Question

此代码不返回任何结果，现在显示错误。所有我想要的是如果我点击一个碰撞，它应该带我到一个新的页面，并显示该用户从两个表中选择的所有信息。如果我从1中选择它有效但在两个中它不会

$query ='SELECT  
 tish_clientinfo.client_id, tish_clientinfo.firstname,
 tish_images.image_name
    FROM  tish_clientinfo 
    INNER JOIN tish_images 
ON tish_clientinfo.user_id = tish_images.user_id
WHERE user_id= '. $_GET['user_id'];

Answer 1

ambigued name字段user_id，也许??试试这个：

$query ="SELECT  
tish_clientinfo.client_id, tish_clientinfo.firstname,
tish_images.image_name
  FROM  tish_clientinfo 
  INNER JOIN tish_images 
ON tish_clientinfo.user_id = tish_images.user_id
WHERE tish_clientinfo.user_id= ". intval($_GET['user_id']).";";

Saludos;）

Mysql php查询返回没有结果

1 个答案: