[英]Return PHP error when using ajaxForm Jquery Plugin
I'm trying to make a login script that uses ajaxForm and the validate plugin, but if PHP provides an error, it doesn't know. 我正在尝试制作一个使用ajaxForm和validate插件的登录脚本,但是如果PHP提供了错误,它将不知道。 This is my Javascript
这是我的Javascript
$(function login() {
$("#login").validate({ // initialize the plugin
// any other options,
onkeyup: false,
rules: {
email: {
required: true,
email: true
},
password: {
required: true
}
}
});
$('form').ajaxForm({
beforeSend: function() {
return $("#login").valid();
},
success: function() {
window.location="index.php";
},
error: function(e) {
alert(e);
}
});
});
Keep in mind I'm new to JS and there's probably a better way to do this. 请记住,我是JS的新手,也许有更好的方法可以做到这一点。 What I need is, if when the form is submitted but the username or password is wrong, I need it to not redirect, and give the error alert, but this does not work currently.
我需要的是,如果提交表单时,但用户名或密码错误,则我需要它不重定向,并给出错误提示,但这当前不起作用。 I've googled this before, and checked here and so far have found nothing.
我之前在Google上进行过搜索,并在此处进行了检查,到目前为止没有任何发现。
edit: using the below code, it still doesn't work: 编辑:使用下面的代码,它仍然不起作用:
JS JS
$(function login() {
$("#login").validate({ // initialize the plugin
// any other options,
onkeyup: false,
rules: {
email: {
required: true,
email: true
},
password: {
required: true
}
}
});
$("#login").submit(function(e) {
e.preventDefault();
$.ajax({
type : "POST",
dataType : "json",
cache : false,
url : "/doLogin",
data : $(this).serializeArray(),
success : function(result) {
if(result.result == "success"){
window.location = "/index.php";
}else if(result.result == "failure"){
$("#alert").html("Test");
}
},
error : function() {
$("#failure").show();
$(".btn-load").button('reset');
$("#email").focus();
}
});
});
});
HTML HTML
<div class="shadowbar">
<div id="alert"></div>
<form id="login" method="post" action="/doLogin">
<fieldset>
<legend>Log In</legend>
<div class="input-group">
<span class="input-group-addon">E-Mail</span>
<input type="email" class="form-control" name="email" value="" /><br />
</div>
<div class="input-group">
<span class="input-group-addon">Password</span>
<input type="password" class="form-control" name="password" />
</div>
</fieldset>
<input type="submit" class="btn btn-primary" value="Log In" name="submit" />
</form></div>
PHP PHP
public function login() {
global $dbc, $layout;
if(!isset($_SESSION['uid'])){
if(isset($_POST['submit'])){
$username = mysqli_real_escape_string($dbc, trim($_POST['email']));
$password = mysqli_real_escape_string($dbc, trim($_POST['password']));
if(!empty($username) && !empty($password)){
$query = "SELECT uid, email, username, password, hash FROM users WHERE email = '$username' AND password = SHA('$password') AND activated = '1'";
$data = mysqli_query($dbc, $query);
if((mysqli_num_rows($data) === 1)){
$row = mysqli_fetch_array($data);
$_SESSION['uid'] = $row['uid'];
$_SESSION['username'] = $row['username'];
$_SERVER['REMOTE_ADDR'] = isset($_SERVER["HTTP_CF_CONNECTING_IP"]) ? $_SERVER["HTTP_CF_CONNECTING_IP"] : $_SERVER["REMOTE_ADDR"];
$ip = $_SERVER['REMOTE_ADDR'];
$user = $row['uid'];
$query = "UPDATE users SET ip = '$ip' WHERE uid = '$user' ";
mysqli_query($dbc, $query);
setcookie("ID", $row['uid'], time()+3600*24);
setcookie("IP", $ip, time()+3600*24);
setcookie("HASH", $row['hash'], time()+3600*24);
header('Location: /index.php');
exit();
} else {
$error = '<div class="shadowbar">It seems we have run into a problem... Either your username or password are incorrect or you haven\'t activated your account yet.</div>' ;
return $error;
echo "{\"result\":\"failure\"}";
}
} else {
$error = '<div class="shadowbar">You must enter both your username AND password.</div>';
return $error;
$err = "{\"result\":\"failure\"}";
echo json_encode($err);
}
echo "{\"result\":\"success\"}";
}
} else {
echo '{"result":"success"}';
exit();
}
return $error;
}
In your login script, you will need to return errors in json format. 在您的登录脚本中,您将需要以json格式返回错误。
For Example 例如
In your login script, if your query finds a row in the database for that user, echo this: 在您的登录脚本中,如果您的查询在数据库中找到该用户的行,则回显此命令:
echo "{\"result\":\"success\"}";
and if it fails: 如果失败:
echo "{\"result\":\"failure\"}";
You then can parse these in JavaScript like so: 然后,您可以像这样在JavaScript中解析这些内容:
$('form').ajaxForm({
beforeSend: function() {
return $("#login").valid();
},
success: function(result) {
if(result.result == "success"){
window.location = "index.php";
}else if(result.result == "failure"){
alert('Failure!');
}
error: function(e) {
alert(e);
}
}
});
Here's an example of an Ajax script I use to log users into my site, you can use this for reference if needed. 这是我用来将用户登录到我的站点的Ajax脚本的示例 ,如果需要,可以将其用作参考。 This is just to help you get an even broader understanding of what I am talking about:
这只是为了帮助您更深入地了解我在说什么:
I return more than just a success and failure for various reasons such as user intuitiveness, but the gist is there. 由于诸如用户直观性之类的各种原因,我返回的不只是成功和失败,而是要旨。
$("#loginForm").bind("submit", function() {
$("#invalid").hide();
$("#disabled").hide();
$("#error").hide();
$("#failure").hide();
$("#blocked").hide();
var email = document.getElementById("email").value;
var password = document.getElementById("password").value;
if(email != "" && password != ""){
$.ajax({
type : "POST",
dataType : "json",
cache : false,
url : "/ajax/functions/login",
data : $(this).serializeArray(),
success : function(result) {
if(result.result == "success"){
window.location = "/account";
}else if(result.result == "failure"){
$("#invalid").show();
$(".btn-load").button('reset');
$("#email").focus();
}else if(result.result == "disabled"){
$("#disabled").show();
$(".btn-load").button('reset');
$("#email").focus();
}else if(result.result == "blocked"){
$("#blocked").show();
$(".btn-load").button('reset');
$("#email").focus();
}
},
error : function() {
$("#failure").show();
$(".btn-load").button('reset');
$("#email").focus();
}
});
}else{
$("#error").show();
$(".btn-load").button('reset');
$("#email").focus();
}
return false;
});
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