I'm struggling with just getting data from an ajax created radio button to correctly post the data at the click of a submit button, and I'm getting frustrated trying to figure out where I've gone wrong.
Am I wrong in thinking that once the ajax radio buttons have been properly created in my HTML driven website from my ajax backend shown here
while(odbc_fetch_row($rs)) {
$rowCarDeets = odbc_result($rs, 'CarDescript');
$response = $response . "<input type='radio' name='CarDescription' value='" . $rowCarDeets . "'>". $rowCarDeets . "</input></br>";
}
That all I need to do is grab it with a $_POST['CarDescription']? I have very little experience working with this, so if there is something that is supposed to be very obvious I need to add, please don't hesitate to suggest it just in case I did forget.
Just as a note, everything else in the form (not provided via ajax, simply taking user input) that these radio buttons are in get posted perfectly.
in $_POST, you should be able to just get the checked value by using the element's name attribute.
so: $_POST['CarDescription']
Related
I am using CakePHP 2.3.6. In a page, I have a checkbox. I want, when I select the checkbox(click on it), an ajax post request will be sent to a page, some data will be saved in the db, and after that a successful notification will appear. And when I deselect it(click on it again), an ajax post request will be sent to that page again, some data will be saved in the db, and after that a successful notification will appear. So, my view file is like this :
.
.
.
<form>
<label><input type="checkbox" class="item_id" />Save the data</label>
</form>
.
.
.
<?php $this->Js->get('.item_id')->event('onclick',$this->Js->request(array('controller'=>'items','action'=>'saveData',$item['Item']['id'],$another_id),array('async'=>true)));?>
In the saveData() function :
if($this->request->is('post')){
//save the data in the db
}
But, when I click on the checkbox, nothing happens.
I put the php code for the ajax call at the end of that specific view file, not the default layout. And, when I see the source code of the page from the browser("view source" option), I don't see any jQuery code that was supposed to be generated by CakePHP.
What is the problem ? What should I do here ? I need to save some data in my db and show some notification.
Thanks.
Have you included the Js helper in the controller like
public $helpers = array('Js' => array('Jquery'));
replace jquery with your javascript file.
Also try including this line just before the ending tag:
echo $this->Js->writeBuffer(); // Write cached scripts
to print the cache.
here's the situation:
I have a page generated from a custom post type in Wordpress.
This page has a custom metabox with a "thank you" message.
In this page there is a contact form generated with Contact Form 7.
What I would like to archive is that when someone submits the post, a ajax function replaces the form with the content in the metabox.
Now the content in the metabox has also some script (tracking codes) that I don't want to load before the the form is submitted.
Any suggestions?
Thanks in advance!
It is possible to do what you want but since you are using CF7 it requires quite a lot of customisation.
It would be way easier to achieve the desired result if you could hardcode the php form into the page.
Let me know if that's an option for you and maybe I can give you an hand.
I want to show my users data from my mysql database without reloading the page.
I am using Codeigniter. I have a dropdown menu like the following, - when the page loads I want it to show all data by default. But when any user selects any name it will query to database and show results immediately without having the page reloaded.
I already have a controller, model and view to display all data from database but I don't know how to get the dropdown menu work after a person has selected a value, to fetch data from database and show immediately without reloading the page.
I have some basic knowledge of PHP but I have no idea about AJAX. Would you please kindly give me an example or idea on how to do this?
I am not expecting you to write the code for me, I am just asking for an example or a guideline. :)
Thanks in Advance.
<form>
<select name="info">
<option value="">Select a person:</option>
<option value="11080101">John</option>
<option value="11080102">Bon Jovi</option>
</select>
</form>
It sounds like with what you are aiming to achieve this isn't really a codeigniter related question but more a HTML, JQUERY question.
Using jQuery To Manipulate and Filter Data over at Net-tuts shows a jquery solution to sorting and filtering data. Their solution is based on a table but the principals are there so a modification can get it to do what you want it to do.
well i can give you the flow of how can you try it
trigger a jquery event on selecting an dropdown menu item. if you don't have any idea how to do it try reading the documentation of jquery for using selectors .
once you have the selected element , extract its value and send an ajax call to your path
like this (for jquery ajax visit jqapi it has all the documentation for jquery ajax functions and its derivatives)
$.post('/user/data/' + id , function(response) {
console.log(response)
})
or
$.post('/user/data/', 'id=' + id , function(response) {
console.log(response)
})
and now you have your data in reponse so you can do whatever you wish to do with it
I am trying to use CodeIgniter and jQuery-ui dialog to create a modal window with form to update user information.
The process should be like:
1. Press a button on a view page.
2. A modal window pops up.
3. Inside the window is a form that a user can fill.
4. If the user filled something before, the information should be shown in corresponding field
5. Click the update button on the modal window to save the changes to database.
Can anyone provide a good sample of this process?
I used ajax to pass the data but it didn't work when I was trying to update the data to the database. It would be nice if an example of how to pass data from ajax to php and how php handle that.
Thanks,
Milo
well the jquery bit for post(), get(), ajax() works the same in any measure you would normally use it.. key difference here is with CI you can't post directly to a file-name file-location due to how it handles the URI requests. That said your post URL would be the similar to how you would access a view file normally otherwise
ie: /viewName/functionName (how you've done it with controllers to view all along. post, get, ajax doesnt have to end in a extension. I wish I had a better example then this but I can't seem to find one at the moment..
url = '/home/specialFunction';
jQuery.get(url, function(data) {
jQuery("#div2display").html(data);
});
in the case of the above you notice despite it not being a great example that. you have the url with 2 parameters home and specialFunction
home in this case is the controller file for home in the control folder for the home file in views the specialFunction is a "public function" within the class that makes the home controller file. similar to that of index() but a separate function all together. Best way I have found to handle it is through .post() and a callback output expected in JSON cause you can form an array of data on the php side json_encode it and echo out that json_encode and then work with that like you would any JSON output. or if your just expecting a sinlge output and not multiples echoing it out is fine but enough of the end run output thats for you to decide with what your comfortable doing currently. Hopefully all around though this gives you some clairity and hopefully it works out for you.
I'm requesting an ASP.net MVC view into a live box and the view contains form fields that have been marked up with attributes to be used by JQuery's unobtrusive validators plug-in.
The client script is not however working and my theory is that its because the validation framework is only being triggered on page load which has long since passed by the time the MVC view has been loaded into the live box.
Thus how can I let the validation framework know that it has new form fields to fix up?
Cheers, Ian.
var $form = $("form");
$form.unbind();
$form.data("validator", null);
$.validator.unobtrusive.parse(document);
// Re add validation with changes
$form.validate($form.data("unobtrusiveValidation").options);
You may take a look at the following blog post. And here's another one.
Another option, rather trick, which worked for me. Just add following line in the beginning of the partial view which is being returned by ajax call
this.ViewContext.FormContext = new FormContext();
Reference
For some reason I had to combine bjan and dfortun's answers...
So I put this in my view:
#{
this.ViewContext.FormContext = new FormContext();
}
And this execute this after the ajax call finishes:
var form = $("#EnrollmentForm");
form.unbind();
form.data("validator", null);
$.validator.unobtrusive.parse(document);
form.validate(form.data("unobtrusiveValidation").options);
I had a similar issue. I had a form that was using Ajax requests to re-display a part of the form with different form fields. I used unobtrusive validation by manually doing it on the client side using the
#Html.TextBoxFor
for my text boxes. For some reason the validation works when attempting to submit with invalid fields (i.e., the text boxes get outlined in red and the appropriate error messages display with the content I put in the
data_val_required
attribute, for example.
However, after I click a button that makes an Ajax request to modify the form with different fields and then submit again, only the red outline on the invalid fields display, but no error messages are rendered.
bjan's trick worked for me, but I still can't see what was causing the issue. All the HTML necessary to carry out the client-side validation was there I just can't figure out why the error message attribute values wouldn't display.
All I can think of is that the jQuery validation code doesn't make a second attempt to check the form fields after a submit was made.