PHP mysql_num_rows()期望参数1是资源

Question

这是我的代码......如果(mysql_num_rows($ res)> 0){我在线获取错误} {

$con= mysqli_connect("localhost","celento","password");
mysql_select_db("celento");
// Make sure we connected succesfully
if(! $con)
{
    die('Connection Failed'.mysql_error());
}
$username= $_POST['username'];
$password= $_POST['password'];


 $sql=("SELECT * FROM login WHERE username="'.$username.'" AND password="'.$password.'" LIMIT 1") or die(mysql_error());;

 $res = mysql_query($sql);

 if (mysql_num_rows($res)> 0){
             echo "Success";}
              else
             {echo"Fail";
 }
?>      

---

请帮我修复错误.提前致谢.

Answer 1

你搞砸了单引号.此外,不需要(以及)何时将其存储为字符串.它应该是

$sql="SELECT * FROM login WHERE username='".$username."' AND password='".$password."' LIMIT 1";

注意:您使用的是已弃用的API.mysql_*PHP v5.5之后不再支持这些函数.最好切换到mysqli或PDO.这是一个很好的教程,可以帮助您入门.