mysql_fetch_row和mysql_fetch_array之间的冲突

Question

我在PHP中有一个函数，它给我一些来自DB的值

$select = mysql_query("SELECT * FROM sfide WHERE accepted = 0");

if($select){
    if(mysql_fetch_row($select) != 0){
        $sep = "~";
        $return = "";
        $i = 0;
        while($row = mysql_fetch_array($select)){
            if($i > 0){
                $return = $return.",";
            };
            $return = $return.$row['id'].$sep.$row['from_'];
            $i += 1;
        }
        return $return;
    }else{
        return "NO";
    }
}else{
return "E, 000, SELECT, ".mysql_error();        
}

问题在于，如果我喜欢上面的代码，它什么都不返回，相反，如果我复制了它的选择变量：

$select = mysql_query("SELECT * FROM sfide WHERE accepted = 0");
$num_select = mysql_query("SELECT * FROM sfide WHERE accepted = 0");
if($select){
    if(mysql_fetch_row($num_select) != 0){
        $sep = "~";
        $return = "";
        $i = 0;
        while($row = mysql_fetch_array($select)){
            if($i > 0){
                $return = $return.",";
            };
            $return = $return.$row['id'].$sep.$row['from_'];
            $i += 1;
        }
        return $return;
    }else{
        return "NO";
    }
}else{
return "E, 000, SELECT, ".mysql_error();        
}

为什么？

Answer 1

而不是

mysql_fetch_row($select) != 0

使用

mysql_num_rows($select) != 0

将返回int供您比较。