ | 108 | 5 | 0.99 | 2005-05-29 07:25:16 | | 110 | 5 | 1.99 | 2005-05-31 19:46:38 | | 111 | 5 | 3.99 | 2005-06-15 22:03:14 | | 112 | 5 | 2.99 | 2005-06-16 08:01:02 | | 113 | 5 | 4.99 | 2005-06-17 15:56:53 | | 115 | 5 | 4.99 | 2005-06-20 18:38:22 | | 116 | 5 | 4.99 | 2005-07-06 09:11:58 | | 117 | 5 | 2.99 | 2005-07-08 20:04:43 | | 118 | 5 | 4.99 | 2005-07-09 01:57:57 | | 119 | 5 | 5.99 | 2005-07-09 07:13:52 | | 120 | 5 | 1.99 | 2005-07-09 08:51:42 | +------------+-------------+--------+---------------------+ 11 rows in set (0.00 sec)
- The third exercise checks for payment_id and amount values where the amount is either 1.98, 7.98, or 9.98. You can structure that question as the following query:
mysql> SELECT payment_id -> , amount -> FROM payment -> WHERE amount IN (1.98,7.98,9.98);It would return the following seven rows from the payment table:
+------------+--------+ | payment_id | amount | +------------+--------+ | 1482 | 7.98 | | 1670 | 9.98 | | 2901 | 1.98 | | 4234 | 7.98 | | 4449 | 7.98 | | 7243 | 7.98 | | 9585 | 7.98 | +------------+--------+ 7 rows in set (0.00 sec)
- The fourth exercise checks for the first_name and last_name of customers where the last_name contains an ‘A’ in the second position and a ‘W’ after the ‘A’ character. You can structure that question as the following query:
mysql> SELECT first_name -> , last_name -> FROM customer -> WHERE last_name LIKE '_A%W%';The trick to the
WHEREclause is that the ‘%’ looks for zero to many characters in between two strings.It would return the following nine rows from the customer table:
+------------+------------+ | first_name | last_name | +------------+------------+ | JILL | HAWKINS | | ERICA | MATTHEWS | | LAURIE | LAWRENCE | | JEANNE | LAWSON | | KAY | CALDWELL | | JOHN | FARNSWORTH | | SAMUEL | MARLOW | | LAWRENCE | LAWTON | | LEE | HAWKS | +------------+------------+ 9 rows in set (0.10 sec)
As always, I hope this helps those looking for a solution.
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