Question

How does one programmatically select the top row of a JQGrid. I want to have the top row already selected when it is opened on the page. My grid is sorted by a descriptive column so the first row s id could be any number. I know the method to use I just don t know how to get the rowid for the top (first) row. The method is:

jQuery("#mygrid").setSelection(rowid, true);

Answer 1

Or, without using the jqGrid API, you should be able to retrieve the top row by navigating the DOM:

var top_rowid = $( #mygrid tbody:first-child tr:first ).attr( id );

Answer 2

The answer above was close, but the case was off. It should be:

$("#mygrid").getDataIDs()[0];

That should work properly.

Answer 3

jqGrid supports a setSelection method it just needs to be called correctly:

var grid = jQuery("#mygrid"),
    ids = grid.jqGrid("getDataIDs");
if(ids && ids.length > 0)
    grid.jqGrid("setSelection", ids[0]);

Answer 4

 $("#mygrid").getDataIDs()[0]; // SO now requires 30 characters, so....

Answer 5

Complete code when table have header row:

var top_rowid = $( #mygrid tr:nth-child(2) ).attr( id ); 
$("#mygrid").setSelection(top_rowid, true);

Answer 6

I ve used the following:

var grid = $( #list );
grid.jqGrid({
    ...
    gridComplete: function() {
        var ids = grid.jqGrid("getDataIDs");
        if(ids.length > 0) { 
            grid.jqGrid("setSelection", ids[0]);
        }
    },
    ...
});

Answer 7

When you have a header row, try this:

var top_rowid = $( #mygrid tbody:first-child tr:nth-child(2) ).attr( id );

Answer 8

If you have header row try this :

$( #tb_par tbody:first-child tr:nth-child(2) ).trigger("click");

If not than :

$( #mygrid tbody:first-child tr:first ).trigger("click");

It will directly trigger click event of the JqGrid.

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