下面是选自 "Websites" 表的数据：
+----+--------------+---------------------------+-------+---------+
| id | name | url | alexa | country |
+----+--------------+---------------------------+-------+---------+
| 1 | Google | https://www.google.cm/ | 1 | USA |
| 2 | 淘宝 | https://www.taobao.com/ | 13 | CN |
| 3 | 菜鸟教程 | http://www.runoob.com/ | 4689 | CN |
| 4 | 微博 | http://weibo.com/ | 20 | CN |
| 5 | Facebook | https://www.facebook.com/ | 3 | USA |
+----+--------------+---------------------------+-------+---------+
下面是 "access_log" 网站访问记录表的数据：
mysql> SELECT * FROM access_log;
+-----+---------+-------+------------+
| aid | site_id | count | date |
+-----+---------+-------+------------+
| 1 | 1 | 45 | 2016-05-10 |
| 2 | 3 | 100 | 2016-05-13 |
| 3 | 1 | 230 | 2016-05-14 |
| 4 | 2 | 10 | 2016-05-14 |
| 5 | 5 | 205 | 2016-05-14 |
| 6 | 4 | 13 | 2016-05-15 |
| 7 | 3 | 220 | 2016-05-15 |
| 8 | 5 | 545 | 2016-05-16 |
| 9 | 3 | 201 | 2016-05-17 |
+-----+---------+-------+------------+
9 rows in set (0.00 sec)
SQL EXISTS 实例
现在我们想要查找总访问量(count 字段)大于 200 的网站是否存在。
我们使用下面的 SQL 语句：
SELECT
Websites
.
name
,
Websites
.
url
FROM
Websites
WHERE
EXISTS
(
SELECT
count
FROM
access_log
WHERE
Websites
.
id
=
access_log
.
site_id
AND
count
>
200
)
;
执行以上 SQL 输出结果如下：
SELECT
Websites
.
name
,
Websites
.
url
FROM
Websites
WHERE
NOT
EXISTS
(
SELECT
count
FROM
access_log
WHERE
Websites
.
id
=
access_log
.
site_id
AND
count
>
200
)
;
执行以上 SQL 输出结果如下：
