I'm trying to show errors in form validation.
But these messages are always visible.
<form name="user-form" method="POST" action="{{route('registrationUser')}}">
<div class="col-lg-8">
Логин <input type="text" name="log" ng-model="mobile" required pattern="[a-zA-Z ]*"><br>
<span class="help-block" ng-show="errors.log[0]"><p>Только английский</p></span>
</div>
Пароль <input type="text" name="pass" ng-minlength="8" ng-pattern="regex" required pattern="[a-zA-Z]{8,32}" ><br>
<div ng-show="user-form.pass.$error.pattern">Name doesn't match pattern!</div>
<button ng-click='SaveUser()' name="Регистрация">Регисрация</button>
</form>
you can add another condition to ng-show
<div ng-show="user-form.pass.$error.pattern && buttonclicked">Name doesn't match pattern!</div>
and set that condition to true in your controller
$scope.SaveUser= function () {
$scope.buttonclicked = true;
};
so error messages will be visible only when user clicks on submit button
Related
I need to code an AJAX request with jQuery (3.2.1) and display email input value after the form was sended.
When I click on submit button, i have this following message, I didn't understand what is wrong with my code.
<form action="process.php" method="post" id="myForm">
<div class="form-line">
<input type="email" name="emailaddress" id="emailaddress" />
</div>
<div id="emailMsg"></div>
<div class="form-line">
<input type="password" name="psw" id="password" />
<button type="submit" id="submit"><span>OK</span></button>
</div>
<div id="passMsg"></div>
</form>
And this is what i have done on jQuery code
$("#myForm").on('submit',function(e) {
e.preventDefault();
var email = $("#emailaddress").val();
$.post("../process.php", {
emailaddress:emailaddress
}, function(response) {
$("#connexion-rollover").html(email);
});
});
I have also create this PHP file to add this code
<?php
echo $_POST['emailaddress'];
?>
I hope that my first message on this site was clear.
How do I get this cookies code to display properly? Both in Google Chrome and Firefox I get a dropdown suggestion from the browser. I would like to only have my own php "setcookie" to display.
PHP:
index.php:
<form action="index.php" method="POST">
<div class="row">
<div class="form-group">
<input type="email" class="form-control fill-bar" name="email" aria-describedby="emailHelp" placeholder="Email"
value="<?php
if(isset($_COOKIE["username"])) {
if(isset($_POST["email"])) {
echo($_POST["email"]);
}
} ?>" required>
</div>
<div class="form-group">
<input type="password" class="form-control fill-bar" id="passwd" name="password" placeholder="Password" value="<?php if(isset($_COOKIE["password"])) { echo $_COOKIE["password"]; } ?>" required>
</div>
<p><input type="checkbox" name="remember" class="mr-1">Remember Me</p>
Forgot your password?
<br><br><br>
<input type="submit" class="btn d-inline btn-primary register_button mr-3 w-100" id="login" name="login" value="Login">
<p class="text-center mt-3">No account yet? Register Here</p>
<br>
</div>
</div> <!-- end row -->
</form>
page-2.php: (inside if isset login button clicked)
if(!empty($_POST["remember"])) {
setcookie ("email",$_POST["email"],time()+ 3600);
setcookie ("password",$_POST["password"],time()+ 3600);
echo "Cookies Set Successfuly";
} else {
setcookie("email","");
setcookie("password","");
echo "Cookies Not Set";
}
Instead of my "setcookie", both Google and Firefox displayes these, I would like to have them not appear.
In your form, add the autocomplete attribute to your inputs and set it to off, like the following:
<form action="index.php" method="POST">
...
<div class="form-group">
<input type="password" class="form-control fill-bar" id="passwd" name="password" placeholder="Password" value="<?php if(isset($_COOKIE["password"])) { echo $_COOKIE["password"]; } ?>" autocomplete="off" required>
</div>
...
</form>
That should be enough to prevent autocomplete. However, browsers are not required to follow that attribute. You can check which browsers currently comply with the autocomplete attribute here. In the case of a browser not respecting the autocomplete attribute, you could try using methods to mangle the input names (such as jQuery plugins or by using PHP to generate random names and store those names in the user's session, and verify the form versus the session on submit).
I am trying to submit a form using Ajax, the problem is the validation file I made is not called at all (I checked it by putting a bit of js alert inside the PHP validation file). I searched the internet and tried a lot of solutions but none worked. Is there something wrong I am making in my code?
Here is my HTML:
<form method="post" id="signup">
<div class="form-group">
<input type="fname" name="fname" id="fname" value="" class="form-control" placeholder="Your First name" />
</div>
<div class="form-group">
<input type="lname" name="lname" id="lname" value="" class="form-control" placeholder="Your Last name" />
</div>
<div class="form-group">
<input type="email" name="email" id="email" value="" class="form-control" placeholder="Your Email" />
</div>
<div class="form-group">
<input type="password" name="password" id="password" value="" class="form-control" placeholder="Your Password" />
</div>
<div class="form-group submit-btn">
<input id="signupsumbit" type="submit" name="submit" value="Sign Up" class="btn btn-success btn-lg"/>
</div>
</form>
And here is the ajax call:
$(function (){
$('#signup').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'POST',
data: $('#signup').serialize(),
url: 'login.php',
success: function (data) {
alert("data is"+data);
}
});
});
});
Here is a bit of my PHP code:
include("connection.php");
echo "<script>alert(2)</script>";
if (isset($_GET['logout']) and $_GET['logout']==1 and isset($_COOKIE[id]))
{
setcookie("email", '', time()-60*60*24);
setcookie("password", '', time()-60*60*24);
setcookie("id", '', time()-60*60*24);
$message = "You have been logged out! Come back soon!";
}
if(mysqli_connect_error()) // if failed to connect to database
{
die("Could not connect to database"); //stop the whole script from runnning
}
if (isset($_POST['submit']) and $_POST['submit'] == 'Sign Up')
{
echo '<script>console.log(1)</script>'; //
$error = '';
$success = '';
// email validation
$email = $_POST['email'];
if (empty($_POST['email']))
{
$error .="<br />Please enter your email";
}
else
{
if (!filter_var($email, FILTER_VALIDATE_EMAIL))
{
$error .="<br />Please enter a valid email address";
}
}
.
.
.
.
?>
And of course, the PHP file I made validation at is called login.php.
As you may notice, I put two alerts in my PHP code, I alerted 2 to make sure that the file is called, and alerted 1 to make sure that the form was posted. Odds are, when I submit the form the normal way (with no AJAX), both of alerts work, but when I submit it with AJAX, none of them is alerted.
My Console shows this:
Any help would be appreciated. Thanks.
Testing your script, i found that the submit button is not sent to your login.php script. If you e.preventDefault() and then later trigger the submit, javascript was submitting the form but not the button. To fix your issue create a hidden field with name submit and value Sign Up and remove the name attribute from your submit button (don't need that) and it should work fine.
<form method="post" id="signup">
<input type="hidden" name="submit" value="Sign Up"> <!-- note this hidden field -->
<div class="form-group">
<input type="fname" name="fname" id="fname" value="" class="form-control" placeholder="Your First name" />
</div>
<div class="form-group">
<input type="lname" name="lname" id="lname" value="" class="form-control" placeholder="Your Last name" />
</div>
<div class="form-group">
<input type="email" name="email" id="email" value="" class="form-control" placeholder="Your Email" />
</div>
<div class="form-group">
<input type="password" name="password" id="password" value="" class="form-control" placeholder="Your Password" />
</div>
<div class="form-group submit-btn">
<input id="signupsumbit" type="submit" value="Sign Up" class="btn btn-success btn-lg"/><!-- note the removed name attribute -->
</div>
</form>
I have a website with two form authentication in different pages, have different input name and link to different pages . The problem is that when I save my authentication to a browser (chrome) of a form , the browser fill in the fields with the same data in the other form . How is it possible?
First form
<form action="" method="POST" enctype="multipart/form-data">
<div class="form-group">
<label for="exampleInputEmail1">Email</label>
<input type="email" name="private_email" class="form-control" id="email1" value="" placeholder="Enter email" required>
</div>
<div class="form-group">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" name="private_password" value="" id="password1" placeholder="Password" required>
</div>
<input type="submit" name="login" class="btn btn-default" value="Login">
</form>
Second Form (It is a form of a cms)
<form action="http://escuolainsieme.it/shop/login" method="post" id="login_form" class="box">
<h3 class="page-subheading">Sei già registrato?</h3>
<div class="form_content clearfix">
<div class="form-group form-ok">
<label for="email">Indirizzo email</label>
<input class="is_required validate account_input form-control" data-validate="isEmail" type="text" id="email" name="email" value="">
</div>
<div class="form-group">
<label for="passwd">Password</label>
<span><input class="is_required validate account_input form-control" type="password" data-validate="isPasswd" id="passwd" name="passwd" value=""></span>
</div>
<p class="lost_password form-group">Hai dimenticato la password?</p>
<p class="submit">
<input type="hidden" class="hidden" name="back" value="my-account"> <button type="submit" id="SubmitLogin" name="SubmitLogin" class="button btn btn-default button-medium">
<span>
<i class="icon-lock left"></i>
Entra
</span>
</button>
</p>
</div>
</form>
Login.php
<?php session_start(); // Starting Session
$error = ''; // Variable To Store Error Message
if (isset($_POST['private_login'])) {
if (empty($_POST['private_email']) || empty($_POST['private_password'])) {
$error = "<div class='alert alert-danger'>Compila tutti i campi</div>";
} else {
$email = mysqli_real_escape_string(conn(), $_POST['private_email']);
$password = mysqli_real_escape_string(conn(), $_POST['private_password']);
$cls_utente = new utente();
if ($cls_utente->check_user($email, $password) == 1) {
$_SESSION['login_user'] = $email; // Initializing Session
$_SESSION['is_logged'] = true;
} else {
$error .= "<div class='alert alert-danger'>Email o password errati</div>";
}
}}?>
You can try using autocomplete="false" or autocomplete="off" to disable it, not sure if it will work but give it a try.
As far as i am concerned it's not possible to make the browser 'realize' that they are different forms and they should not be auto-filled with the same data.
Have a look at these 2 answers, answer1 answer2 for more information how browser detects the forms.
You must add another one column in your user table, example if you add type column the value set default super admin,moderator, user. Now you can check the login authentication with this column.If user-name and password and type is equal redirect to particular page.so you can redirect different page depends upon the user type..
i have a registration page
<form class="register" action="regist.php" method="post" >
<h3>Register</h3>
<div class="column">
<div>
<label>First Name:</label>
<input type="text" name="f_name"/>
<span ></span>
</div>
<div>
<label>Last Name:</label>
<input type="text" name="l_name"/>
<span class="error">This is an error</span>
</div>
<div>
<label>mobile:</label>
<input type="text" name="mobile"/>
<span class="error">This is an error</span>
</div>
</div>
<div class="column">
<div>
<label>Username:</label>
<input type="text"name="user"/>
<span >This is an error</span>
</div>
<div>
<label>Email:</label>
<input type="text" name="email"/>
<span class="error">This is an error</span>
</div>
<div>
<label>Password:</label>
<input type="password" name="pass"/>
<span class="error">This is an error</span>
</div>
</div>
<div class="bottom">
<div class="remember">
</div>
<input type="submit" value="Register" name="submet" />
You have an account already? Log in here
<div class="clear"></div>
</div>
</form>
that use jquery bpopup
$(function() {
// Binding a click event
// From jQuery v.1.7.0 use .on() instead of .bind()
$('#my-button').bind('click', function(e) {
// Prevents the default action to be triggered.
e.preventDefault();
// Triggering bPopup when click event is fired
$('#element_to_pop_up').bPopup();
});
}
so when some one click register the form will pop up every thing work fine the problem is that i cant do validation to the form because when i click register after filling the form
the pop up will disappear so i cant make the validation on the same form
i tried
1 to make another pop up that show the error but the data wont pass to the new popup
2 to make the action $_SERVER['self_request']
can some one help me withe this please i have a project for my college i nee to do it in 3 dayes
what i need is a validation for the form in the same page or another popup before send it to the MySQL script
Make an php page ( here called validation.php ) that uses the GET function to get the info...
Then set some ID's on the input fields...
And then make an function like this:
function checkAll(){
var firstName = document.getElementById("f_name").value();
var lastName = document.getElementById("l_name").value();
$.get( "validation.php?f_name=" + firstName + "&l_name=" + lastName, function( response ) {
// console.log( response ); // server response
response = response.trim();
if(response == 1){
alert("Everything is alright");
} else{
alert(response);
return false;
}
});
}
And change your form to this:
<form class="register" action="regist.php" method="post" onSubmit="checkAll();">
Now the key is that in the validation.php ,if everything works fine then you can echo value 1, and when something is not alright, you echo the error code ( or any code, and then with an switch clause you can dynamically add the error to your form).
The return false prevents the form from submitting...