JQuery :: Form Not Submitted When Click Method Used
Jun 21, 2010
I have some function to fire when I click a submit button. When I do that a form does not submit anything, it just fire my function etc. Here is my code:
$("#hm-submit-form").click(function () {
$("#nav ul li.products").delay(200).fadeIn();
$("#nav ul li.agel-tour").delay(600).fadeIn();
$("#nav ul li.compensation").delay(1000).fadeIn();
$("#thanks-box").delay(200).fadeIn();
$("#thanks-box").delay(1500).fadeOut();
$('#target').submit(); //this I guess should submit a form
I am having trouble finding a code example that uses javascript to get values from a form submitted on a previous page using the "POST" method. I've found some examples on how to parse the query string, but unfrotunately my form on the previous page has WAY too many charcters being submitted to be able to use the "GET" method (there is some upper limit here, I think it may be 255 or some number similar).
By the way this ONLY needs to work in IE, but it is better if it works in all browsers.
Using jQuery with validate plugin.I have to click twice on the submit button to get it working.First time I click on submit form = nothing happens. No error codeSecond time = form successfully submited in ajax.Here the code :
I am re-developing my website and want to use some Ajax/jQuery in it (so I am learning as I go) and I am not sure if this is possible so that is my first question:I have a registration form and I would like to know if after the person registering has completed their first and surname and the 2 sections of their postcode if I can use jQuery to run a PHP script to see if they are already registered before waiting for the form to be submitted and running a script then.I am thinking onkeyup (or similar) after the last field is completed but I don't know how that would work needing to also use 3 other field values.If this is possible, can anyone give me any ideas of examples that do this or how I can go about passing all the variables and running the scripts at the correct time.
What I want to do is replace a form with the submitted content. As an example, The form is submitted using the jquery.form plugin. The content is saved to the DB and a rendered comment is passed back to the client, and the success callback then replaces the form with the content. Simple enough. However, if server-side validation fails, I want to render the form, along with validation error messages, and insert that back into the client page, i.e. replace the first form with the second (which also contains the submitted content). Also straightforward, except that, on a truly successful submission, I'll need to do some other things on the page. If validation fails I only need to display the form with errors. I can think of two approaches. One is to search the returned data for some string (e.g., "<form") to decide if the other tasks are to be run.
The other is to send back the data encapsulated in a JSON object, eg: { 'status': 'FAIL', 'data': '<form ...' messages: '...' }
In my application,there are several autocomplte field is available.When I type "pank" and then it list out the several matching in the autocomplete field and select one and then hit the enter then it submitted the entire form, which will cause the error message. Because there are several more mandatory field are there, which i need to fill up.I was trying to restrict in this way, but i believe this is the wrong way.
I have email, password and some other fields, and I'm using $.post to send data.
In Ajax form submission, I don't want to submit a particular field. The serializeArray() method returns all the fields in the form. So, I tried something like this to prevent the password field being serialized.
This works great but the ajax submission doen't work.
I have a page which displays a list of resources in the database. For each of these resources I have a comment form. I am using jQuery form plugin to submit the form through ajax. After submitting I want to display the comment which was just submitted w/o reloading the page. But how do I know which form was submitted? I attach the resource ID as a GET variable to the action url of the form. If I can get the action url then I think I'm saved.
How can I target a specific <div> to display the result of a form? I would like to be able to specify the target div on a per-form basis, and have the target independent of the <div> that contained the actual form.
So for example I may have a form in <div id="left"> <form>.....</form></div>, that I want to post the data to somepage.php, and load somepage.php into <div id="right">
I have a form which has a few submit buttons, all named SubmitButton so I can pass the clicked button's value through to the submission processor and then act according to which button is pressed. However on the jquery form submit even, I also want to check certain things before I fire the submission off. As the button isn't a "conventional" input I can't check the document object to see which one was clicked, so I need to check the form contents that are about to be submitted, can I do this from the submission event or will I have to bind to each button individually?
I use unordered list item (<ul>) to replace select type <select> (select) in my form. I have problem on bind the value after submit the form. Because <ul> is not a form element. May i know how should I overcome it?
What i intend to do is similar effect as google contact book: 1. User click ADD button, then <ul> and <textbox> will be appended into the form. 2. After form submitted, user can see what are the value he has entered earlier. I can bind the textbox, post the unorder list item with hidden element. but i have no clue how to bind the unordered list item..
Kindly refer to my code as below. It is working except cannot bind unordered list item... p/s: I am using PHP + JQuery.
I have a form I'm going to present to visitors. The goal is that when a visitor enters a Zip Code, for the form to compare that Zip Code submitted to Zip Codes that I'll hide in the form. I'm not sure if it's better to have the hidden zip code values as comma separated or just let them have their own individual hidden input fields. When the form is submitted, I need it to check and find out if the submitted zip code matches any of the hidden zip codes. If any of the hidden values match the submitted value, I want to be able to have a hidden div be shown, but if the form is submitted and there are no zip codes that match, show another div.
I have a script that is supposed to: Make sure that the user selects a performance from a select field Make sure that the user hasn't selected a performance that contains the string "SOLD OUT" in its value Here is the code I am using:
I am trying to create a form that writes text to an HTML canvas when submitted. Eventually, the function that writes the text will be more complex. The problem is the text only appears briefly, because the function is only called once when the form is submitted. I want the function to be called continuously after the form is submitted.How do I do this? I have had very little experience with JS.A lame (failed) attempt...
Am creating a framed chat application and when the user types a message in the form field and clicks the submit button, the message gets sent to the display frame, but the message stays in the form field. How can i get the form to submit the message AND reset the form field to blank too?
I have a form, It has multiple submit buttons (Save/Quit) I want my form calidation to run when I save and not when I quit. I still want to submit the form when I quit.
i have tried using submit() command to submit the form but the form does not get submitted. I wonder why.I am using lotus designer to develop the form:
var f= document.forms[0]; var answer = confirm('Do you want to referback?') if(answer ){
I am relatively new to web programming (1.5 years) and this question just occurred to me.
I use jquery form validation to validate forms on my site. If my browser has javascript disabled, what is stopping someone from submitting crap through my form?
I tried disabling javascript and submitting a form on my site and it posted.
Is there a way to make it so if the user does not have javascript enabled the form cannot post?
I'm trying to submit form data to mysql database. The POST form action calls a javacript/ajax httpRequest. My php script has a redirect at the end of it on completion of data submission. My problem is after submitting the form, the redirection works, but no data is entered into database. I'm wondering if there is a problem with my httpRequest.
