How to programmatically select top row of JQGrid?

How does one pro开发者_运维技巧grammatically 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);

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

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

That should work properly.

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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');

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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]);

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

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Complete code when table have header row:

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

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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]);
        }
    },
    ...
});

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When you have a header row, try this:

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

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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.