Question

我使用code列出了json列表，并且在我希望{6}每6秒附加数据来自db并具有相同的列表json格式列表之后也获得了正确的答案它也可以正常工作而没有子网格系统，但我得到一个父轮询数据正确但我没有得到子网格数据它为空（为子网格获取数据后端罚款）。

Oleg评论后第二次更新代码我附上了代码，请找到它

    var $grid = $("#list11"),
    mainGridPrefix = "s_";

jQuery("#list11").jqGrid({
    url: 'server.php?getList',
    datatype: "json",
    data: firstListJson,
    height: 200,
    colNames: ['Inv No', 'Date', 'Client'],
    colModel: [{
        name: 'id',
        index: 'id',
        width: 55
    }, {
        name: 'invdate',
        index: 'invdate',
        width: 90
    }, {
        name: 'name',
        index: 'name',
        width: 100
    }],
    rowNum: 10,
    gridview: true,
    rowList: [10, 20, 30],
    pager: '#pager11',
    loadonce: false,
    sortname: 'id',
    viewrecords: true,
    idPrefix: mainGridPrefix,
    sortorder: "desc",
    multiselect: false,
    caption: "Subgrid With polling",
    jsonReader: {
        repeatitems: false,
        id: 'id'
    },
    beforeProcessing: function(data) {

        var rows = data,
            l = data.length,
            i, item, subgrids = {};

        for (i = 0; i < l; i++) {
            item = rows[i];
            if (item.subGridData) {
                subgrids[item.id] = item.subGridData;
            }
        }
        //alert(subgrids);
        data.userdata = subgrids;
    },
    subGrid: true,
    subGridRowExpanded: function(subgridDivId, rowId) {


        var subGridID = $("#" + subgridDivId + "_t");
        var $subgrid = $("<table id='" + subgridDivId + "_t'></table>"),
            pureRowId = $.jgrid.stripPref(mainGridPrefix, rowId),
            subgrids = $(this).jqGrid("getGridParam", "userData");

        $subgrid.appendTo("#" + $.jgrid.jqID(subgridDivId));
        $subgrid.jqGrid({
            datatype: "local",
            data: subgrids[pureRowId],
            colNames: ['Emp ID', 'Name', 'Age'],
            colModel: [{
                name: 'id',
                index: 'id',
                width: 55
            }, {
                name: 'name',
                index: 'name',
                width: 90
            }, {
                name: 'age',
                index: 'age',
                width: 100
            }],
            rowNum: 10,
            rowList: [10, 20, 30],
            sortname: 'id',
            viewrecords: true,
            sortorder: "desc",
            multiselect: false

        });
    }
});
jQuery("#list11").jqGrid('navGrid', '#pager11', {
    add: false,
    edit: false,
    del: false
});





Below code
for json list of list



var firstListJson = [{
    "id": "01",
    "invdate": "2014-07-24",
    "name": "John",
    "subGridData": [{
        "id": "01",
        "name": "Krishna",
        "age": "28"
    }, {
        "id": "01",
        "name": "Jai",
        "age": "28"
    }, {
        "id": "01",
        "name": "Suresh",
        "age": "28"
    }]
}, {
    "id": "02",
    "invdate": "2014-07-24",
    "name": "Hill",
    "subGridData": [{
        "id": "01",
        "name": "Mani",
        "age": "28"
    }, {
        "id": "01",
        "name": "Raj",
        "age": "28"
    }, {
        "id": "01",
        "name": "Main",
        "age": "28"
    }]
}];

Below code
for polling code

function pollData() {
    var pollingListUrl = 'server.php?getPollList';
    $.ajax({
        type: "POST",
        url: pollingListUrl,
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        async: true,
        cache: false,
        success: function(data) {
            var $mygrid = $("#list11");
            $mygrid.jqGrid("addRowData", "id", data);
            $mygrid.trigger("reloadGrid", [{
                current: true
            }]);
        },
        error: function(x, e) {
            alert("error occur");
        }
    });
}

Answer 1

我在您的代码中看到3个错误：

您使用datatype: "xml"，但beforeProcessing的代码面向data object 而非XML-Document的代码。

顺便提一下，我强烈建议您在网格和子网格中添加gridview: true以提高性能。
您使用$mygrid.jqGrid("addRowData", "ID", data);但网格不包含列"ID"。您应该使用"id"代替。
您在"部分数据中使用了不需要的subGridData。你发布的代码中的颜色（我的意思是美化突出显示）甚至会显示出相同的颜色。

而不是

var firstListJson=[
    {"id":"01","invdate":"2014-07-24","name":"John",
        "subGridData":"[{
                "id":"01","name":"Krishna","age":"28"},
                {"id":"01","name":"Jai","age":"28"},
                {"id":"01","name":"Suresh","age":"28"}
            ]"},
    {"id":"02","invdate":"2014-07-24","name":"Hill",
        "subGridData":"[{
                "id":"01","name":"Mani","age":"28"},
                {"id":"01","name":"Raj","age":"28"},
                {"id":"01","name":"Main","age":"28"}
            ]"}
];

应该使用

var firstListJson=[
    {"id":"01","invdate":"2014-07-24","name":"John",
        "subGridData": [{
               "id":"01","name":"Krishna","age":"28"},
               {"id":"01","name":"Jai","age":"28"},
               {"id":"01","name":"Suresh","age":"28"}
            ]},
    {"id":"02","invdate":"2014-07-24","name":"Hill",
        "subGridData": [{
                "id":"01","name":"Mani","age":"28"},
               {"id":"01","name":"Raj","age":"28"},
               {"id":"01","name":"Main","age":"28"}
            ]}
];

如果subGridData的值为object（项目数组）而不是string。

如何在轮询ajax时填充子网格然后在jqgrid中应用现有网格？

1 个答案: