Question

我有两个表，我从中选择1我想获取用户信息，2我想获取属于用户的所有图像。但查询不检索图像，但在查询窗口中我得到它们。同样在脚本中，如果我决定从图像表中选择图像，但是当我进行加入时，它不起作用。我知道有些事情做得不好。请任何帮助将不胜感激贝娄是代码

$query='SELECT 
tish_clientinfo.lastname, tish_clientinfo.address,
tish_clientinfo.firstname,
tish_images.image_name
  FROM  tish_clientinfo 
  INNER JOIN tish_images 
ON tish_clientinfo.user_id = tish_images.user_id
 WHERE user_id= '. intval($_GET['user_id']);
$result = $con->prepare($query);
$result->execute();

Answer 1

可能是服务器对user_id感到困惑。试试这个

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

Answer 2

你刚刚在user_id上的where子句之后遇到一个小问题就像我在下面的回答中所做的那样指定表

$query='SELECT 
tish_clientinfo.lastname, tish_clientinfo.address,
tish_clientinfo.firstname,
tish_images.image_name
  FROM  tish_clientinfo 
  INNER JOIN tish_images 
ON tish_clientinfo.user_id = tish_images.user_id
 WHERE tish_clientinfo.user_id= '. intval($_GET['user_id']);
$result = $con->prepare($query);
$result->execute();

查询在mysqlyog查询窗口中工作，但不在PHP脚本中？

2 个答案: