UI Grid Angular，网格渲染但不显示数据

Dayan 发表于 Dev

大雁

我在这里想念什么？网格呈现无错误，空白行...我检查了JSON到现在为止都很好了$scope.gridOptions.data = response['data'];好像它呈现了空数组并且从没有到达实际数据...

  app.controller('MainCtrl', ['$scope', '$http', function ($scope, $http) {
        $scope.myData = [];
        $scope.loading = true;

        $scope.gridOptions = {
            enableSorting: true,
            columnDefs: [
              { name: 'Id', field: 'PK_Inspection' },
              { name: 'Employee Id', field: 'EmployeeId' },
              { name: 'Employee Name', field: 'EmployeeName' },
              { name: 'Equipment Id', field: 'EquipmentId' },
              { name: 'Equipment Name', field: 'EquipmentName' },
              { name: 'Sequence', field: 'SequenceName' },
              { name: 'Details', field: 'SequenceDetails' },
              { name: 'Type', field: 'SequenceTypeName' },
              { name: 'Shift', field: 'ShiftName' },
              { name: 'Status', field: 'StatusName' }
            ],
            data:[]
        };

        $http.get('/Home/GetAllData')
            .then(function (response) {
                $scope.gridOptions.data = response['data'];
            })
            .catch(function (err) {
                $scope.loading = false;
                console.log("Error Receiving Data.");
            })
            .finally(function () {
                $scope.loading = false;
                console.log($scope.gridOptions.data);

            });

    }]);

数据传递到$scope.gridOptions.data：

[
    {
        "PK_Inspection": 1,
        "EmployeeId": "4433112",
        "EmployeeName": "",
        "EquipmentId": "1122113",
        "EquipmentName": "",
        "SequenceName": "UNIT 1",
        "SequenceDetails": "Bent, Dented, Broken, Torn or Deformed Parts.",
        "SequenceTypeName": "Visual Inspection",
        "ShiftName": "Day",
        "StatusName": "OK"
    },
    {
        "PK_Inspection": 2,
        "EmployeeId": "4433112",
        "EmployeeName": "",
        "EquipmentId": "1122113",
        "EquipmentName": "",
        "SequenceName": "UNIT 2",
        "SequenceDetails": "Charge, Water Levels, Vent Caps in place, Power Disconnect works.",
        "SequenceTypeName": "Visual Inspection",
        "ShiftName": "Day",
        "StatusName": "OK"
    }
]

这是HTML：

<div ng-controller="MainCtrl">
    <i class="fa fa-spinner fa-spin text-center" ng-show="loading"></i>
    <div id="mainGrid" ui-grid="gridOptions" ui-grid-edit class="myGrid"></div>
</div>

屏幕截图

在此处输入图片说明

大雁

我想通了，看来我的问题是两件事的结合。

传入的JSON是一个字符串，我不确定100％是否需要通过使用转换为对象JSON.parse然后将其传递给，$scope.gridOptions.data但这可能是我在上述原始问题中发布的代码的问题。

经过更多研究，我在Angular UI Grid官方文档中找到了一个很好的深入示例。遵循这种技术，我能够正确呈现数据。

var rowCount = 0;
var i = 0;
$scope.refreshData = function () {
    $scope.loading = true;
    $scope.myData = [];

    $http.get('/Home/GetAllData')
        .success(function (response) {
            var jsonObj = JSON.parse(response);
            rowCount = jsonObj.length;

            jsonObj.forEach(function (row) {
                row.id = i; i++;
                $scope.myData.push(row);
            });
            $scope.loading = false;

        })
        .error(function() {
            $scope.loading = false;
            console.log("Error retrieving data.");
        });
 };