Dynamic columns in grid

[jsam123Member]

Hi Peter stoev , I have gone thru the Jquery-grid-getting-started. I have not understood the //prepare the data part ,where the data is hard coded in an array. If i am getting the data from the mysql server in json format ,then how should the data be prepared. Kindly explain.

[Peter StoevKeymaster]

Hi jsam123,

The “Getting Started” help topics shows how to create a simple Grid bound to Array. If you want to bind the Grid to other data sources, you may look at the other help topics like the “Data Sources” help topic, the PHP Integration help topics, etc. In addition, there are multiple samples available online for binding the Grid to JSON, XML, TAB, CSV, JSONP – http://www.jqwidgets.com/jquery-widgets-demo/demos/jqxgrid/index.htm and http://www.jqwidgets.com/jquery-widgets-demo/demos/php/index.htm?(arctic)#demos/php/listbox.htm

Best Regards,
Peter Stoev

jQWidgets Team
http://www.jqwidgets.com

[markus_1Participant]

Hi Peter,

changing the columns of a jqxgrid works fine, even with columngroups.
Extract of my code:
$(‘#jqxgrid’).jqxGrid(‘beginupdate’);
$(‘#jqxgrid’).jqxGrid({ columngroups: columngroups1 });
$(‘#jqxgrid’).jqxGrid({ columns: columns1 });
$(‘#jqxgrid’).jqxGrid(‘endupdate’);

But if I use the parentgroup option in the columngroups, dynamic changing of the columns doesn’t work. Can you please check this?

best regards
Markus

[Author]

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