Komentarze do wpisu: Zend Framework + Doctrine + jQuery Grid

Przez: Adriano

Adriano — Mon, 19 Jul 2010 19:51:08 +0000

Thank’s save my life in my service …

Przez: Kamil Adryjanek

Kamil Adryjanek — Mon, 19 Jul 2010 19:29:33 +0000

Send, enyoj!

Przez: Adriano

Adriano — Mon, 19 Jul 2010 18:19:57 +0000

Good day! My frendy I need your help, I sorry my english I m Brazilian…

I need of your exemple for studante moore !!

Przez: thantang

thantang — Wed, 07 Jul 2010 08:47:39 +0000

You can send me your’s code,plz.
I need it !
Thanks.

Przez: karimulla

karimulla — Thu, 27 May 2010 08:27:09 +0000

can you send me the source code please!

Przez: Shahab

Shahab — Thu, 29 Apr 2010 18:26:36 +0000

@Kamil Adryjanek: could you please send me the code on my id shahab_qadeer@hotmail

Przez: maddog

maddog — Wed, 14 Apr 2010 07:48:47 +0000

That’s another problem.
I want connect flexigrid and my search form / no internal flexigrid search/. Flexigrid should reload only when i pushed send button from my valided search form.

Przez: Kamil Adryjanek

Kamil Adryjanek — Wed, 14 Apr 2010 05:50:44 +0000

And what is the problem?? You just want to add a record to database? You can create simple form and add ajax submit handler to it to send data from your form via AJAX. For example:
$(’#yourForm’).submit(function() {
$.ajax({
type: ‘POST’,
url: ‘yourAjaxHandlerUrl’,
data: formData,
error: function() {
…
}
success: function() {
…
}
});

In success action you sholud reload the grid and reset the form. In error handler you can display some errors from PHP validation (you can also add JS validation).
It should help.

Przez: maddog

maddog — Tue, 13 Apr 2010 08:30:13 +0000

Idea?
Steps:
1. User open page. There is form with some inputs, button submit and empty grid.
2. User fill inputs and send data.
3. Next starts zend validation. If validation on requesting data is ok, datas are sending to grid.

Przez: Kamil Adryjanek

Kamil Adryjanek — Mon, 12 Apr 2010 15:28:22 +0000

@Chris: sent.
@maddog: what do you mean exactly? Some dialog?