如何在MVC中选择JQGrid的selectedrow名称值 [英] How to select the selectedrow name value of JQGrid in MVC

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问题描述



我有以下代码.我认为:

Hi,

I have the following code. In my view:

        $(document).ready(function () {
                    $("#ListOfPeople").jqGrid({
                        url: '/Absence/PeopleList/',
                datatype: 'json',
                type: 'GET',
//                mtype: 'POST',
//                postData: {
//                    PersonRoleKey: @*@Model.personviewmodel.id*@
//                },
                colNames: ['Name'],
                colModel: [
                            { name: 'Name', index: 'Name', width: 200, align: 'center', sortable: true, editable: false }

                           ],
                rowNum: 10,
                sortname: 'Name',
                sortorder: 'asc',
                caption: 'List of People',
                hiddengrid: false,
                pager: '#nameofpeoplestoolbar',
                pgbuttons: true,
                pginput: false,
                scroll: true,
                scrollOffset:1,
                autowidth:true,
                sortable: true, // This and following line sorts the data
                loadonce: true,
                                viewrecords:true

                            });
});


如何获得Selected JQGrid ID并将其传递给控制器​​,该控制器根据传递的值显示数据?

谢谢


How can I get the Selected JQGrid ID to pass it it to a controller where it dispalys the data based on the passed value?

Thank you

推荐答案

(文档).ready(功能(){
(document).ready(function () {


(" ).jqGrid({ 网址:' /Absence/PeopleList/', 数据类型:' json', 类型:' GET'// mtype:'POST', // postData:{ // PersonRoleKey:@ * @ Model.personviewmodel.id * @ // }, colNames:[' 名称'], colModel:[ {名称:' 名称',索引:' 名称',宽度: 200 ,对齐:' 居中',可排序: true ,可 false } ], rowNum: 10 , 排序名称:' 名称', 排序顺序:' asc', 标题:' 人员名单, hiddengrid: false , 传呼机:' #nameofpeoplestoolbar', pgbuttons: true , pginput: false , 滚动:, scrollOffset: 1 , 自动宽度:, sortable: true // 数据 loadonce: true , viewrecords:为真 }); });
("#ListOfPeople").jqGrid({ url: '/Absence/PeopleList/', datatype: 'json', type: 'GET', // mtype: 'POST', // postData: { // PersonRoleKey: @*@Model.personviewmodel.id*@ // }, colNames: ['Name'], colModel: [ { name: 'Name', index: 'Name', width: 200, align: 'center', sortable: true, editable: false } ], rowNum: 10, sortname: 'Name', sortorder: 'asc', caption: 'List of People', hiddengrid: false, pager: '#nameofpeoplestoolbar', pgbuttons: true, pginput: false, scroll: true, scrollOffset:1, autowidth:true, sortable: true, // This and following line sorts the data loadonce: true, viewrecords:true }); });


如何获得Selected JQGrid ID并将其传递给控制器​​,该控制器根据传递的值显示数据?

谢谢


How can I get the Selected JQGrid ID to pass it it to a controller where it dispalys the data based on the passed value?

Thank you


尝试以下功能

Try following function

function getSelectedRow() {
var grid = jQuery("#JQGrid1");
var rowKey = grid.getGridParam("selrow");
if (rowKey)
alert("Selected row primary key is: " + rowKey);
else
alert("No rows are selected");
}



要获取更多帮助,请找到链接有用.


HTH



For more help find this link useful.


HTH


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