sql语句中mysql会话变量值的优先级 [英] Precedence of a mysql session variable value in an sql statement

问题描述

在SQL语句中使用会话变量时，标准行为是什么?

What is the standard behaviour of a session variable when used in an SQL statement.

案例1 :
在下面的示例中，会话变量的行为符合预期.

Case 1:
In the following example, session variable is behaving as expected.

mysql> set @m1=0, @m2=0, @m3=0;
Query OK, 0 rows affected (0.00 sec)

mysql>
mysql> select
    ->  @m1 := 55 m1, @m2 := 42 m2, @m3 := 66 m3,
    ->  @m1, @m2, @m3,
    ->  @b1 := greatest( @m1, @m2, @m3 ) b1,
    ->  @b2 := ( ( @total := @m1 + @m2 + @m3 )
    ->           - ( @b1 + least( @m1, @m2, @m3 ) )) b2,
    ->  @total total;
+----+----+----+------+------+------+------+------+-------+
| m1 | m2 | m3 | @m1  | @m2  | @m3  | b1   | b2   | total |
+----+----+----+------+------+------+------+------+-------+
| 55 | 42 | 66 |   55 |   42 |   66 |   66 |   55 |   163 |
+----+----+----+------+------+------+------+------+-------+
1 row in set (0.00 sec)

mysql>
mysql> set @m1=0, @m2=0, @m3=0;
Query OK, 0 rows affected (0.00 sec)

mysql> select
    ->  @m1 := m1+3, @m2 := m2+3, @m3 := m3+3,
    ->  @m1, @m2, @m3,
    ->  @b1 := greatest( @m1, @m2, @m3 ) b1,
    ->  @b2 := ( ( @total := @m1 + @m2 + @m3 )
    ->           - ( @b1 + least( @m1, @m2, @m3 ) )) b2,
    ->  @total total
    -> from
    ->  ( select 55 m1, 42 m2, 66 m3  union all select 48, 63, 92 ) marks_list
    -> ;
+-------------+-------------+-------------+------+------+------+------+------+-------+
| @m1 := m1+3 | @m2 := m2+3 | @m3 := m3+3 | @m1  | @m2  | @m3  | b1   | b2   | total |
+-------------+-------------+-------------+------+------+------+------+------+-------+
|          58 |          45 |          69 |   58 |   45 |   69 |   69 |   58 |   172 |
|          51 |          66 |          95 |   51 |   66 |   95 |   95 |   66 |   212 |
+-------------+-------------+-------------+------+------+------+------+------+-------+
2 rows in set (0.00 sec)

上面的示例未使用任何聚合函数.

The above example did not use any of aggregate functions.

案例2 :
当使用诸如count，sum，group by之类的聚合函数执行类似的语句时，结果模式将完全不同.

Case 2:
When a similar statement was executed with aggregate functions like count, sum, group by, the results pattern was entirely different.

请在以下位置找到示例: SQL小提琴

Please find an example on: SQL Fiddle

在发布此查询之前，我试图了解 SQL查询操作顺序 .

Before posting this query, I tried to understand the SQL-Query-Order-of-Operations.

推荐答案

如果我理解正确，那么您正在询问select语句中表达式的求值顺序. MySQL 文档很清楚，您不能依赖于此:

If I understand correctly, you are asking about the order of evaluation of expressions in select statements. MySQL documentation is quite clear that you cannot depend on this:

在下面的语句中，您 可能认为MySQL首先会评估@a然后执行 作业第二:

In the following statement, you might think that MySQL will evaluate @a first and then do an assignment second:

SELECT @a, @a:=@a+1, ...;

但是，涉及用户的表达式的求值顺序 变量未定义.

However, the order of evaluation for expressions involving user variables is undefined.

我认为问题是您期望按顺序分配值.这是不正确的.就是说，我必须承认，在使用变量时，确实确实会时不时地做出相同的顺序评估假设，主要是出于懒惰和事实，它通常是有效的.

I think the issue is your expectation that the values are assigned in order. This is incorrect. That said, I must admit that I do make the same sequential evaluation assumption from time-to-time when using variables, primarily out of laziness and the fact that it does generally work.

如果要保证顺序评估，请使用case技巧:

If you want to guarantee sequential evaluation, then use the case trick:

select (case when (@m := 55) is null then null
             when (@m := 42) is null then null
        . . .
        end)

case保证对条件进行顺序评估，直到满足条件为止.与null的比较可确保对所有内容进行评估.

The case guarantees sequential evaluation of the conditions until one is true. The comparison to null ensures that all get evaluated.

这篇关于sql语句中mysql会话变量值的优先级的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！