jqgrid formatter not working and no response - jqgrid

Question

i have a js file and the function name called viewLineBtn.
in my server code, i have created a list of object
List<GridModelClass> addmodelResult = new List<GridModelClass>();
addmodelResult.Add(new GridModelClass { name = "AddTestApprove", label = "Approve", width = "40", hidden = false, formatter = "viewLineBtn" });
however, the viewLineBtn can not be recognize. How to solve it?

Answer 1

It seems that you returns colModel as JSON data return from the server. As the result you have formatter property which have the type string (like formatter: "viewLineBtn") instead of function (like formatter: viewLineBtn, where viewLineBtn defined before as a function).
To fix the problem you can change how you defined formatter function. Instead of the usage
function viewLineBtn (cellValue, options, rowObject) {
// do something here
return htmlFragmentOfCell;
}
you should use
$.extend($.fn.fmatter, {
viewLineBtn: function (cellValue, options, rowObject) {
// do something here
return htmlFragmentUsedInCell;
}
});
In the case you can use formatter: "viewLineBtn" in colModel.
I would recommend you to define always custom unformatter (unformat property) if you defines custom formatter. To do this in case of usage string value of formatter you should define unformatter in the following way:
$.extend($.fn.fmatter, {
viewLineBtn: function (cellValue, options, rowObject) {
// do something here
return htmlFragmentUsedInCell;
}
});
$.extend($.fn.fmatter.viewLineBtn, {
unformat: function (cellValue, options, elem) {
// do something here
return htmlFragmentFromDomElem;
}
});