Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Post preview - Passing data with AJAX and Fancybox

I'm trying to do a post preview, which will appears in a new Fancybox iframe. Since couple of weeks I'm looking for some help or best practices, but I can't find it.

My main problem is to pass the data from form (before updating database) to Fancybox window. My AJAX skills are very poor, so maybe my problem isn't so hard.

Please check the code:

$('.preview2').fancybox({
fitToView    : false,
width        : 905,
height        : 505,
autoSize    : false,
closeClick    : false,
openEffect    : 'none',
closeEffect    : 'none',
ajax: {
    type: "POST",
    cache : false,
    url: "preview.php",
    data: $('#postp').serialize()
}
});

And a calling link:

<a class="preview2" data-fancybox-type="iframe" href="preview.php" id="preview2">Preview</a>

I'm almost sure everything is fine with preview.php file, just posting the variables and printing it in right places.

Can someone check Fancybox / AJAX part?

like image 731
kacper Avatar asked Jan 14 '13 09:01

kacper


2 Answers

As I mentioned in my comments, your preview button should submit the form via ajax to get the POST preview values (we'll use ajax instead of iframe) so :

<a class="preview2" data-fancybox-type="ajax" href="preview.php" id="preview2">Preview</a>

Then you would need to bind the preview button to a manual on("click") method to submit the form via ajax firstly ....and then post the results in fancybox secondly like :

$(document).ready(function () {
  $('.preview2').on("click", function (e) {
    e.preventDefault(); // avoids calling preview.php
    $.ajax({
      type: "POST",
      cache: false,
      url: this.href, // preview.php
      data: $("#postp").serializeArray(), // all form fields
      success: function (data) {
        // on success, post (preview) returned data in fancybox
        $.fancybox(data, {
          // fancybox API options
          fitToView: false,
          width: 905,
          height: 505,
          autoSize: false,
          closeClick: false,
          openEffect: 'none',
          closeEffect: 'none'
        }); // fancybox
      } // success
    }); // ajax
  }); // on
}); // ready

See DEMO (feel free to explore the source code)

like image 97
JFK Avatar answered Sep 30 '22 13:09

JFK


I don't like the solution, so I will post my own investigation.

Why writing code with 1. .on("click", ... 2. e.preventDefault 3. $.ajax 4. this.href just to call fancybox on success, that already has all this functions inside?

If you decide to use ajax instead of iframe (like in accepted answer) you could simply add type property to fancybox() call, cause it's beening checked, and pass all other

$('.preview2').fancybox({
    type: "ajax",
    ajax: {
        data:  $('#postp').serialize() 
     // url: "someurl.php" not needed here, it's taken from href
     //                    if you need it, e.g. you have a button, not a link
     //                    use "href" property that overrides everything
     //                    not attribute, cause it's invalid
    }
 // href: "url_to_add_or_override_existing_one",
 // all other effects
 // afterLoad: function () { // other cool stuff }
});

EDIT As @JFK pointed it's not enough, you have to get form data each time you click the link, so beforeLoad() needed instead of data. Finnaly:

$('.preview2').fancybox({
    type: "ajax",
    beforeLoad: function() {
        this.ajax.data = $('#postp').serialize();
    }
});

You can remove all data-* atrributes too

FIDDLE

KISS

like image 42
vladkras Avatar answered Sep 30 '22 14:09

vladkras