PHP/MySQL：没有找到匹配项时返回什么？

我想使用PHP来找出是否没有找到匹配项。我尝试了以下操作，但“is_resource（）”始终返回true

$result = mysql_query('...');
if(is_resource($result)){
// Results are found
}

mysql\u num\u rows（）

if (mysql_num_rows($result)>0) {
    //Results are found
}

这就是您想要的：
所以只要您能够正确访问数据库，$result始终是一个资源。mysql_num_rows（）假设查询本身已成功运行。我想说试试这样的东西：

if($result === FALSE) // Query failed due to not having proper permissions on the table
    die('Invalid query: ' . mysql_error());
else if(mysql_num_rows($result) >0)) // We have more than 1 row returned which means we have data
    // INPUT RESULTS PROCESSING HERE
else // No rows were returned therefore there were no matches
    echo 'No rows returned';

if ( $stmt = mysql_query("...") )
{
    // Do some things
}
else
{
    // Do some other things
}

希望这有点帮助（）
如果需要，请查看此处了解更多信息：
如果查询失败，mysql\u查询将返回false，因此您可以像这样检查代码：

if($result === FALSE) // Query failed due to not having proper permissions on the table
    die('Invalid query: ' . mysql_error());
else if(mysql_num_rows($result) >0)) // We have more than 1 row returned which means we have data
    // INPUT RESULTS PROCESSING HERE
else // No rows were returned therefore there were no matches
    echo 'No rows returned';

if ( $stmt = mysql_query("...") )
{
    // Do some things
}
else
{
    // Do some other things
}

或者您可以像上面提到的那样使用mysql_num_行。
但您应该真正了解它是一个内置的数据库类。学习并使用它。你的生活会轻松得多