Foundation alert show only when success - javascript

Hello so I have 2 php files
index.php
<form action="send.php" method="POST">
<div data-alert class="alert-box success radius">
Message sent.
×
</div>
<input type="text" name="txt_message">
<input type="submit" value="Send">
</form>
<script>
<script src="js/vendor/jquery.js"></script>
<script src="js/foundation.min.js"></script>
$(document).foundation();
$('.alert-box > a.close').click(function() { $(this).closest('[data-alert]').fadeOut(); });
</script>
send.php
if(!empty($_POST['txt_message'])) {
echo "<script>alert('Message will be sent shortly...').location.href='index.php';</script>";
} else {
echo "<script>alert('Message not sent').location.href='index.php';</script>";
}
What I want to do is to show only the success data-alert when message is sent.. Any thoughts? Thank you.

You must hide the data-alert on page loading $('.alert-box').hide(),
add an ID to your form <form id="form" action="send.php" method="POST">
and match the submit event $('#form').submit({$('.alert-box').show();});
On this event you can check the field validity before to show the alert box

Related

Submitting form on same loaded page if the page is loaded via ajax

So i have a PHP page that is loaded via AJAX using the .load method.
This is then displayed on my page, however i need to be able to submit the form that is located on my second page that im loading using .load, submit this form and process the data on the same page without it reloading.
I'm very new to javascript and AJAX so i've no idea if i'm evening doing this correct using .load()
Any help would be appreciated.
page 1 is the following:
<button onclick="Test()" type="button">Load Content</button>
<div class="box" id="box" name="box">
</div>
<script>
function Test(id) {
$( "#box" ).load( "test.php?id=" + id );
}
</script>
The second page which is test.php houses the following
<form id="enrolemployee" action="" method="post">
<div class="form-group">
<label>Test</label>
<input type="text" class="form-control" id="Test" name="Test" placeholder="Test">
</div>
<input type="hidden" name="token" value="<?php echo Token::generate(); ?>">
<button type="submit" class="btn btn-outline-custom">Submit</button>
</form>
Now to submit the form on test.php i use a custom token class to generate a random hash per form submit so i can validate the data on the same page and process it through my database.
is this possible to do? or do i need to have my form then post the data through a different page?
Basically i need the entire thing to stay on PAGE 1, process the form like it normally should whilst the persons page does not reload and is always static on page 1, this should never redirect to test.php
Thankyou.
You can do it like
index.php
<button onclick="Test('10')" type="button">Load Content</button>
<div class="box" id="box" name="box">
</div>
<script src="//code.jquery.com/jquery-latest.js"></script>
<script>
function Test(id) {
$("#box").load("test.php?id=" + id);
}
</script>
test.php
<form id="enrolemployee" action="" method="post">
<div class="form-group">
<label>Test</label>
<input type="text" class="form-control" id="Test" name="Test" placeholder="Test">
</div>
<input type="hidden" name="token" value="">
<button type="submit" class="btn btn-outline-custom">Submit</button>
</form>
<script>
$("#enrolemployee").on('click', e => {
// Avoid reloading page
e.preventDefault();
console.log(<?php echo $_GET['id']; ?>);
})
</script>

How to submit form without submit button

I'm trying to submit form without submit button. I used this code to submit: document.forms['form_id'].submit(); but this code submitin over and over again. I want submit just one time
<form method="POST" id="form_id">
<input type="hidden" id="lan" value="" name="number">
</form>
<script type="text/javascript">
document.forms['form_id'].submit();
</script>
the form is submitting it self everytime the javascript is loaded, which is everytime you load the page. The default behavior of a form is to submit to the location you are currently at, if not defined, you are continously submitting the form to the page that has the form and that again submits the form.
If you want this submit only to happen once, when a visitor is (re)directed to this page for instance, you should submit to a different one by using the action attribute of your form.
If you want the submit to happen on te request of a user, wrap your submit in a function that is called by an onclick event on a button, or any other event.
<script>
function submittheform(){
document.forms['form_id'].submit();
}
</script>
<form method="POST" id="form_id" action="someHandlerUrl">
<input type="hidden" id="lan" value="" name="number"/>
<input type="button" onclick="submittheform()" value="submit the form"/>
</form>
you could use this script in the PHP building your page;
<?php
if(isset($_POST['number'])){
$number = $_POST['number'];
$out = '<h1>The form was submitted</h1>';
}else{
$out = '
<h1>The form needs to be submitted</h1>
<form method="POST" id="form_id">
<input type="hidden" id="lan" value="" name="number">
</form>
<script type="text/javascript">
document.forms[\'form_id\'].submit();
</script>
';
}
echo $out;
?>
When the form is submitted and the number value is present,
it shows a page without the script and form.
<form method="POST" action="" id="mailform">
<input type="email" name="mail" placeholder="EMAIL" id="mail">
<div id="sub" onclick="mailsubmit()">Click Me to Submit</div>
</form>
<script type="text/javascript">
function mailsubmit(){
document.getElementById("mailform").submit();
var mailid = document.getElementById("mail").value;
return mailid?alert(mailid):alert("You did not submit your Email");
}
</script>

Validating and linking to another page

I have created a login page.In whicn iam unable to link it with the next page after clicking submit button.I want to validate and redirect to the next page.ie home.php.Kindly help me find out what am i missing.
signin.php
<!DOCTYPE html>
<html>
<head>
<script>
function val()
{
var a=document.signin.user.value;
var b=document.signin.password.value;
if ( a == "admin" && b == "rec"){
alert ("Login success");
window.location = "home.php";
return false;
}
else{
alert("login failed")
}
}
</script>
<meta charset="UTF-8">
<title>LIBRARY </title>
</head>
<body>
<div class="body"></div>
<div class="grad"></div>
<div class="header">
<div>REC<span>LIBRARY</span></div>
</div>
<br>
<br>
<br>
<div class="login">
<form name="signin" method="post" onsubmit="val();">
<input type="text" placeholder="username" name="user"><br>
<input type="password" placeholder="password" name="password"><br>
<input type="submit" id="mybutton" value="login"></form>
</div>
</body>
</html>
The problem is the onsubmit event is not having false returned to it, so it posts the form normally, after your JavaScript has finished. Even in the case of successful login and the redirect is executed, the form will still submit and it will override the redirect.
Firstly, Move your return false; to the end of the function, so that it always executes.
Secondly, change your onsubmit="val();" to onsubmit="return val();". This means the onsubmit event will always be returned false and will not try to post the form.
Side note: this is by no means a secure system. Any visitor can simply observe the HTML source to find the password, or just navigate directly to home.php. For a secure system, you will need to do the authentication on the server side (in the PHP).
You could use preventDefault() Event method without using onsubmit=val() like below.
document.getElementById("signin").addEventListener("submit", function(e){
e.preventDefault()
// actual code to validate
});
or
can try some dirty work on server side directly to hide the validation part
<?php
ob_start();
session_start();
loginForm();
$userinfo = array(
'admin'=>'0b2c082c00e002a2f571cbe340644239'
);
if(isset($_POST['username'])){
if($userinfo[$_POST['username']] == md5($_POST['password'])){
$_SESSION['username'] = $_POST['username'];
header('Location: home.php');
exit();
}else{
?>
<script type="text/javascript">
alert("Oops.... User name or Pasword is worng, Please try again");
</script>
<?php
}
}
function loginForm()
{
?>
<form name="login" action="" method="post">
Username: <input type="text" name="username"> <br><br>
Password: <input type="password" name="password"><br>
<input type="submit" value="login">
</form>
<?php
}
?>

Specify dynamically where to output ajax success result for multiple forms

I have two questions:
How can I create a second form in the same page and ajax handle the right form depending on which submit button got clicked?
Now let's suppose I have multiple forms in the same page. I want to change where the output goes depending on which submit button is clicked.
Source code:
Ajax:
<script type="text/javascript">
$(document).ready(function(){
$("#myform").validate({
submitHandler: function(form) {
// do other stuff for a valid form
$.post('post.php', $("#myform").serialize(), function(data) {
$('#results').html(data);
});
}
});
});
</script>
Html:
<form id="myform" action="" method="POST">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results"></div> <!-- post.php output goes here.-->
<!-- until here everything is working fine -->
<!-- now if I want to add a second form I don't know what id I should use -->
I would use a class on your forms to bind to, and then each form can have an element that holds the output id which you select -
javascript
<script type="text/javascript">
$(document).ready(function(){
$(".myforms").validate({
submitHandler: function(form) {
// do other stuff for a valid form
$.post('post.php', $(form).serialize(), function(data) {
var output = $(form).data('output');
$("#"+output).html(data);
});
}
});
});
</script>
html
<form class="myforms" action="" method="POST" data-output="results1">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results1"></div>
<form class="myforms" action="" method="POST" data-output="results2">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results2"></div>
<form class="myforms" action="" method="POST" data-output="results3">
<input type="text" name="a"><input type="submit" value="submit">
</form>
<div id="results3"></div>

Javascript/PHP auto submit keeps submitting form

This code:
<?php
if(isset($_GET['submit']))
{
?>
<head>
<script type="text/javascript">
document.getElementById('up').submit(); // SUBMIT FORM
</script>
</head>
<?php
}
?>
form:
<form name="up" id="up" action="" method="post">
<textarea name="text" rows="40" cols="100"></textarea>
<input type="submit" name="ingameban" value="Save in-game banlist (Upload to server and make new bans take effect)" style="height: 64px; width: 550px;" />
</form>
Keeps looping all the time, the same result as smashing the reload button.
It has to submit the form when the url states ?submit=submit
What to do to fix this?
Thanks
Your approach is right, but the problem is that submit=submit in the URL is copied to the new URL used to submit the form. Because in your form you have:
<form name="up" id="up" action="" method="post">
Since action is empty, the exact same URL is used, so submit=submit stays in the URL. Instead, provide the proper URL in action. Then submit=submit won't be copied to the new URL:
<form name="up" id="up" action="/my-url" method="post">
How about setting an input hidden field, which you mark as true when you submit.
Check this field before submitting again.
Try this:
<?php
if(isset($_GET['submit']))
{
if(isset($_GET['submitted']) && $_GET['submitted'] == 'false') {
?>
<head>
<script type="text/javascript">
document.getElementById('submitted').value = 'true';
document.getElementById('up').submit(); // SUBMIT FORM
</script>
</head>
}
<?php
}
?>
<body>
<form method="get" id="up">
<input type="hidden" id="submitted" name="submitted" value="false" />
...
</form>

Categories

Resources