I have a form that has two different submit-buttons that should submit to different pages.
HTML:
<form id="campaign" method="post" enctype="multipart/form-data">
<input type="submit" value="Promote!" name="ap_promote" onsubmit="promote('?page=campaigns&id=<?php echo $campaigns_id?>&edit=true&test=2')" id="ap_promote">
</form>
Javascript:
function promote(action)
{
if (confirm('Are you sure you want to promote this campaign?'))
{
document.getElementById('campaign').action = action;
document.getElementById('campaign').submit();
}
else
{
return false;
}
}
As you see, it should send the form to ?page=campaigns&id=#&test=2. The problem is that it doesn't show any confirmation box and it just sends the form to itself, and not to the specified url.
Buttons don't have onsubmit event, it's a form event. Since you plan to have different actions per depending on clicked button, you can use combination of button onclick and form onsubmit events. Check it out:
<form id="campaign" method="post" enctype="multipart/form-data" onsubmit="return promote()">
<input type="submit" value="Promote!" name="ap_promote" onclick="this.form.action='one'" id="ap_promote" />
<input type="submit" value="Promote!" name="ap_promote" onclick="this.form.action='two'" id="ap_promote" />
</form>
And JS code becomes as simple as:
function promote() {
return confirm('Are you sure you want to promote this campaign?');
}
Change onsubmit="promote(..." to onsubmit="return promote(..." in your button click handler
I hope this will help you.
You should use the button instead of submit button
<form id="campaign" method="post" enctype="multipart/form-data">
<input type="button" value="Promote!" name="ap_promote" onclick="promote('?page=campaigns&id=<?php echo $campaigns_id?>&edit=true&test=2')" id="ap_promote">
</form>
Related
I have created a global function to add a loading effect to my submit buttons in my forms :
// add loading effet for forms
$('form').not('.form-ajax').on('submit', function() {
btnLoad($(this).find('button[type="submit"]'));
});
It just shows a loader on the submit button, and disable it.
It works, but sometimes I want to show confirm before submitting :
<form method="post"
action="/delete/post"
onsubmit="return confirm('Do you want to delete this post ?');"
>
<input type="hidden" value="2" name="id" />
<button type="submit">
Delete post
</button>
</form>
So it shows the loader on my button, but the form is not submitting if the user click no on confirm dialog.
Can I catch it easily ? To show loader on if form is really submitted ?
it'll help u:
<form method="post"
action="/delete/post"
onsubmit="return validate(this);"
>
<input type="hidden" value="2" name="id" />
<button type="submit">
Delete post
</button>
</form>
<script>
function validate(form) {
// validation code here ...
if(!valid) {
alert('Please correct the errors in the form!');
return false;
}
else {
return confirm('Do you really want to submit the form?');
}
}
</script>
I am using sweatalert2, when the form submitted sweatalert2 is coming but it closing with any press button (OK button) it close very fast. but when I use in a simple button without the form, it's working properly.
const sweatalert = () => {
return Swal('Good job!','You clicked the button!','success')
}
<!-- Include sweet alert 2. -->
<script src="https://cdnjs.cloudflare.com/ajax/libs/limonte-sweetalert2/7.32.4/sweetalert2.all.min.js"></script>
<form action="" method="post" enctype="multipart/form-data">
<input
type="submit"
name="addrecord"
class="btn btn-primary"
onclick="sweatalert()"
style="margin-left:100px;"
value="Add Record"
/>
</form>
<!-- Just here for visual purpose. -->
<hr/>
<p>Out side the form is OK.</p>
<input
type="submit"
name="addrecord"
class="btn btn-primary"
onclick="sweatalert()"
style="margin-left:100px;"
value="Add"
/>
This is happening because the form is submitting when you click the button. This means the page will refresh, and remove the alert generated. You can use e.preventDefault() to stop the submission from happening. You also need to make sure you pass the event through in your onclick callback:
onclick="sweatalert(event)"
See example below:
function sweatalert(e) {
e.preventDefault(); // stop default functionality of submission (ie. page refresh and data post)
swal(
'Good job!',
'You clicked the button!',
'success'
)
}
<script src="https://unpkg.com/sweetalert/dist/sweetalert.min.js"></script>
<form action="" method="post" enctype="multipart/form-data">
<button type="submit" name="addrecord" class="btn btn-primary" onclick="sweatalert(event)" style="margin-left:100px;">Add Record</button>
</form>
However, do note: This will stop the submission of the form occurring. If you want to submit the form and then display the alert after submitting your data to PHP you should use AJAX. If you use jQuery you can use $.post or $.ajax methods to send your data to PHP, and then use a callback function which will be called if your data was submitted successfully.
Swal returns a promise. Hence you need to prevent the default submit event and use .then() where you can submit() the form:
const sweatalert = (ele, evt) => {
evt.preventDefault();
Swal('Good job!','You clicked the button!','success').then((result) => {
ele.closest('form').submit(); // get the closest form and submit
});
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/limonte-sweetalert2/7.32.4/sweetalert2.all.min.js"></script>
<form action="" method="post" enctype="multipart/form-data">
<input
type="submit"
name="addrecord"
class="btn btn-primary"
onclick="sweatalert(this, event)"
style="margin-left:100px;"
value="Add Record"
/>
</form>
I have an upload file form, and i try to upload the file when it's selected,so i tried something like this:
<form enctype="multipart/form-data" method="POST" onsubmit="return
UploadFile(this);">
<input id="upfile" type="file" onchange="this.form.submit();"/>
</form>
The form.submit() works , but of course i need to do some validation on submit,so i tried to run a function:
function UploadFile(file){
alert('Bleah');
return false;
}
On normal circumstances it should return false, and the form shouldn't reload the page,but this doesn't happens.
If i add a submit input into the form, it works as expected:
<form enctype="multipart/form-data" method="POST" onsubmit="return
UploadFile(this);">
<input type="submit" name="upload" value="Upload">
<input id="upfile" type="file"/>
</form>
Can anyone explain me what is wrong please?
Try this:
function UploadFile(file) {
if (file.value === '') {
alert("Invalid File");
} else {
alert('Form will be submitted now!');
document.getElementById('myForm').submit();
}
}
<form enctype="multipart/form-data" method="POST" id="myForm">
<input id="upfile" name="upfile" type="file" onchange="UploadFile(this);" />
</form>
To upload the file when it's selected, you must call UploadFile() function on the input change, not on the form change tag. If you submit on input change, the page gets reloaded.
So, you'd better use something like this:
$('#upfile').onchange(function(){
if(UploadFile(this.parent('form'))){
this.parent('form').submit();
}
})
And you won't need onchange and onsubmit inside the tags any more.
Solution:
<form id="formname" enctype="multipart/form-data" method="POST" action="test.html">
<input id="upfile" type="file" onchange="sendForm()"/>
</form>
<script>
function sendForm() {
var field = document.getElementById("upfile");
if (field) {
console.log("the is a file and the form will be sent");
document.forms["formname"].submit();
}
}
</script>
OLD--
I dont understand, how would you like to submit the form without a submit button? or at least, handle the submission in javascript "object.addEventListener("keydown", myScript);"
--
ok, I read it once again and I understand the question
You need to handle this on javascript and detect the selection of the file. Look at this thread:
how to check if a file is selected using javascript?
I have been battling for the past two days with the evil onbeforeunload function in JavaScript. I have a function that warns the user when they are about to close a page.
However before the page close I would like to submit the form using JavaScript's .submit().
This is my code:
function setPopUpWindow(submitForm){
window.onbeforeunload = function () {
if (submitForm == false ) {
//alert("It worked"); --This code gets called so I know it works
document.getElementById("CancelScripting").submit();
//return "Unsaved Data would be lost";
}
}
}
In my html I have two buttons, one is (supposed to) trigger the .submit() and the other will just ignore it.
<body>
<form action=tett.html id="popUpForm" method=POST>
<script>setPopUpWindow();</script>
<input type="submit" id="submit_button" onclick="setPopUpWindow(true);">
<input class=b1 type=submit id="CancelScripting" style="visibility:hidden" value="CancelScripting" >
</body>
The `setPopWindow value for the second input is not defined so it would be false.
For some reason the submit is not working well.
------------------------Edit to my question-----------------------------------------------
I would like to submit the form even if the user leaves the page by closing the X button on their window. This is the reason why I have the hidden button... Looks like people misunderstood my question.
The only thing you can do is to ask the user if they really want to leave the page:
<head>
<script type="text/javascript">
var submitForm = false;
window.onbeforeunload = function () {
if(submitForm == false){
return 'You have an unfinished form ...';
}
}
function setPopUpWindow(type){
submitForm = true;
}
</script>
</head>
<body>
<form action="" method="post" name="SubmitForm" id="SubmitForm">
<input type="submit" id="submit_button" onclick="setPopUpWindow(true);">
</form>
</body>
I think that what you want to do is submit the form rather than the button by doing something like:
document.forms["formId"].submit();
where formId is the id of the form.
Also, I dont see anywhere in your code where your form is but your buttons should be inside of form tags.
For example, it should look like this:
<body>
<script>setPopUpWindow();</script>
<form id="formId" action="" method="post">
<input type="submit" id="submit_button" onclick="setPopUpWindow(true);">
<input class=b1 type=submit id="CancelScripting" style="visibility:hidden" value="CancelScripting" >
</form>
</body>
Here's the problem: I have a simple form with three buttons and some hidden input fields. Depending on the button pressed (different name="" values), the action does something different.
I am now trying to add a confirmation dialog box to this form by doing this:
<form method="POST" action="/action" onsubmit="return confirmFormSubmit(this);">
<input type="submit" name="one" value="This">
<input type="submit" name="two" value="That">
<input type="submit" name="three" value="Something else">
</form>
<script type="text/javascript">
function confirmFormSubmit(obj)
{
window.event.preventDefault();
jConfirm('Are you sure you want to do this?', 'Awaiting confirmation', function(r) {
if (r == true) {
obj.form.submit();
} else {
return false;
}
});
}
</script>
When I click OK, the action happens, but the input button is not submitted.
Doing 'document.location = obj.form.action;' is not an option because that will not submit the POST parameters.
How can I make the damn thing submit the input fields and not just call the action?
I think that it is because the onsumit method overrides the action in your form declaration.
I would actually change the button of the form and make it a button linked to a javascript method that performs required tests and submit values to the right action.
<form method="POST" action="/action">
<a href="javascript: confirmFormSubmit(this)">
<input type="button" name="three" value="Something else">
</a>
</form>
something like this should be working