PHP不显示MYSQL查询的结果 [英] PHP not displaying result from MYSQL query

问题描述

我有以下代码

$sql = "SET @uid := (SELECT ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);";
$sql = "UPDATE channels SET Used = 1 WHERE ID = @uid;";
$sql = "SELECT * FROM channels WHERE ID IN = @uid;";
$result = mysqli_multi_query($conn, $sql)
                 or die( mysqli_error($sql) );
if (mysqli_num_rows($result) > 0) {
  $text = '';
  while($row = mysqli_fetch_assoc($result)) {  
      $Channel_Location = $row['Channel_Location'];
      $text =  $text . $Channel_Location;

    }       
}

现在我遇到的问题是php is not显示由MYSQL查询返回的结果，该结果存储在稍后要存储在虚拟页面中的代码中的会话中，它出现以下错误

Now the issue i'm having is the php isnt displaying the result returned by the MYSQL query which is stored in a session later on in the code to be displayed on a dummy page it comes up with the following error

警告:mysqli_num_rows()期望参数1为mysqli_result

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result

我的SQL查询完全可以满足我的需要，所以我真的不想更改它.我只需要一些有关如何使PHP回显@uid的建议 有谁愿意帮助我解决这个问题?如果是这样，谢谢.

The my SQL query does exactly what I need it to I just need to, so I don't really want to change it. I just need some advice on how i'd get the PHP to echo the @uid is there anyone willing to help me solve the issue? if so thankyou.

推荐答案

在$ sql中有3个查询，因此应使用multi_query函数 http://php.net/manual/en/mysqli.multi-query.php

You have 3 queries in your $sql so you should use multi_query function http://php.net/manual/en/mysqli.multi-query.php

您可以将第一个查询更改为:

And you can change your first query to:

SET @uid = 0;
SELECT @uid := ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);

更新，您可以尝试对这段代码进行修改，并添加所有已注释的改进.

Update You can try this fragment of your code modified with all commented improvements.

$sql = 'SET @uid = 0;';
$sql .= 'SELECT @uid:= ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);';
$sql .= 'UPDATE channels SET Used = 1 WHERE ID = @uid;';
$sql .= 'SELECT * FROM channels WHERE ID IN = @uid;';
if (mysqli_multi_query($conn, $sql)) {
   do {
       $result = mysqli_store_result($conn);
   } while(mysqli_next_result($conn));
   if (mysqli_num_rows($result) > 0) {
     $text = '';
     while($row = mysqli_fetch_assoc($result)) {  
       $Channel_Location = $row['Channel_Location'];
       $text =  $text . $Channel_Location;
     }       
   }
} else {
  die( mysqli_error($conn) );
}

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