2011-03-26 60 views
0

您好请帮我看看关于这个查询,我从不同的表中获取数据我面临的问题是,在表中有类似的员工拥有的colum名称,用户也有名称。查询工作完美,但我wordering如何,我可以显示此数据为从内部连接查询获取数据

$data["employee.name"] 
$data["user.name"] 

这里是查询:

SELECT task.employee_id , task.user_id , task.service_id, user.name, 
     user.pic_path , employee.name ,employee.pic_path 
FROM task 
INNER JOIN employee ON employee.pno = task.employee_id 
INNER JOIN user ON user.pno = task.user_id 
INNER JOIN service ON service.service_id = task.service_id "; 

回答

4
SELECT user.name AS username, employee.name AS employeename 

你明白了吧。

+0

是的,谢谢,我知道了 – umar 2011-03-26 19:00:58

+0

如果你同意,接受为正确答案;)干杯。 – 2011-03-26 19:07:00

2

有两个步骤:您需要定义列别名在SQL语句中的两列的至少一个

  1. SELECT t.employee_id, 
         t.user_id, 
         t.service_id, 
         u.name AS user_name, 
         u.pic_path, 
         e.name AS employee_name, 
         e.pic_path 
        FROM TASK t 
        JOIN EMPLOYEE e ON e.pno = t.employee_id 
        JOIN USER u ON ur.pno = t.user_id 
        JOIN SERVICE s ON s.service_id = t.service_id 
    
  2. 然后,你需要更新PHP逻辑使用的列别名:

    $empname = $data["employee_name"]; 
    $username = $data["user_name"];