我希望将我的 json 存储在控制器中的字符串变量中

Question

我有一个从服务中获取的 JSON，我想使用该 json 存储在我的字符串变量中。 所以我的分页有效。

var app = angular.module('Techapp', ['ngTable']);

app.controller('ngTableCtrl', function ($scope, ngTableParams) {
    var data = [{ Technology: "Core JAVA", CourseType: "Training Plan", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Material", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Plan", JobLevelPrefered: "JL 3+", ContactPerson: "Shanthi_A" },
                { Technology: "Core JAVA", CourseType: "Training Material", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Material", JobLevelPrefered: "JL 3+", ContactPerson: "Muthumari_T" },
                { Technology: "Core JAVA", CourseType: "Checklists", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Plan", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Plan", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Material", JobLevelPrefered: "JL 3+", ContactPerson: "Vatsala_Agrawal" },
                { Technology: "Core JAVA", CourseType: "Training Plan", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Plan", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Material", JobLevelPrefered: "JL 3+", ContactPerson: "Selvam_K01" },
                { Technology: "Core JAVA", CourseType: "Checklists", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" },
                { Technology: "Core JAVA", CourseType: "Training Plan", JobLevelPrefered: "JL 3+", ContactPerson: "Kavitha_arjun" }
               ];

    $scope.tableParams = new ngTableParams(
    {
        page: 1, // show first page
        count: 5 // count per page
    },  
    {
        total: data.length, // length of data
        getData: function($defer, params) {
            $defer.resolve(data.slice((params.page() - 1) * params.count(),  params.page() * params.count()));
        }
    });
});

app.controller('CertificationCtrl', function ($scope, ngTableParams, $http) {
    var url = "http://localhost:58402/Search.svc/abc/ListsJava/Certification";
    $http.get(url)
      .success(function (data) {
         var myjson = JSON.parse(data);
         $scope.Title = JSON.parse(myjson)
      });
    var data = [];
    $scope.tableParams = new ngTableParams(
    {
        page: 1, // show first page
        count: 5 // count per page
    }, 
    {
        total: data.length, // length of data
        getData: function ($defer, params) {
            $defer.resolve(data.slice((params.page() - 1) * params.count(), params.page() * params.count()));
        }
    });
});

就像在 ng-TableCtrl 中，json 在字符串变量中。所以我的分页工作正常。但是在我的 CertificationCtrl 中，因为我从服务中获取我的 json，所以我无法应用分页。请帮忙

谢谢

喜欢

Answer 1

如果 http://localhost:58402/Search.svc/abc/ListsJava/Certification 已经在返回您的 JSON 数据，您就不能像这样在 success 回调中调用您的分页代码吗：

app.controller('CertificationCtrl', function ($scope, ngTableParams, $http) {
  var url = "http://localhost:58402/Search.svc/abc/ListsJava/Certification";
  $http.get(url)
    .success(function (data) {
      $scope.tableParams = new ngTableParams(
         {
           page: 1, // show first page
           count: 5 // count per page
         }, 
         {
           total: data.length, // length of data
           getData: function ($defer, params) {
             $defer.resolve(data.slice((params.page() - 1) * params.count(), params.page() * params.count()));
         }
    });
  });
});

更新

请参阅jsfiddle 上的现场演示。不要担心 Javascript 部分的第 1 到 10 行；我用它来设置要发送到 jsfiddle echo 服务的测试数据。将第 11 行的 url 变量替换为您的 URL。

另外，请注意我使用$http.post 进行演示，并且还删除了您所有的JSON.parse() 代码。