Hi,
First of all, congratulations to Jqwidget team for writing such an amazing widgets code.Please accept my apology for this basic question as i am new to json.
I am trying to implement JQ data grid using mysql & PHP as described in “Bind Grid to MySQL Database” section on
http://www.jqwidgets.com/jquery-widgets-demo/demos/php/index.htm?(arctic)#demos/php/grid.htm
What i am trying to achieve is to display extra field(“Company_id”) as hyperlink along with the grid against each individual record so that i could navigate to edit company page.But the data grid class and div is not allowing me to display anything.Please guide me how can i retrieve and display extra field outside the grid box like this:
http://sdahosting.co.uk/vm7.jpg
I would appreciate you help in this regard.