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Question  posted by oVox on Dec 06, 2007 11:37
open dhtmlx forum
Submit Grid

Hi!, this is my source code:

<form name="main" method="post">
<input type="hidden" name="behavior"    value="load" />

<div id="gridList" style="width:100%;height:660px;background-color:white;overflow:hidden">
<div id="title_flt_box" style="display:none">
<input id="Filter1" type="100%" style="border:1px solid gray;" onkeydown="doSearch(arguments[0]||event)" onclick="(arguments[0]||window.event).cancelBubble=true;">
</div>
</div>
</form>

My grid is IN a form, but when i submit this form all the checboxes (checked) doesn't go through post, how can i do this?
The hidden input "Filter1" doesn't go post too.....the only data is the hidden input "behavior"...
Please help.
My Version is dhtmlxGrid_pro_v14_71022

Regards, oVox.
Answer posted by Support on Dec 07, 2007 02:05
Please be sure that dhtmlxgrid_form.js included in your page, without this file grid will work correctly but integration with form will not work

>>The hidden input "Filter1" doesn't go post too....
Not related to grid component, it just miss "name" attribute.

Answer posted by oVox on Dec 07, 2007 06:25

Ok, you were right, Filter1 now is OK by POST but grid (at least checkboxes) doesn'tgo by POST, can anybody Help?

My Includes:

dhtmlxcommon.js
dhtmlxgrid.js
dhtmlxgrid_srnd.js
dhtmlxgridcell.js
dhtmlxgrid_excell_link.js
dhtmlxgrid_excell_calck.js
dhtmlxgrid_excell_acheck.js
dhtmlxgrid_excell_calendar.js
dhtmlxgrid_excell_clist.js
dhtmlxgrid_form.js

My grid:

  mygrid = new dhtmlXGridObject('gridList');
  mygrid.setImagePath("<?=$url_library?>dhtmlx/dhtmlxGrid/codebase/imgs/");
  mygrid.setHeader("-,Nombre,Valor sugerido,Descripción");
  
  mygrid.setInitWidths("25,*,*,*");
  mygrid.setColAlign("left,left,left,left");
  mygrid.setColTypes("ch,ed,ed,ed");
  
  mygrid.setColumnMinWidth(50,0);
  mygrid.setEditable (true);
  mygrid.enableKeyboardSupport(true);

  mygrid.setOnColumnSort(updatedSort); 

  mygrid.setSkin("light");
  mygrid.setOnRowDblClickedHandler(doOnRowDblClicked);
  mygrid.init();
  mygrid.enableSmartRendering(true);
  mygrid.setOnLoadingEnd(setCounter)

 

Answer posted by Support on Dec 10, 2007 02:53
Your code seems correct, but you are using "Smart Rendering" mode which can cause some side effects on form integration.
If problem caused by smart rendering usage ( you can try to comment enableSmartRendering command ) - please contact us directly at support@dhtmlx.com - we will send latest code which was adapted for smart rendering mode.