我想在jqgrid的列中显示我从' / Index / CaseBatchDetailList'获得的所有数据的图像。我有我的jquery,
$(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>";
}
}
],
...
});
我刚试过img src的例子。我需要在不同的列中显示不同的图像。作为初学者,我不知道如何操纵这个。请告诉我如何实现这一目标。
答案 0 :(得分:1)
当我看到您的评论表明您在列表中有图像源时,您可以在格式化程序中执行此操作:
formatter: function (cellvalue, options, rowobject) {
return "<img src='"+cellvalue+"'>";
}