Question

我有一个使用Google Civic Information API来拉动政府代表的功能，当我提交一个硬编码的URL时，它可以按预期工作，但是当我尝试使用PARAMS传递相同的查询时，它仅返回一小部分。两种测试功能的响应都为200。

具有在URL中硬编码的查询参数的功能：

$scope.submit = function() {
  $http({
    method: 'GET',
    url: 'https://www.googleapis.com/civicinfo/v2/representatives?key=my-google-api-key&address=114%20Grand%20St%20Albany%20NY%2012202&fields=officials'
  }).then(function successCallback(response) {
      // this callback will be called asynchronously
      // when the response is available
      console.log(response);
    }, function errorCallback(response) {
      // called asynchronously if an error occurs
      // or server returns response with an error status.
      console.log("it didn't work.");
    });      

};

控制台记录了26位官员的答复。

具有通过PARAMS传递的查询参数的功能：

$scope.submit = function() {
  $http({
    method: 'GET',
    url: 'https://www.googleapis.com/civicinfo/v2/representatives',
    params: {
      key: 'my-google-api-key',
      address: '114%20Grand%20St%20Albany%20NY%2012202',
      fields: 'officials'
    }
  }).then(function successCallback(response) {
      // this callback will be called asynchronously
      // when the response is available
      console.log(response);
    }, function errorCallback(response) {
      // called asynchronously if an error occurs
      // or server returns response with an error status.
      console.log("it didn't work.");
    });      

};

控制台仅记录8位官员。

我在$ http，$ http.get和$ httpParamSerializer的AngularJS API引用中上下浮动，我一生都无法弄清楚自己可能会丢失什么。我在做什么错了？

Answer 1

尝试将114 Grand St Albany NY 12202而不是114%20Grand%20St%20Albany%20NY%2012202作为地址参数传递。

它应该像这样：

$scope.submit = function() {
  $http({
    method: 'GET',
    url: 'https://www.googleapis.com/civicinfo/v2/representatives',
    params: {
      key: 'my-google-api-key',
      address: '114 Grand St Albany NY 12202',
      fields: 'officials'
    }
  }).then(function successCallback(response) {
    // this callback will be called asynchronously
    // when the response is available
   console.log(response);
  }, function errorCallback(response) {
    // called asynchronously if an error occurs
    // or server returns response with an error status.
    console.log("it didn't work.");
  });      

};

我认为，通过传递114%2520Grand%2520St%2520Albany%2520NY%252012202作为参数，它将获得两次URL编码，并且您将丢失信息。

使用参数与硬编码URL时来自http.get的不同响应

1 个答案: