Question

您好我如何在角度为Http的情况下将此对象作为param传递？因为我的wcg public void CreateNewAccount（Users us）

$scope.RegisterUser = function(){
var us = {
  UserName:$scope.userName,
  Password:$scope.password,
  UserRoleID:null,
  Company:$scope.company,
  Terms:$scope.terms,
  ID:null,
  BuyerID:app.buyerId
};

$http.get(
    app.wcf+'/CreateNewAccount'angular.toJson({us:us}))
    .then(
    function(resp){
     app.Logger(resp.data);
    },
    function(err){
      app.Logger(err);
})};

Answer 1

您需要将对象params传递给$http.get(url, config)方法的配置。

$http.get(app.wcf + '/CreateNewAccount', {params: us})
  .then(function(resp){
     app.Logger(resp.data);
  },
  function(err){
    app.Logger(err);
})};

也就是说，您不应该将此数据作为GET请求传递，尤其是在查询字符串中没有用户名和密码。

Answer 2

对于get，您应该使用参数：

$http({method:'GET', url:'url', params:us})

Answer 3

我得到这个接缝来调用OPTIONS Remote Address:192.168.58.182:80 Request URL:http://192.168.58.182/ESService/ESService.svc/CreateNewAccount Request Method:OPTIONS Status Code:405 Method Not Allowed Request Headersview source Accept:*/* Accept-Encoding:gzip, deflate, sdch Accept-Language:en-US,en;q=0.8,pl;q=0.6 Access-Control-Request-Headers:accept, content-type Access-Control-Request-Method:POST Connection:keep-alive Host:192.168.58.182 Origin:http://localhost:8100 Referer:http://localhost:8100/ User-Agent:Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/40.0.2214.93 Safari/537.36 Response Headersview source Access-Control-Allow-Headers:Content-Type, Accept Access-Control-Allow-Methods:POST,GET,OPTIONS Access-Control-Allow-Origin:* Access-Control-Max-Age:1728000 Allow:POST Content-Length:1565 Content-Type:text/html; charset=UTF-8 Date:Fri, 30 Jan 2015 14:10:34 GMT Server:Microsoft-IIS/7.5 X-Powered-By:ASP.NET

$ http角度传递对象？

3 个答案: