所以我有這樣的錯誤:警告:mysqli_fetch_array()預計參數1被mysqli_result當我有2個參數
警告:mysqli_fetch_array()預計參數1 被mysqli_result,串在C中給出:\ wamp \ www \ Foredeck \ login.php在 第101行。
我在互聯網上做了一些研究,但仍然不明白錯誤在哪裏。我想我在這裏有兩個參數。那麼,怎麼了?
我是新來的PHP的方式。
<?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>";
}
?>
應該是'mysqli_fetch_array($結果);'。 '$ query'持有查詢字符串不是結果資源 –
'mysqli_fetch_array($ result);'not'$ query' –
仍然得到相同的錯誤@Anant –