Question

你好，请帮我解决这个问题，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colum名称，如员工有，用户也有名称。查询工作完美，但我担心如何将此数据显示为

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

这是查询：

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 ";

Answer 1

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

你明白了。

Answer 2

有两个步骤：

您需要为SQL语句中的两列中的至少一列定义列别名：

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

然后，您需要更新PHP逻辑以使用列别名：

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

从内部联接查询中获取数据

2 个答案: