php - 警告 : mysqli_fetch_array() expects parameter 1 to be mysqli_result when I have 2 parameters

Question

所以我有这个错误:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in C:\wamp\www\Foredeck\login.php on line 101 .

我在互联网上做了一些研究，但仍然不明白错误在哪里。我想我这里有两个参数。那怎么了？

顺便说一句，我是 PHP 新手。

<?php
include("bdconnect_Foredeck.php");
$link = mysqli_connect($host, $login, $pass, $dbname);
$msg  = '';

if (isset($_POST['login']) && !empty($_POST['username']) && !empty($_POST['password'])) {

  $Identifiant = $_POST["username"];
  $MotPasse    = $_POST["password"];
  $query       = "
    SELECT * 
    FROM   admin 
    WHERE  identifiant = '$Identifiant' 
           AND mdp_admin = '$MotPasse'";
  $result      = mysqli_query($link, $query);
  $row         = mysqli_fetch_array($query); //Line 101//

  if ($row) {
    $_SESSION['valid']   = true;
    $_SESSION['timeout'] = time();
    $msg                 = 'Connexion Réussite';

    if ($_POST['username'] == 'Isabelle' && $_POST['password'] == 'Isabelle1') {

      $_SESSION['username'] = $_row['Identifiant'];
      echo "<script type='text/javascript'>alert('Connexion Réussite');
window.location='foredeck_superadmin.php'; </script>";
      header("refresh:3 location: foredeck_superadmin.php");

    } else {

      $_SESSION['username'] = 'foredeckadmin';
      echo "<script type='text/javascript'>alert('Connexion Réussite');
window.location='foredeck.php'; </script>";
      header("refresh:3 location: foredeck.php");

    }

  }

} else {
  $msg = 'Identifiant ou Mot de Passe incorrecte';
  $msg = "<script type='text/javascript'>alert('$msg')</script>";
}
?>

Answer 1

使用 $row=mysqli_fetch_array($result); 代替 $row=mysqli_fetch_array($query);

由于 mysqli_fetch_array() 函数将结果行作为关联数组获取，并且您正在传递类型为 string 的 $query。所以，它的抛出错误。