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Home/ Questions/Q 1081753
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Editorial Team
  • 0
Editorial Team
Asked: May 16, 20262026-05-16T22:12:31+00:00 2026-05-16T22:12:31+00:00

When I am using rowSelect and sending the values of column 1 and 2

  • 0

When I am using rowSelect and sending the values of column 1 and 2 its not sending the value selected for second rowSelect …instead only sending all the value of column 1 but not column 2 . …

my approach:

jQuery("#viewt").click( function(){

    var grid = jQuery("#inventoryInq");
    var ids =grid.jqGrid('getGridParam','selarrrow');
    if (ids.length>0) {
        var names = [];
        for (var i=0, il=ids.length; i < il; i++) {
            var name = grid.jqGrid('getCell', ids[i], 'sku');
            names.push(name);
        }

        $.ajax({
            type: "POST",
            url: "/cpsb/transactionHistory.do",
            data:{
                method:"getTransactionHistory",
                lpn:JSON.stringify(ids),
                sku:JSON.stringify(name)
            },
            dataType: "json",
            success: function(msg){
                alert(msg);
            }

        });
    }

});

I have set key=true for column 1 .

update:1

 jQuery("form#viewform").submit( function(){

         var grid = jQuery("#inventoryInq");
         var id =grid.jqGrid('getGridParam','selrow');
           if (id) { 
             var ret = grid.jqGrid('getRowData',id);
             } 

             url:"/cpsb/transactionHistory.do?method=getTransactionHistory&lpn="+ret.licensePlateNumber+"&sku="+ret.sku;  


          });
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T22:12:31+00:00Added an answer on May 16, 2026 at 10:12 pm

    It seems to me, that your error is very easy: you should replace JSON.stringify(name) to JSON.stringify(names).

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