SQL LEFT JOIN 关键字
SQL LEFT JOIN 关键字
LEFT JOIN 关键字从左表(table1)返回所有的行,即使右表(table2)中没有匹配。如果右表中没有匹配,则结果为 NULL。
SQL LEFT JOIN 语法
SELECT column_name(s)
FROM table1
LEFT JOIN table2
ON table1.column_name=table2.column_name;
FROM table1
LEFT JOIN table2
ON table1.column_name=table2.column_name;
或:
SELECT column_name(s)
FROM table1
LEFT OUTER JOIN table2
ON table1.column_name=table2.column_name;
FROM table1
LEFT OUTER JOIN table2
ON table1.column_name=table2.column_name;
注释:在某些数据库中,LEFT JOIN 称为 LEFT OUTER JOIN。
演示数据库
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 | | 7 | stackoverflow | http://stackoverflow.com/ | 0 | IND | +----+---------------+---------------------------+-------+---------+
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 LEFT JOIN 实例
下面的 SQL 语句将返回所有网站及他们的访问量(如果有的话)。
以下实例中我们把 Websites 作为左表,access_log 作为右表:
实例
SELECT Websites.name, access_log.count, access_log.date
FROM Websites
LEFT JOIN access_log
ON Websites.id=access_log.site_id
ORDER BY access_log.count DESC;
FROM Websites
LEFT JOIN access_log
ON Websites.id=access_log.site_id
ORDER BY access_log.count DESC;
执行以上 SQL 输出结果如下:
注释:LEFT JOIN 关键字从左表(Websites)返回所有的行,即使右表(access_log)中没有匹配。
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