Code example of linking to a page from a cell inside a grid.
Tony,
I am trying to use the formatter(showlink) option:
My code:
There is a field in the sub grid row that determines what column to hide.
That still doesn't work.
this still doesn't work?
I solved my problem by added a second jsonReader into the deifinition of the subGrid.
OK, I solved the sytax
For those that have been following this thread:
To toggle the subgrid indicator on/off based on field(s) in your main query.
$("#mygrid").jqGrid({
...,
Thanks once again.
Thanks Tony,
That worked!
Jim P said:
In Firebug on FF I get this error on the code you ask me to run.
From running this code:
afterInsertRow: function(rowid,rowdata,rowelem){
Anyone,
I'm completely lost here.
I'm a complete nubie to jqgrid.
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