Logic Value of Assignment Expression

Question

I'm doing the leetcode SQL problems and kind of confused with the code below:

select distinct Num from (
 select
 Num,
 case
 when @prevNum = Num then @count := @count + 1
 when (@prevNum := Num) is not null then @count := 1
 end n
 from Logs, (select @prevNum := NULL) r
 order by Id
) a where n >= 3

Two questions:

First when (@prevNum := Num) is not null what is the logic value of assignment expression in SQL? I have googled MySQL and Oracle reference book, this forum and I got nothing about it. And this condition is really confusing.

Second the n after case end statement, I know n stands for the alias of case statement, but there are two variables in case end statement here, how do SQL know that n stands for @count but not @prevNum?

Thanks

The question is as follows:

Write a SQL query to find all numbers that appear at least three times consecutively.

+----+-----+
| Id | Num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+

For example, given the above Logs table, 1 is the only number that appears consecutively for at least three times.

• user1822: Which DBMS are you using? @prevNum = .. is non-standard SQL (i.e. invalid for most DBMS). And what exactly is the "leetcode SQL problem"?
• Tony Chen: I think they are using MySQL because they support limit but not top, and I have updated the question.

Answer 1

PROPOSED SOLUTION

Using the IF() function to check for consecutive num values

SET @x = 0;
SET @n = 0;
SELECT num,consecutive FROM
(SELECT
 id,num,
 (@x:=IF(num=@n,@x+1,1)) consecutive,
 (@n:=num) inc
FROM logs) A
WHERE consecutive >= 3;

YOUR SAMPLE DATA

DROP DATABASE IF EXISTS tonychen;
CREATE DATABASE tonychen;
USE tonychen
CREATE TABLE logs
(
 id INT NOT NULL AUTO_INCREMENT,
 num INT NOT NULL,
 PRIMARY KEY (id)
);
INSERT INTO logs (num) VALUES (1),(1),(1),(2),(1),(2),(2);
SELECT * FROM logs;

YOUR SAMPLE DATA LOADED

mysql> DROP DATABASE IF EXISTS tonychen;
Query OK, 1 row affected (0.25 sec)
mysql> CREATE DATABASE tonychen;
Query OK, 1 row affected (0.00 sec)
mysql> USE tonychen
Database changed
mysql> CREATE TABLE logs
 -> (
 -> id INT NOT NULL AUTO_INCREMENT,
 -> num INT NOT NULL,
 -> PRIMARY KEY (id)
 -> );
Query OK, 0 rows affected (0.31 sec)
mysql> INSERT INTO logs (num) VALUES (1),(1),(1),(2),(1),(2),(2);
Query OK, 7 rows affected (0.05 sec)
Records: 7 Duplicates: 0 Warnings: 0
mysql> SELECT * FROM logs;
+----+-----+
| id | num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+
7 rows in set (0.00 sec)
mysql>

PROPOSED SOLUTION EXECUTED

mysql> SET @x = 0;
Query OK, 0 rows affected (0.00 sec)
mysql> SET @n = 0;
Query OK, 0 rows affected (0.00 sec)
mysql> SELECT num,consecutive FROM
 -> (SELECT
 -> id,num,
 -> (@x:=IF(num=@n,@x+1,1)) consecutive,
 -> (@n:=num) inc
 -> FROM logs) A
 -> WHERE consecutive >= 3;
+-----+-------------+
| num | consecutive |
+-----+-------------+
| 1 | 3 |
+-----+-------------+
1 row in set (0.00 sec)
mysql>

GIVE IT A TRY !!!

• Tony Chen: Thank you for working it out so quickly. I have tried that on my MySQL and it perfectly works. However, I do know that there are other answers and I'm just confused about the code I posted, which is specifically about the case... when... then clause. Can you do me a favor and answer that?