douwo3665 2012-01-30 12:52
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mysql_num_row():提供的参数不是有效的MySQL结果资源

I don't know why this error appears: mysql_num_row():supplied argument is not a valid MySQL result resource

$sqlCheckLevel = 'SELECT * FROM levelPerClass WHERE clsId = \''.$class.'\' AND lvlId = \''.$rowLevel['lvlId'].'\'';
$resCheckLevel = mysql_query($sqlCheckLevel);
print_r($rowCheckLevel = mysql_fetch_assoc($resCheckLevel));
//prints 'Array ( [clsId] => 15 [lvlId] => 18 )'    

if(mysql_num_rows($rowCheckLevel) == 0) {
    //etc

What is wrong?

EDIT: I feel stupid. It needs to be

 if(mysql_num_rows($resCheckLevel) == 0) {
  • 写回答

1条回答 默认 最新

  • drcmg28484 2012-01-30 12:55
    关注

    Use like,

    mysql_num_rows($resCheckLevel) instead

    mysql_num_rows($rowCheckLevel)

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