I was creating a form with validation
here is the js file is : http://pastebin.com/Ahy4dbsS
php file : http://pastebin.com/5B1hMK49
the problem is when I submit the form without clicking on any text boxes,
no validation happens, so i changed the js file so that the first text box will be focused on page load.
but when that text box is validated (ie. click outside of the form) and form can be submitted without validation. how can i validate all the text boxes on page load?
Better use a variable ex: var flag=0; and for every condition increment the flag as flag=flag+1; do this in all conditions finally make a condition:
if(flag>0){return false}else{return true}
Related
i made a plugin to upload a image file which is already exist in database.image of plugin given below.
This modal have one <form> tag.
Now i wants to implement this code to my form.so Form Have Already one form So how can i manage Both Form.
if user upload file than above form will skip and this form is submitted and above data will be lost.
My Form is like this
remember you should provide your work
$("#html_form_id").submit(function() {
$( "#modelForm_id" ).submit();
return false;
});
Description is ambiguous not clear if you want to submit first form, second form or both.
I've a jQuery code which show/hide some disabled fields, based on an user select option:
$('.fieldcontent').not('.info').hide();
$('#selector_cs').change(function() {
$('.fieldcontent').customFadeOut(100);
$('.' + $(this).val()).customFadeIn(900);
$('input').prop('disabled',false);
$('textarea').prop('disabled',false);
$('select').prop('disabled',false);
});
});
The big headache is: if one or more fields are fading in, these fields doesn't pass my php validation, nor submitting the form.
If javascript is disabled on all browsers, the form works perfectly.
PHP validation is
if(!isset($_POST['products'])) {
$products[2] = clean_var($_POST['products']);
}
else {
$error = 1;
$products[3] = 'color:#FF0000;';
}
for all fields
Is there any php /jquery solution (no ajax please, 'cause i won't make the whole form again, and don't know anything about ajax)?
Thanks in advance for help
EDIT: Just detected another error: If the jquery script fadein another section of the form, PHP doesn't validate it anymore. Why? Never had problems like this with php-jquery.
If you leave them enabled, but change the property type to hidden then PHP should see the fields when you submit them.
Update
How about the following scenario:
5 fields shown
start-hiding two fields (A+B)
'add' two type=hidden fields with the same name as A+B and copy their current value from A+B to `A+`B
PHP should see these two hidden fields instead of the fields that are currently fading out.
If the user did not submit while fading out, then when the fields have faded out remove the `A+`B fields and set the property of the two faded fields to hidden
Inverse this for showing them fields again.
Do take note that this is not a problem caused by PHP but caused by how/when a browser interprets a form field to be a valid form field to be send when submitting the form
I have been on this site all day and can't seem to find what I need - no duplicate post intended.
I have a form containing a link which calls a modal popup containing a different form.
###### PARENT PAGE #############
<form...>
...
link
...
</form>
**** MODAL POPUP ****
<form...>
...
<input type="submit" />
</form>
**** END MODAL POPUP ****
### END PARENT PAGE #############
When I submit the form in the modal popup, the parent page is refreshed to show the updated info in the corresponding section of the page; except that the first form is not submitted and when the page refreshes to update the necessary section, the contents of the first form is lost.
I have tried using ajax to refresh only the necessary section of the page but that doesn't work as the sections that need refreshing use php variables with contents from mysql.
The system does what it needs to do and I don't mind the refresh. But I need a way to keep the user data entered into the first form.
Is it possible to submit the first form at the same time as the second to the same php page or any other way of preserving the user data in the first form on page reload without submitting it.
You cannot do this with pure php. You'll need javascript and write it in a way that when you hit submit on the modal it 'puts' the information back into the parent form.
One way is to make the modal form submit button not an actual submit button.
You might even be able to get away with taking the filled out section dom elements in the modal injected back into the parent form. Some jquery plugins already do this. For example colorbox
Here is a working example using only ONE <form> tag and jquery colorbox. http://jsbin.com/olalam/1/edit
I am not a php developer, so I'll suggest an alternative approach.
Before you refresh the page, you can serialize the form and store the data locally (e.g. in a cookie) then restore the data back into the form. Granted, that will require a bit more JS code, but should get you what you want.
UPDATE: Since you mentioned that you might need a little assistance on the JS front, here is some guidance:
Grab the jquery.cookie plugin here.
Grab the jquery.deserialize plugin here.
Use the following code as a starting point.
.
// the name of the cookie
var cookieName = 'myCookieName';
function beforeSubmit() {
// serialize the form into a variable
var serializedForm = $('#id-of-form').serialize();
// store the serialized form
$.cookie(cookieName, serializedForm, { path: '/' });
}
function afterRefresh() {
// read the cookie
var serializedForm = $.cookie(cookieName);
// de-serialize the form
$('#id-of-form').deserialize(serializedForm, true);
}
HTH
yes this question is asked some times. but I can't find an answer for my problem.
I've posted my question here because it is more php / javascript related then drupal I guess.
So basically I have a form that is validated and on the submit part I have a ( drupal hook ) function in php where I can add / modify code.
This function catches the form fields and stores them in the database.
This means that the fields are filled correctly so I want to implement something here that opens a new tab or window for the user without closing the current one.
So I know you can't call javascript from php to use something like window.open() but is there anyway I can make it so that this window.open() is called?
I'm also open for a bit different approach if you have any.
You can use an anchor element to both submit a form and open a new window:
<script>
function submitMyForm()
{
// Validate form fields here
// ...
// If form is valid, submit it
myForm.submit();
}
</script>
Submit Form
I need to dynamically add form elements to an HTML form as soon as the Submit button is clicked but before the POST data is sent to a server. The new elements must be "read" from a PHP file on my server.
HISTORY:
Currently my HTML form has "hidden" fields that are submitted to another server for processing. I have no control over the other server. My problem is that anyone can edit these hidden fields.
How can I dynamically add form elements to the POST data as soon as the form is submitted?
You can try it this way:
First disable the submit by changing the submit button type from 'submit' to 'button' (or whatever)
Put in onclick on that button to a javascript routine (here i use submit_form()).
Create an empty div within your form. (here i call it with id = 'dynamic')
Using jquery, this is the submit_form().
I think you will need to give it some time for these elements to bind properly before submitting. Maybe a short time delay before $("#myForm").submit();
Here is the code for the submit_form() function:
function submit_form()
{
$("#dynamic").append("<input type='hidden' name='input1' value='whatever'>");
$("#dynamic").append("<input type='hidden' name='input2' value='whatever'>");
$("#myForm").submit();
}
You can post the data to your server and after it post again to the external server with the new elements attached.
Your job is done on server side.
See also:
php server-to-server post?
If you need any control over what is submitted to the other server, you have to do that yourself. Make the form submit to your own server, then validate it, add your data and re-submit it to the other server.
You can use the CURL extension in PHP to post data from your server.