最近刷到面試題：Mysql 中 exists 和 in 的區別，先說下答案。
下面將主查詢的表稱爲外表；子查詢的表稱爲內表。exists 與 in 的區別如下：
- 子查詢使用 exists，會先進行主查詢，將查詢到的每行數據循環帶入子查詢校驗是否存在，過濾出整體的返回數據；子查詢使用 in，會先進行子查詢獲取結果集，然後主查詢匹配子查詢的結果集，返回數據
- 外表內表相對大小情況不一樣時，查詢效率不一樣：兩表大小相當，in 和 exists 差別不大；內表大，用 exists 效率較高；內表小，用 in 效率較高。
- 不管外表與內表的大小，not exists 的效率一般要高於 not in，跟子查詢的索引訪問類型有關。
建表、造數據，驗證一下以上答案。
# 建表 student1
drop table if exists student1;
create table student1(
sid int primary key auto_increment,
sname varchar(40)
);
# 建存儲過程給表 student1，插入1000條數據
drop procedure if exists addStudent1;
create procedure addStudent1()
BEGIN
declare idx int;
set idx = 1;
while idx <= 1000 DO
insert into student1 values(null, concat('student-', idx));
set idx = idx + 1;
end while;
end;
call addStudent1();
select * from student1;
# 建表 student2
drop table if exists student2;
create table student2(
sid int primary key auto_increment,
sname varchar(40)
);
# 建存儲過程給表 student2，插入100000條數據
drop procedure if exists addStudent2;
create procedure addStudent2()
BEGIN
declare idx int;
set idx = 1;
while idx <= 100000 DO
insert into student2 values(null, concat('student-', idx));
set idx = idx + 1;
end while;
end;
call addStudent2();
select * from student2;
in 與 exists 的查詢 SQL
select count(1) from student1 where sname in (select sname from student2);
select count(1) from student1 where exists (select sname from student2 where student2.sname = student1.sname);
select count(1) from student2 where sname in (select sname from student1);
select count(1) from student2 where exists (select sname from student1 where student2.sname = student1.sname);
執行時間：
[SQL] select count(1) from student1 where sname in (select sname from student2);
受影響的行: 0
時間: 0.092s
[SQL]
select count(1) from student1 where exists (select sname from student2 where student2.sname = student1.sname);
受影響的行: 0
時間: 0.076s
[SQL]
select count(1) from student2 where sname in (select sname from student1);
受影響的行: 0
時間: 14.820s
[SQL]
select count(1) from student2 where exists (select sname from student1 where student2.sname = student1.sname);
受影響的行: 0
時間: 15.144s
結論：student2 大表在內適用 exists，所以第 2 條 SQL 比第 1 條快；student1 小表在內適用 in，所以第 3 條 SQL 比第 4 條快。
not in 與 not exists 的查詢 SQL
select count(1) from student1 where sname not in (select sname from student2);
select count(1) from student1 where not exists (select sname from student2 where student2.sname = student1.sname);
select count(1) from student2 where sname not in (select sname from student1);
select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);
執行時間：
[SQL]
select count(1) from student1 where sname not in (select sname from student2);
受影響的行: 0
時間: 0.079s
[SQL]
select count(1) from student1 where not exists (select sname from student2 where student2.sname = student1.sname);
受影響的行: 0
時間: 0.075s
[SQL] select count(1) from student2 where sname not in (select sname from student1);
受影響的行: 0
時間: 15.797s
[SQL]
select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);
受影響的行: 0
時間: 15.160s
結論：not exists 性能高於 not in
給 student1、student2 sname 字段，加上索引，上述結論仍然成立。
create index idx_1 on student1(sname);
create index idx_2 on student2(sname);
執行時間：
[SQL] select count(1) from student1 where sname in (select sname from student2);
受影響的行: 0
時間: 0.022s
[SQL]
select count(1) from student1 where exists (select sname from student2 where student2.sname = student1.sname);
受影響的行: 0
時間: 0.014s
[SQL]
select count(1) from student2 where sname in (select sname from student1);
受影響的行: 0
時間: 0.379s
[SQL]
select count(1) from student2 where exists (select sname from student1 where student2.sname = student1.sname);
受影響的行: 0
時間: 0.373s
[SQL]
select count(1) from student1 where sname not in (select sname from student2);
受影響的行: 0
時間: 0.006s
[SQL]
select count(1) from student1 where not exists (select sname from student2 where student2.sname = student1.sname);
受影響的行: 0
時間: 0.006s
[SQL]
select count(1) from student2 where sname not in (select sname from student1);
受影響的行: 0
時間: 0.455s
[SQL]
select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);
受影響的行: 0
時間: 0.418s
再細看一下，not in 與 not exists 查詢索引使用情況
not in，子查詢使用了 index_subquery 訪問類型
EXPLAIN EXTENDED select count(1) from student2 where sname not in (select sname from student1);
SHOW WARNINGS;
not exists，子查詢使用了 ref 訪問類型
EXPLAIN EXTENDED select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);
SHOW WARNINGS;
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