CODEWITHSUNDEEP
c#jqueryrazormodel-view-controllerjqgrid
Kavitha P.
Kavitha P.

Reputation: 155

How to display an image in the column of jqgrid

I want to display a image in the column of jqgrid for all data I get from '/Index/CaseBatchDetailList'. i have my jquery like,

 $(function () {
var outerwidth = $('#TableCaseDtlsView').width();
$("#CaseBatchDetailList").jqGrid({
    url: '/Index/CaseBatchDetailList',
    datatype: 'json',
    mtype: "GET",
    colNames: ["Serial Number", "Denomination", "Note Image", "Back Image"],
    colModel: [
                    { name: "serialno", sortable: true, width: "120px", align: "center" },
                    { name: "denomination", sortable: true, width: "120px", align: "left" },
                    {
                        name: "frontimageurl", sortable: false, width: "290px", align: "center", fixed: true,
                        formatter: function () {
                            return "<img src = /F:/MCP_Images/635101007551068098_324.jpg>";
                        }
                    },
                    {
                        name: "backimageurl", sortable: false, width: "290px", align: "center", fixed: true,
                        formatter: function () {
                            return "<img src = /F:/MCP_Images/635101007551068098_324.jpg>";
                        }
                    }
    ],
 ...
});

i just tried with the example for img src. I need to display the different images in different columns. As iam the beginner i dont know how to manipulate this.Kindly tell me how to achieve this.

Upvotes: 0

Views: 3350

Answers (1)

Jai
Jai

Reputation: 74738

As i saw your comment that you are having the image source in the list, then within your formatter you can do this:

formatter: function (cellvalue, options, rowobject) {
    return "<img src='"+cellvalue+"'>";
}

Upvotes: 1

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