jqgrid load large data set without pagination

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On example of this demo you can see the time of loading 1500 rows for your grid in case of usage of gridview: true.

The most performance problem of your example are inside of loadComplete function. If you do need to make some modifications on the grid you should use jQuery to manipulate the grid contain. The best performance you can archive if you use DOM elements of the grid directly like in the example

loadComplete: function() {
    var i=0, indexes = this.p._index, localdata = this.p.data,
        rows=this.rows, rowsCount = rows.length, row, rowid, rowData, className;

    for(;i<rowsCount;i++) {
        row = rows[i];
        className = row.className;
        //if ($(row).hasClass('jqgrow')) { // test for standard row
        if (className.indexOf('jqgrow') !== -1) {
            rowid = row.id;
            rowData = localdata[indexes[rowid]];
            if (rowData.amount !== "200") {
                // if (!$(row).hasClass('ui-state-disabled')) {
                if (className.indexOf('ui-state-disabled') === -1) {
                    row.className = className + ' ui-state-disabled';
                }
                //$(row).addClass('ui-state-disabled');
            }
        }
    }
}

You can see the corresponding example live here.

In the implementation of loadComplete function I use the fact, that jqGrid having loadonce:true parameter use internal parameters _index and data which can be used to access the contain of the grid. In the example I disabled the rows which not contain “200” in the amount column.

UPDATED: The answer describes how to use rowattr callback to simplify the solution and to have the best performance (in case of gridview: true of cause).

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