使用AJAX请求OAuth2令牌

时间:2018-03-05 20:33:26

标签: javascript jquery ajax oauth-2.0

我试图使用AJAX来请求令牌。这只会在localhost上用于发出请求。我有代码:

$.ajax({
  url: "TOKEN URL HERE",
  beforeSend: function(xhr) {
    xhr.setRequestHeader("grant_type", "client_credentials");
    xhr.setRequestHeader("client_id", "ENTER CLIENT ID");
    xhr.setRequestHeader("client_secret", "ENTER CLIENT SECRET");
    xhr.setRequestHeader("Content-Type", "application/json");
    xhr.setRequestHeader("Accept", "application/json");
  },
  dataType: "json",
  //content-Type: "application/json",
  type: "POST",
  success: function(response) {
    token = response.access_token;
    expiresIn = response.expires_in;
  },
  error: function(errorThrown) {
    alert(errorThrown.error);
  }
});

但是,它不起作用。这是正确的方法还是我的参数不正确? (或者是使用AJAX / JQuery / JS无法实现的OAuth2令牌请求)

2 个答案:

答案 0 :(得分:0)

确定,grant_type,client_id,client_secret应该通过标头而不是有效负载传递吗?

尝试从标题中删除它们(左接受,仅限内容类型),然后使用其余参数添加“data”属性。

类似的东西:

$.ajax({
    url: "TOKEN URL HERE",
    beforeSend: function(xhr) {
        xhr.setRequestHeader("Content-Type", "application/json");
        xhr.setRequestHeader("Accept", "application/json");
    },
    dataType: "json",
    data: {
        client_id: "ENTER CLIENT ID",
        client_secret: "ENTER CLIENT SECRET",
        grant_type: "client_credentials"
    },
    type: "POST",
    success: function(response) {
        token = response.access_token;
        expiresIn = response.expires_in;
    },
    error: function(errorThrown) {
        alert(errorThrown.error);
    }
});

答案 1 :(得分:0)

Piotr P的答案对我不起作用。我从中获取令牌的客户端在作为数据的一部分发送时不接受客户端ID和凭据。我不得不改用基本身份验证。这是对我有用的ajax调用。

$.ajax({
  "type": "POST",
  "url": "https://some.domain.com/oath/token",
  "headers": {
    "Accept": "application/json",
    "Authorization": "Basic " + btoa(username + ":" + password)
  },

  "data": {
    "grant_type": "client_credentials"
  },


  "success": function(response) {
    token = response.access_token;
    expiresIn = response.expires_in;
  },
  "error": function(errorThrown) {
    alert(JSON.stringify(errorThrown.error()));
  }
});