Question

我需要将参数传递给下面的webmethod。我有值Param，需要传递给webmethod。

var param='test'
$http({
method:"POST",
url:'/sites/Demo/_layouts/15/demo/Demo.aspx/mywebmethod',   
data: JSON,
headers: {
'Content-Type': 'application/json'
}
})
.then(function(data) 
{
$scope.Jobject = data.data.d;
});

请告知。

Answer 1

您可以传递params关键字参数或data关键表单正文数据GET方法

var params = {
  name: 'This is parameter query'
}
var data = {
  name: 'This is body data'
}
$http({
  method: 'POST',
  url: '/sites/Demo/_layouts/15/demo/Demo.aspx/mywebmethod',
  data: data,
  params: params
}).then(function successCallback(response) {
    $scope.Jobject = response.data.d;
  }, function errorCallback(response) {
    console.log(response);
  });

或只是使用

$http.post('/sites/Demo/_layouts/15/demo/Demo.aspx/mywebmethod', data).then(function successCallback(response) {
        $scope.Jobject = response.data.d;
      }, function errorCallback(response) {
        console.log(response);
      });

Answer 2

您可以使用$http.post

$http.post('/sites/Demo/_layouts/15/demo/Demo.aspx/mywebmethod', params)

有关详细信息，请查看official documentation

Answer 3

$http({
 method:"POST",
 url:'/sites/Demo/_layouts/15/demo/Demo.aspx/mywebmethod',   
 data: JSON,
 headers: {
 'Content-Type': 'application/json'
 },
 params: {name: param}
})
.then(function(data) 
{
  $scope.Jobject = data.data.d;
});

假设您要将params字符串作为name

发送

将参数传递给webmethod $ http post

3 个答案: