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我不知道为什么会出现这个错误：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

怎么了？

编辑： 我觉得很愚蠢。它需要是

 if(mysql_num_rows($resCheckLevel) == 0) {
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1 回答 1

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使用喜欢，

mysql_num_rows($resCheckLevel)反而

mysql_num_rows($rowCheckLevel)

于 2012-01-30T12:55:18.247 回答