Hi Barry,

The method I mentioned for Pete was setting a column with the needed id. If you are talking about the problem with the select input (Post returns value as it should but then when we load value, it displays value and not label 🙁 ) I had to do some serverside juggling to display the results correctly.

Cheers,
P.

Hi,

Just off the top of my head (and keep in mind I to am still new to jqGrid) but I think the best way would maybe be with userdata (see docs). You could then do something along the line of $(“#grid_id”).setGridParam(caption:”Output from userdata”,);

Hopefully some of the real experts will get back to you on this one to 😉

Cheers,
P.

Hi diptendu.

This is how I got mine to work. I'm sure there may be better ways of doing this but here goes:

Thanks all,

Got this working finally!

Cheers,
P.

Same here,

Linux, FF3.

Cheers,
P.

Mike said:

…

Other bug detected in my own code : don’t use “echo” or “print” in the PHP program. it destroys the structure of json string awaited by jqgrid.

Best solution here is to create the desired array and then return the result of json_encode($dataarray).

Cheers,
P.

HI.

This seems to be similar to what I need. I have a few select boxes that need to change dynamically based on the previous selection, example, First choose country, next selct box shows only cities of that country, then choose city, next shows prvinces of that city… and so on.

Tony, I didn't quite understand what you meant. I have a field in the colum of edittype select say name:country.

onInitializeForm : function (form_id) {