PHP不显示MYSQL查询的结果 - PHP not displaying result from MYSQL query

PHP not displaying result from MYSQL query

本文关键字：结果查询 MYSQL 显示 PHP | 更新日期: 2023-09-27

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没有显示MYSQL查询返回的结果，该查询存储在稍后的会话中，代码将显示在一个伪页面上，它会出现以下错误

警告：mysqli_num_rows（）要求参数1为mysqli_result

我的SQL查询正是我需要的，我只需要，所以我真的不想改变它。我只需要一些关于如何让PHP回显@uid的建议有人愿意帮我解决这个问题吗？如果是的话，谢谢。

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

您可以将第一个查询更改为：

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

更新您可以尝试用所有注释的改进修改代码的这一片段。

$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) );
}