I have a page (run with classic asp) that uses drop down boxes and this function to refresh the page for form submit:
<script>
function dept_onchange(frmSelect) {
frmSelect.submit();
};
</script>
<SELECT onchange="return dept_onchange(frmSelect)">
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I've literally tried everything. Read 26 tutorials, interchanged code, etc. My validation functions all work. My AJAX functions work (tested manually using servlet URL's). The second servlet validates the reCaptcha form that's generated on my webpage. After the form is validated, even if everything's correct, nothing happens upon clicking submit. I even have an alert pop up if with the captcha result, just for middle-layer debugging purposes.
I want to do all of my validation clientside; none serverside. However, going to be tough if I can't get my god damn form to submit. I've been puzzled by this for close to 36 hours straight. I can't see, and I'm going to get some rest and hope that there is some useful insight on my problem when I return.
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Does the "submit" button always submit form data in non-ajax way?What are the best practices for submitting forms using xhr?
View 1 Replies View RelatedI am currently using jquery .post ajax code to submit a form.
The problem is that when I click on submit it takes me to the php file page. Which is blank.
I don't want this... this functions just like if I made the form in html meaning when submit it takes you to the .php file well opens it in a page.
I want to use ajax that when you click submit it submits the form and then shows a message on screen.
When I click the submit button my message gets shown on the screen saying it submittedsuccessful then with at least 2 seconds it sends the user to the php file meaning opens the page up in the browser not a new tab or new window but sends the user to the php file. Which I don't want it to happen. So the page is blank.
I used the .post and serialized the form. Is this the right thing to do?
I don't want to send the user to the php file. Just use ajax and check if it updated the database and if so send either 1 or 0 depending if it went threw then the form page willreceivethat data and then fade in a message saying either error or success depending what went on.
I would like to know how you can use jquery ajax to submit a form without refreshing the page. I do know about the serialization function but would like to see an example of what is submit().
I currently tried to get my form working but but can't get it to work meaning to submit something from the form.
I have a general idea of how this works but i need further explanation. I have a form that needs to post into one list and then post into another list as well.Here is the ajax code and html code:
View 2 Replies View RelatedFor example, I have commenting system where I want to use ajax to display and submit comments.
How do I check if user is logged in, if logged in then display the form, otherwise forward it to login page.
Once they submit the form, how do you update the page to reflect the changes?
Do you update the partial page or just reload the whole page with new comment?
dt_save is a <form>, the submit trigger works as usual but when doing an Ajax request the code interrupts without any error. I put in 2 alert's just to check, only the first alert box appears when submitting the form. Also I can't find my request in Firebug's Network window (Firefox addon).
$('#dt_save').submit(function(){ alert('json request'); $.ajax({ type: "POST", url: "json.php?action=extendable_select_insert", dataType: "json", data: "into=empfaengerliste&col=empfaenger&value=" + encodeURIComponent(document.getElementByID("data[empfaenger]_t").value), success: function(data) {} }); alert('done'); });
PS: Of course the requested PHP-File exists!
I'm not sure if this is a php issue or jquery issue. When I complete the form and hit submit it puts the successful message up like its supposed to but also with the fields still filled in the form which it shouldn't be doing and it doesn't actually post in the database so I'm not sure if its a php issue or jquery issue.
<?php
How can I submit the contents of my form? This code submits NO data. I want to submit the data in my form. I want the values of my form to be submitted.
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I have one-field form on my website, which is supposed to be filled with barcode number. It was functional until few days age when the need emerged for the auto-submit. As you can see, I have to make script which will automatically submit form when data is filled in barcode field.
View 9 Replies View RelatedI have a form and i am using jquery validation plugin to validate form.now i want to after validation submit form using ajax.
here is the example.
Code:
<form id="frmRegister" action="" method="Post" >
<input type="text" id="username" name="username" value="" />
<input type="submit" id="btnRegister" name="btnRegister" value="Register" />
</form>
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I am adapting a small cart for a website of mine - I am using the cart below
[url]
I have tweaked it to my liking but I am wondering how exactly the form is setup to post? I would like to set it up so that it posts directly into a database on submit, and then displays confirmation that it has been submitted. At the moment it just echo's "TOTAL PRICE:" XXX it's obviously submitting the form somewhere, but where is it going and how can I change it?
Here we are (a part of index.php) :
My problem is that when I submit my form,writePersonneSGBD.php seems to be quite well executed , because my record is created as wished in my database, but there is no way it returns anything (out) and display something in my div container. instead of that,and I really dont understand why, after the end of the function, index.php is executed again and so a new form is displayed again in the div "container"
I'm using the submit and ajax functions to check a form. When the form is submitted the submit event executes and within that event the data (a code the user has to fill in) is sent to the server to check if it's correct. If it is correct, the submit event supposes to return true (so the form submits), if it's incorrect the submit event supposes to return false (so submitting the form cancels).
Here's my code that should do this:
$(function(){
var submit = false;
$('#login').submit(function(){
$('span.error').fadeOut('slow');
$.ajax({
url: '/ajax/login',
type: 'POST',
data: 'code=' + $('#code').val(),
success: function(response){
if (response != ''){
$('span.error').fadeIn('slow').text(response);
submit = false;
}else{
submit = true;
}}})
return submit;
})});
But here's the problem: While jQuery does the ajax request, the script continues. So it will return false - even if the code is validated - because the var submit has been set to false at line 4. When I submit the form again, the form is validated because 'submit' is true (but that's set to true the first time). How can I ensure the submit function waits for the ajax request to finish?
my issue is both a php and a javascript (AJAX) one its one or the other. I have two combo boxes one dependant on the other, populated by AJAX but when I submit the form the secondary combo is not being registered as a php $_POST['var'] in the following action page. The secondary combo is contained and altered using divs and inner.HTML. Is this an expected behaviour or is my issue else where?
View 3 Replies View RelatedI load into a div on index.php page via AJAX another php page. Once that page loads, I lose all the function of that page. For example, I click on the search or submit button and nothing happens.
I am trying to create a site using mostly PHP and MySQL and I have integrated some AJAX functions into the site to load new PHP pages into a DIV called "dynamic" on the index.php page without having to reload the entire page. This is working great.
However, when I use the menu with the following hyperlink <a href="javascript: addContent('sample.php', 'dynamic')"> calling the AJAX code to load the sample.php page inot the the DIV dynamic, the sample.php pages loads into the DIV but the PROBLEM is that the submit, and or search functions on the sample.php page stop working.
I believe that this has something to do with using the AJAX code because it is Asynchronous. Perhaps I need to use live or event delegation.
Since I am new to webdevelopment and trying to learn best practices if anyone knows a good way to load PHP forms into a DIV on perhaps a index.php page and then have the form controls still work.
Also I have been using NuSphere PhpED for an IDE which has helped me step through PHP and MySQL code to solve problems real well. However, I would like to get a recommendation on a IDE to use with Java Script and AJAX. Part of my problem with resolving this issue is that I can't step through the code to see what is not happening..
I am trying to get a form to call a javascript function on submit to send an ajax request. The site sits on one server and the php target file on another. I get a status of 0. From what I read, this is because the php file sits on a different server. I do don't believe I can get around the files sitting on different servers.I can set the php file as the form action, but I do not want to redirect, just send a request to the other server to modify some values in a DB. If I can not send an ajax request between servers, I was thinking an alternative might be to set the php file as the form action, run the updates, and then have the page redirect me back. Is there a better way to get around this problem
View 1 Replies View RelatedThis is a ? about Ajax, PHP and form processing. When I submit the form further below, nothing happens I didnt set the form-"action" to anything. I left it out, to make the form submit to the same page, so I can validate on the same page too (Which is not made yet, as youll see. The form does work fine when I used in a "normal php script", but when I include it using Ajax, the form doesnt submit after it has been sent to the page.
Anyone knows what is going on, and what I can do to make it work? I guess it must be the action attribute of the form which is wrong, but I dont understand that since i made the form submit to itself, by leaving out the action attribute entirely.
I have two links on a page, which calls a javascript function, opening a xmlhttprequest: I have only shown one of the forms, as they are almost similar, but there are two forms, in the real page, which displays depending on the link selected.
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i am trying to build my form to submit without reloading the screen. I am using jquery and it inserts the data but it also reloads the screen and two database records get inserted. It appears that one is being inserted by jquery and the other is being inserted by the normal form submission. How do I remove/override the normal form submission? My html code is as follows:
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im trying to learn how jquery ajax function works by validating some form after submit is clicked.
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the alert didnt pop until i added the registration_ok = false; right before, so im assuming it has something to do with the ajax functions's scope.... why does the ajax function not affect the registration_ok variable outside the function? and how should i solve this problem? any better way of doing it?
I would like to submit a form depending on the success data of an ajax post.
Below is my jquery code; as you see #theform is the main form and before submitting the form I need to check the availability of the the date and time and the room using$.ajax. However it doesn't submit the main form if the date, time and the room is available.
required=["txtCal_Event_CalendarID","txt_TreatmentRoom","txtTreatmentID","txtTreatmentTypeID","datepicker1","datepicker2","timepicker1","timepicker2"];
emptyerror="Please fill out this field.";
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I have a problem where if a form submission (set up to submit via AJAX) fails validation, the next time the form is submitted, it doubles the number of post requests - which is definitely not what I want to happen. I'm using the jQuery ValidationEngine plugin to submit forms and bind validation messages to my fields. This is my code below. I think my problem is that I need to unbind from the validationEngine plugin when the form fails, but I can't figure out how to do this.
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On my form I have 3 submit buttons which handle different things.I am looking for a way to stop or continue form execution with a confirm box on the third submit button and the third only.I can't use onsubmit because that will trigger on all three buttons.
View 2 Replies View RelatedIn the following .submit function, I am attempting to grab the value of the selected option in the facilityCodes dropdown list when I click the submit button and then during the submit function,select the facilityCode again in the dropdown list when the page reloads and then disable the list so that the user cannot change it.However,the situation is when I reload the page after the submit button is clicked the dropdown defaults to the first option and the list is enabled.I apparently am not understanding how.submit works so that I'm selecting the option I'm defining in my code below and then disabling the list AFTER the page reloads due to an error on the page. My question is how can I accomplish this?Here is my code:
$(function() {
$("#ARTransferForm").submit(function() {
var msgsCount = 0;[code]....
The following will submit the form data to popup by clicking the submit button. I want it will submit the form automatically to the popup, there is no submit button in this page. Basically this page should not show up.
<script type="text/javascript">
function submitmyform(f) {
f.target = 'foo'
window.open('',f.target,'menubar=no,scrollbars=no, width=800,height=800');
f.submit();
return false;
}
</script>
<form name="myform" action="popup.asp" target="_blank" method="post"
onsubmit="return submitmyform(this);">
<input type="hidden" name="item" value="item"/>
<input type="submit" value="submit to popup"/>
</form>