I'm attempting to submit additional data to the server when I submit a form. Here is the form:
<form class="class" action="{{=URL('default','function')}}" method="post">
<input type="text" id="first" name="first" class="span3"/>
<button id="button" type="submit" class="btn">Submit</button>
</div>
</form>
I've tried to do submit the additional data a couple different ways:
<script>
$('#button').submit(function(){
var cost = +($("#cost").text().replace('$',''));
$.getScript("{{=URL('default','function')}}"+"?cost="+cost);
});
</script>
and
<script>
$('#button').submit(function(){
var cost = +($("#cost").text().replace('$',''));
$.post("{{=URL('default','function')}}",{cost:cost});
});
</script>
However, only the form data appears in request.vars. How can I submit the additional script variable "cost" with the rest of the form?
Add an id to your form or to a div in your form (calling it '#addinput' here), then append a hidden input once your button is clicked:
<script>
$('#button').click(function(){
var cost = +($("#cost").text().replace('$',''));
$('#addinput').append('<input type="hidden" name="id" value="'+cost+'" id="id">')
});
</script>
Related
Here is my form:
<form id="myForm">
<input id="htmlString" type="text" name="htmlField" ><br>
<input type="Submit" value="Submit" >
</form>
And need to fill it from console.
just to use it in my app,
Will inject javascript with data to local html file.
I tried to make the form without a submit button like so:
<body>
<form id="myForm">
<input id="htmlString" type="text" name="htmlField" ><br>
</form>
<script>
htmlString.oninput = function(){
///do some stuff
}
</script>
</body>
Expecting that :
document.getElementById('htmlString').value="moo" ;
It automatically submit the form, because here oninput used.
But it just stayed filled with inputs and not proceed further.
Tried with other solution:
form = document.getElementById("myForm")
form.submit()
But it just refreshed the page and not submitted the form.
The need is just one filed without else, and inject my string to it with javascript to run functions embedded in the html.
Try making the input button hidden.
<body>
<form id="myForm">
<input id="htmlString" type="text" name="htmlField" ><br>
<input type="Submit" value="Submit" style="display: none" >
</form>
<button onclick="simulateConsole()">Try it</button>
<script>
htmlString.oninput = function(){
if(this.value === "moo") {
myForm.submit();
}
}
// This event will be triggered even if you use console
htmlString.onsubmit = function(){
if(this.value === "moo") {
// do something onSubmit
}
}
function simulateConsole() {
// you can simulate this in console
htmlString.value = "moo";
myForm.submit();
}
</script>
</body>
I hope it helps.
You need to supply an action to the form, otherwise it will just reload the page.
See more here: https://developer.mozilla.org/en-US/docs/Web/HTML/Element/form
I am trying to create multiple forms which have two buttons, each will submit the form to different script, one via ajax and second one will just submit the form.
<?php foreach($objects as $object) : ?>
<div class="card-body">
<form id="edit-form" action="#" method="POST">
<input name="subject" value="<?=$object['subject']?>" type="text" id="title" class="input-xxlarge">
<textarea id="content" name="content" rows="25"><?=$object['content']?></textarea>
<button type="button" id="send-button" class="btn btn-primary">Send</button>
<input type="submit" id="submit-button" value="Submit"/>
</form>
</div>
<?php endforeach; ?>
First I am trying to get the current form, but I have problem with that. console.log shows something only on the first form, if I click on the buttons from other forms then It will do nothing.
$('#send-button').on('click', function(e) {
e.defaultPrevented;
$form = $(this);
$url = $form.attr('action');
$data = $form.serialize(); console.log($form);
console.log($url);
});
Is it because my button has same ID for every form ?
You shouln't use ID's multiple times on the same page. Try to use a class for that case. Also as stated in the comments use e.preventDefault(); to stop the event.
$(this) will result in the #send-button beeing targeted. To access the form you need to find the closest form element like this:
$form = $(this).closest('form');
html:
<form method="POST" action="#">
<input type="text">
<button type="button">send</button>
<input type="submit" value="submit" />
</form>
<form method="POST" action="#">
<input type="text">
<button type="button">send</button>
<input type="submit" value="submit" />
</form>
js:
$("form").each(function() {
var form = this;
$(form).find('button').on('click', function(e) {
e.preventDefault();
console.log(form);
console.log(this);
})
});
this will add events on every form you have on your page, button will submit form via script and submit will just submit it. also here's a fiddle to play with
I know how to submit a form from outside the form, for example:
<form action="Get?id_sec=120" method="post" id="form15" name="form15" style="display:none"></form>
<input type="submit" class="finish-button primary-button button" border="0" value="Limpar pedido" form="form15" onclick="javascript:document.form15.submit();" />
But I want to put a tag with a reference to the form with javascript too, because command form="example" doesn't work in Internet Explorer.
example:
<input class="input-cep" name="pr001" id="cepfrete" type="text" form="form15"/>
or
<input type="radio" name="tipofrete" value="4" form="form15">`
How can I do that?
Hey Vince, thanks, this works. Very useful help! I need just one other thing. How can I put an input and select in the same form in jQuery?
example:
<input type="text" data-form="dataForm" name="external-input-2">
<Select id="selectField_1" name="selectField_1" data-form="dataForm" >
<option value="52" data-form="dataForm">A</option>
</Select>
To submit a form from outside the form:
HTML
<form id="theForm">...</form>
<button id="submitTheForm">Click to Submit</button>
jQuery
$('#submitTheForm').on('click', function() {
$('#theForm').submit();
});
To include external inputs in the form submission:
HTML
<form id="theForm">...</form>
<button id="submitTheForm">Click to Submit</button>
<input type="text" data-form="theForm" name="external-input-1">
<input type="text" data-form="theForm" name="external-input-2">
jQuery
You can append the external inputs as hidden inputs to the form:
$('#submitTheForm').on('click', function() {
var form = $('#theForm');
$('input[data-form="theForm"]').each(function() {
var input = $(this);
var hidden = $('<input type="hidden"></input>');
hidden.attr('name', input.attr('name'));
hidden.val(input.val());
form.append(hidden);
});
form.submit();
});
I'm not sure that I can completely understand your question but if you are asking how to submit a form externally in different situations, her is my answer.
For the future, just put an id on the form like this.
<form id="form15"></form>
Then to submit this form from anywhere, all you have to do is call the following javascript line in an onclick, a function, etc.
document.getElementById("form15").submit();
I currently have a button in HTML with the following code:
<form id="tfnewsearch" method="get" >
<input type="text" id="search_query" name="q" size="21" maxlength="120"><input type="button" id="search_button" name="search" value = "Search"onclick="doSearch(this.form.q)">
</form>
The function 'doSearch()' works only if I click the submit button. What changes do I have to do if it has to work even if I just press the Enter key?
<form id="tfnewsearch" method="get" onsubmit="doSearch()" >
Simply change the onclick to an onsubmit and attach it to your form!
The proper way it to move it to simple JS script
<script type="text/javascript">
var form = document.querySelector('#tfnewsearch'),
query = form.querySelector('[name="q"]');
form.addEventListener('submit', function(){
doSearch(query.value);
});
</script>
I have the following dynamically generated HTML
<div id="1">
<form name = "inpForm">
<input name="FirstName" type="text"/>
<input type="submit" value="Submit"/>
</form>
</div>
<div id="2">
<form name = "inpForm">
<input name="FirstName" type="text"/>
<input type="submit" value="Submit"/>
</form>
</div>
The outer divs have different IDs but the form names are the same. I am using Jquery to perform some validation when the form is submitted. However, when the second form is submitted, I always get the values of the first form.
$(document).ready(function () {
$('form[name="inpForm"]').live('submit', function () {
alert($('input[name="FirstName"]').val());
return false;
});
});
How can I modify myJquery to find the "FirstName" element that matches the current form where the submit was triggered?
Thanks
Add some context:
alert($(this).find('input[name="FirstName"]').val());
Use this (the form-element) as context-argument:
alert($('input[name="FirstName"]',this).val());