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。

只有当第二个连接表的数据明显多于主表时，才会出现这种差异。我有过这样的经历……

我们有一个用户表，有10万个条目，他们的会员数据(友谊)大约有30万个条目。这是一个join语句，目的是获取好友及其数据，但有很大的延迟。但是当成员表中只有少量数据时，它工作得很好。一旦我们将其更改为使用子查询，它就可以正常工作。

但与此同时，连接查询正在处理比主表拥有更少条目的其他表。

所以我认为连接和子查询语句工作得很好，这取决于数据和情况。

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。

现在，许多dbs都可以优化子查询和连接。因此，您只需使用explain检查您的查询，看看哪个更快。如果在性能上没有太大的差异，我更喜欢使用子查询，因为它们简单，更容易理解。

子查询是解决“从A获取事实，以B的事实为条件”这种形式的问题的逻辑正确方法。在这种情况下，在子查询中插入B比进行连接更具逻辑意义。从实际意义上讲，它也更安全，因为您不必担心由于与B的多个匹配而从a获得重复的事实。

然而，实际上，答案通常归结于性能。当给出连接和子查询时，一些优化器会很糟糕，而另一些则相反，这是特定于优化器、特定于dbms版本和特定于查询的。

从历史上看，显式连接通常会胜出，因此已经建立的智慧是连接更好，但优化器一直在变得更好，因此我更喜欢先以逻辑一致的方式编写查询，然后在性能限制的情况下重新构造查询。

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。