Question

我有这张桌子：

------------------------------
|ID | name   | employee_code |
------------------------------
|24 | Robert |    20234      |
------------------------------

和

-------------------------------------
|ID |   job_code   |     team       |
-------------------------------------
|24 | 241124 | Robert, Eduard, Etc. |
-------------------------------------

我想通过员工代码在第二个表中搜索，我尝试这样的事情：

$sql=mysql_query("SELECT * FROM works WHERE (SELECT name FROM employee WHERE employee_code LIKE '%".$_GET['employee_code']."%' AS searchname) team Like %searchname% ");

结果：

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

Answer 1

尝试此查询 -

$employee_code = mysql_real_escape_string($_GET['employee_code']);

$sql=mysql_query("SELECT w.* 
                  FROM employee e
                  JOIN works w 
                  ON w.team LIKE CONCAT('%', e.name ,'%')
                  WHERE employee_code LIKE '%$employee_code%'");

请参阅此SQLFiddle示例 - http://sqlfiddle.com/#!2/8f8b7/1

Answer 2

你应该看一个加入。

select * from table1 inner join table2 using (`ID`) where job_code = ....

然后你有一行将两个表连接在一起

还使用mysql_ *函数，不再维护这些函数请更新到mysqli_ *或PDO。

此外，您需要转义查询，等待在该代码中发生的SQL注入攻击

Answer 3

这可能会告诉你究竟出了什么问题。

$sql=mysql_query("SELECT * FROM works WHERE (SELECT name FROM employee WHERE employee_code LIKE '%".$_GET['employee_code']."%' AS searchname) team Like %searchname% ");
if (!$sql)
    echo mysql_error();

你永远不应该假设你的查询有效，然后继续在另一个命令中使用该资源，而不检查它确实有效。

您不应该做的另一件事就是将用户输入直接放入SQL查询中而不进行任何形式的转义，因为它可以让任何人完全控制您的数据库。

SELECT * FROM works WHERE (SELECT name FROM employee WHERE employee_code LIKE '%".mysql_real_escape_string($_GET['employee_code'])."%' AS searchname) team Like %searchname% "

Answer 4

你的SQL查询错误试试这个

 SELECT * FROM works WHERE works.ID=employee.ID AND 
   employee.employee_code=".$_GET['employee_code']."

Answer 5

SELECT * FROM table1 t1 INNER JOIN table2 t2 ON t1.employee_code = t2.job_code

或

SELECT t1.id, t1.name, t2.team FROM table1 t1 INNER JOIN table2 t2 ON t1.employee_code = t2.job_code

获得更清晰的结果

Mysql选择多个查询

5 个答案: