我正在创建一个登录脚本(刚开始,我将在稍后查找如何加密密码)。但我无法弄清楚如何重定向到某个页面。
我现在有两种可能性,即管理员或普通用户。
我的剧本:
$conn = new Connection;
$username = $_POST['username'];
$userpassword = $_POST['userpassword'];
if(empty($username) && empty($userpassword)){
echo 'Vul een gebruikersnaam en wachtwoord in';
}else if(empty($username)){
echo 'Vul een gebruikersnaam in';
}else if(empty($userpassword)){
echo 'Vul een wachtwoord in';
}else{
//Both filled in, begin logincode:
$getuser = "SELECT * FROM users WHERE username = '".$conn->real_escape_string($username)."'";
$getusercon = $conn->query($getuser);
$getuser = $getusercon->fetch_assoc();
if($userpassword == $getuser['password']){
if($getuser['rights'] == '1'){
$_SESSION['user'] = 'admin';
// header("Location: http://www.mysite.nl/addcompany.php");
// exit();
header("Location: http://www.mysite.nl/addcompany.php");
die();
echo 'Je bent ingelogd als '.$_SESSION['user'].'';
}else{
$_SESSION['user'] = 'user';
echo 'Je bent ingelogd als '.$_SESSION['user'].'';
}
}else{
echo 'Wachtwoord en gebruikersnaam komen niet overeen';
}
}
我的ajax代码:
// Login Ajax Code
$( "#content" ).on("submit", "#loginform", function( event ) {
// Stop normal form behaviour
event.preventDefault();
// Haal de inputvelden op met zijn waardes
var $form = $( this ),
$username = $form.find( "input[name='username']" ).val(),
$userpassword = $form.find( "input[name='userpassword']" ).val(),
url = $form.attr( "action" );
// Post above values to the action of the form
var posting = $.post( url, { username: $username, userpassword: $userpassword} );
// Show result in a div
posting.done(function( data ) {
$( "#loginresult" ).empty().slideDown('fast').append( data );
});
});
我试着用这个:
header("Location: http://www.mysite.nl/addcompany.php");
die();
但我只是看到我的预加载器无限旋转,在我的控制台中我看到以下内容:
[Deprecation] Synchronous XMLHttpRequest on the main thread is deprecated because of its detrimental effects to the end user's experience. For more help, check https://xhr.spec.whatwg.org/.
index.php:1 [DOM] Found 8 elements with non-unique id #example-text-input: (More info: https://www.chromium.org/developers/design-documents/create-amazing-password-forms)
但该ID也在我要重定向的页面上,除非它没有重定向,只是在我的预加载器旋转和这些消息的同一页面上停留。
我该怎么办?
答案 0 :(得分:0)
您可以在PHP响应中发送重定向标记,并在JS上添加。
例如:
window.location.replace("http://stackoverflow.com");
window.location.href = "http://stackoverflow.com";