MSDN文档SQL Server说

Many Transact-SQL statements that include subqueries can be alternatively formulated as joins. Other questions can be posed only with subqueries. In Transact-SQL, there is usually no performance difference between a statement that includes a subquery and a semantically equivalent version that does not. However, in some cases where existence must be checked, a join yields better performance. Otherwise, the nested query must be processed for each result of the outer query to ensure elimination of duplicates. In such cases, a join approach would yield better results.

所以如果你需要

select * from t1 where exists select * from t2 where t2.parent=t1.id

尝试使用join代替。在其他情况下，这没有什么区别。

我说:为子查询创建函数可以消除混乱的问题，并允许您为子查询实现额外的逻辑。因此，我建议尽可能为子查询创建函数。

代码中的混乱是一个大问题，几十年来业界一直在努力避免它。

我只是在考虑同样的问题，但我在FROM部分使用子查询。 我需要连接和查询大表，“从”表有2800万条记录，但结果只有128个这样小的结果大数据!我在它上面使用MAX()函数。

首先，我使用LEFT JOIN，因为我认为这是正确的方式，mysql可以优化等。 第二次只是为了测试，我重写了针对JOIN的子选择。

LEFT JOIN运行时:1.12s SUB-SELECT运行时间:0.06秒

子选择比连接快18倍!只是在chokito广告。subselect看起来很糟糕，但结果…

我认为在引用的答案中没有强调的是重复的问题和可能由特定(使用)案例引起的有问题的结果。

(尽管马塞洛·坎托斯提到过)

我将引用斯坦福大学Lagunita SQL课程的例子。

学生表

+------+--------+------+--------+
| sID  | sName  | GPA  | sizeHS |
+------+--------+------+--------+
|  123 | Amy    |  3.9 |   1000 |
|  234 | Bob    |  3.6 |   1500 |
|  345 | Craig  |  3.5 |    500 |
|  456 | Doris  |  3.9 |   1000 |
|  567 | Edward |  2.9 |   2000 |
|  678 | Fay    |  3.8 |    200 |
|  789 | Gary   |  3.4 |    800 |
|  987 | Helen  |  3.7 |    800 |
|  876 | Irene  |  3.9 |    400 |
|  765 | Jay    |  2.9 |   1500 |
|  654 | Amy    |  3.9 |   1000 |
|  543 | Craig  |  3.4 |   2000 |
+------+--------+------+--------+

应用表

(向特定大学及专业申请)

+------+----------+----------------+----------+
| sID  | cName    | major          | decision |
+------+----------+----------------+----------+
|  123 | Stanford | CS             | Y        |
|  123 | Stanford | EE             | N        |
|  123 | Berkeley | CS             | Y        |
|  123 | Cornell  | EE             | Y        |
|  234 | Berkeley | biology        | N        |
|  345 | MIT      | bioengineering | Y        |
|  345 | Cornell  | bioengineering | N        |
|  345 | Cornell  | CS             | Y        |
|  345 | Cornell  | EE             | N        |
|  678 | Stanford | history        | Y        |
|  987 | Stanford | CS             | Y        |
|  987 | Berkeley | CS             | Y        |
|  876 | Stanford | CS             | N        |
|  876 | MIT      | biology        | Y        |
|  876 | MIT      | marine biology | N        |
|  765 | Stanford | history        | Y        |
|  765 | Cornell  | history        | N        |
|  765 | Cornell  | psychology     | Y        |
|  543 | MIT      | CS             | N        |
+------+----------+----------------+----------+

让我们试着找出申请计算机科学专业的学生的平均绩点(不论大学)

使用子查询:

select GPA from Student where sID in (select sID from Apply where major = 'CS');

+------+
| GPA  |
+------+
|  3.9 |
|  3.5 |
|  3.7 |
|  3.9 |
|  3.4 |
+------+

这个结果集的平均值是:

select avg(GPA) from Student where sID in (select sID from Apply where major = 'CS');

+--------------------+
| avg(GPA)           |
+--------------------+
| 3.6800000000000006 |
+--------------------+

使用连接:

select GPA from Student, Apply where Student.sID = Apply.sID and Apply.major = 'CS';

+------+
| GPA  |
+------+
|  3.9 |
|  3.9 |
|  3.5 |
|  3.7 |
|  3.7 |
|  3.9 |
|  3.4 |
+------+

该结果集的平均值:

select avg(GPA) from Student, Apply where Student.sID = Apply.sID and Apply.major = 'CS';

+-------------------+
| avg(GPA)          |
+-------------------+
| 3.714285714285714 |
+-------------------+

It is obvious that the second attempt yields misleading results in our use case, given that it counts duplicates for the computation of the average value. It is also evident that usage of distinct with the join - based statement will not eliminate the problem, given that it will erroneously keep one out of three occurrences of the 3.9 score. The correct case is to account for TWO (2) occurrences of the 3.9 score given that we actually have TWO (2) students with that score that comply with our query criteria.

在某些情况下，除了性能问题，子查询似乎是最安全的方法。

MSDN文档SQL Server说

Many Transact-SQL statements that include subqueries can be alternatively formulated as joins. Other questions can be posed only with subqueries. In Transact-SQL, there is usually no performance difference between a statement that includes a subquery and a semantically equivalent version that does not. However, in some cases where existence must be checked, a join yields better performance. Otherwise, the nested query must be processed for each result of the outer query to ensure elimination of duplicates. In such cases, a join approach would yield better results.

所以如果你需要

select * from t1 where exists select * from t2 where t2.parent=t1.id

尝试使用join代替。在其他情况下，这没有什么区别。

我说:为子查询创建函数可以消除混乱的问题，并允许您为子查询实现额外的逻辑。因此，我建议尽可能为子查询创建函数。

代码中的混乱是一个大问题，几十年来业界一直在努力避免它。

在大多数情况下，join比子查询快，子查询比子查询快的情况非常罕见。

在join中，RDBMS可以为您的查询创建一个更好的执行计划，并可以预测应该加载哪些数据来处理并节省时间，不像子查询，它将运行所有的查询并加载所有的数据来进行处理。

子查询的好处是它们比join更可读:这就是为什么大多数新学习SQL的人更喜欢它们;这是简单的方法;但是当涉及到性能时，join在大多数情况下更好，尽管它们也不难读。

如果你想用join加速你的查询:

对于“inner join/join”， 不要使用where条件，而是使用“ON”条件。 例如:

     select id,name from table1 a  
   join table2 b on a.name=b.name
   where id='123'

 Try,

    select id,name from table1 a  
   join table2 b on a.name=b.name and a.id='123'

对于“左/右连接”， 不要在“ON”条件下使用，因为如果你使用左/右连接，它将获得任何一个表的所有行。所以，在"开"里也没用。所以，尝试使用“Where”条件