Question

我是棱角分明的新手。我有一个链接http://www.bursamalaysia.com/searchbox_data.json，我希望获得一个名称和ID列表。

我能够从json中的链接获取列表，但我需要过滤列表中不需要的项目。如果id超过4位，则删除full_name，name，short_name和id。例如：如果id：123456，则需要将其与名称，短名称一起过滤掉。

app.js

  abc: {
      name: "Momo",
      value: "kls",
      long: "KLSE",
      searchRef: KLSE_SEARCH_REF,
      searchRefURL: "http://www.bursamalaysia.com/searchbox_data.json",

    },

details.js

$ionicLoading.show();

if ($scope.currentMarket == "abc"){

    $webServicesFactory.getNotParsed($marketProvider[$scope.currentMarket].searchRefURL).then(function success(response){
        response = JSON.parse(response);
        for (var i = 0; i < response[0].length; i++){
            $scope.searchRef.push({
              name: response[0][i].name || response[0][i].full_name,
              symbol: response[0][i].short_name,
              code: response[0][i].id,
              market: $marketProvider[$scope.currentMarket].long
            });
        }
        console.info($scope.searchRef);
        $ionicLoading.hide();
    });
}

HTML

<div class="list">
    <div class="item" ng-repeat="item in searchItems" ng-click="openDetail(item)">
        <p>{{item.symbol}} - {{item.name}}</p>
        <p>{{currentMarket | uppercase}}</p>
    </div>
</div>

Answer 1

你可以使用Array.prototype.filter和Array.prototype.map，这非常优雅。

$ionicLoading.show();
  if($scope.currentMarket == "abc") {
    $webServicesFactory.getNotParsed($marketProvider[$scope.currentMarket].searchRefURL).then(
      function success(response) {
        $scope.searchRef = JSON.parse(response)[0].filter(function(itm) {
            // whatever you want to filter should pass this condition
            return itm.id.toString().length <= 3; 
        }).map(function(itm) {
            // for each item, transform to this
            return {
              name: itm.name || itm.full_name,
              symbol: itm.short_name,
              code: itm.id,
              market: $marketProvider[$scope.currentMarket].long
            };
        });

        $ionicLoading.hide();
      }
    );
  }

确保处理任何错误并使您的代码具有防御性。

Answer 2

如果您需要过滤超过4位数的ID值，那么您可以使用简单条件限制if(response[0][i].id <= 999) 以下示例：

for(var i=0; i<response[0].length; i+=1){
   if(response[0][i].id.toString().length <= 3 ) {
      $scope.searchRef.push(
        {
          name: response[0][i].name || response[0][i].full_name,
          symbol: response[0][i].short_name,
          code: response[0][i].id,
          market: $marketProvider[$scope.currentMarket].long
        }
      );
    }
    }

javascript来过滤不需要的数据

2 个答案: