PHP腳本不能正常工作，並顯示錯誤有關的mysqli

Question

該腳本試圖計算一個名爲qstns表項的個數：

<?php 
    $con=mysqli_connect("localhost","root","","QSTNS"); 
    if (mysqli_connect_errno()) { 
     echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

    //-----------------two variables entered in database------------ 
    $user= mysqli_real_escape_string($con, $_POST['user']); 
    $user1= mysqli_real_escape_string($con, $_POST['user1']); 

    $sql="INSERT INTO student (name,rln,scr)VALUES ('$user1','$user','0')"; 
    if (!mysqli_query($con,$sql)) { 
     die('Error: ' . mysqli_error($con)); 
    } 

    $res=mysqli_query('SELECT COUNT(*) FROM qstns', $con); 
    echo "$res \n"; 
    mysqli_close($con); 
?> 

它產生

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\wamp\www\quiiz_portal\logedin.php on line 27

誰能告訴我是什麼這個PHP腳本錯了嗎？

Answer 1

假設27行如下：

$res=mysqli_query('SELECT COUNT(*) FROM qstns', $con); 

你有你的參數倒退。應該是：

$res = mysqli_query($con, 'SELECT COUNT(*) FROM qstns'); 

一旦你返回的數據$res所有你所要做的就是分析它。

$row = mysqli_fetch_array($res); 
echo $row[0];