PHP错误消息

Question

PHP Error Message

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a6397779/public_html/app/ta.phtml on line 11

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这是PHP代码（ta.phtml文件）：

<?php
include('app/config.php');
$link = mysql_connect($AppConfig['db']['host'],$AppConfig['db']['user'],$AppConfig['db']['password']) or die(mysql_error());
mysql_select_db($AppConfig['db']['database'],$link) or die(mysql_error());
?>

<?php

$this->myData['id'] = $this->player->playerId;
$result = mysql_query("SELECT Club,gold_num,Adventures,total_people_count FROM p_players where id='".$this->myData['id']."'");
while($row = mysql_fetch_array($result))
  {
$Club = $row['Club'];
$goldClub = $row['gold_num'];
$Adventures = $row['Adventures'];
$total = $row['total_people_count'];
  }

?>

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请帮助!!

Answer 1

你应该在mysql_query（$ sql，$ link）中添加第二个参数。 你应该使用PDO而不是mysql扩展。