转载:https://www.runoob.com/sql/sql-exists.html
EXISTS 运算符用于判断查询子句是否有记录,如果有一条或多条记录存在返回 True,否则返回 False。
SELECT column_name(s) FROM table_name WHERE EXISTS (SELECT column_name FROM table_name WHERE condition);
在本教程中,我们将使用 RUNOOB 样本数据库。
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)
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:
实例 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:
EXISTS 可以与 NOT 一同使用,查找出不符合查询语句的记录:
实例 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: