Question

表1

+-------+--------+--------+
| name  |  age   | gender |
+-------+--------+--------+
| mark  |  20    |  male  |
| john  |  22    |  male  |
| jenny |  21    | female |
+-------+--------+--------+

表2

+-------+----------+---------+
| name2 |  status  | address |
+-------+----------+---------+
| john  |  single  |  miami  |
| mark  |  single  | new york|
| jenny |  single  |  jersy  |
+-------+----------+---------+

我想这样显示...
对于马克的例子：
姓名：马克年龄：20
性别：男现状：单身地址：newyork
请使用“LIKE”不等于

Answer 1

基本上，=比LIKE优先使用，因为LIKE用于模式匹配。

SELECT  a.name, a.age, a.gender,
        b.status, b.address
FROM    table1 a
        INNER JOIN table2 b
            ON a.name = b.name2

要进一步了解联接，请访问以下链接：

Visual Representation of SQL Joins

如果你真的想使用LIKE，

SELECT  a.name, a.age, a.gender,
        b.status, b.address
FROM    table1 a
        INNER JOIN table2 b
            ON a.name LIKE b.name2

SQLFiddle Demo (both queries)

mySQL连接表如果表1列值LIKE表2列值

1 个答案: