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HTTP POST请求通过截击发送JsonObject

android-volley http-post http android java

1

Hcetipe · 技术社区 · 6 年前

我正试图通过Android上的Volley库发出httppost请求。

API文档提供了这个curl请求:

curl -H "Authorization: Token $API_KEY" \
    --header "Content-Type: application/json" \
    --compressed \
    --data @- \
    https://api.idbus.com/v1/search <<JSON
      {
        "origin_id": 1,
        "destination_id": 17,
        "date": "2015-08-15",
        "passengers": [
          { "id": 1, "age": 30 },
          { "id": 2, "age": 30 },
          { "id": 3, "age": 1 }
        ]
      }
JSON

但是我想把它转换成一个带有Volley的HTTP请求。

问题是,当我启动我的代码时,当curl请求工作时,我得到一个404错误。

我认为问题来自我将JSON对象传递到请求的方式。但我没有发现错误,希望有人能帮我,提前谢谢!

这是我的密码:

    String url = "https://api.idbus.com/v1/search";
    final JSONObject jsonBody = new JSONObject();
    JSONArray passengers = new JSONArray();

    try {
        passengers.put(getPerson(1, 30));
        passengers.put(getPerson(2, 30));
        passengers.put(getPerson(3, 1));
    } catch (JSONException e){
        e.printStackTrace();
    }
    try {
        jsonBody.put("origin_id", 1);
        jsonBody.put("destination_id", 17);
        jsonBody.put("date", "2015-08-15");
        jsonBody.put("passengers", passengers);
    } catch (JSONException e){
        e.printStackTrace();
    }

    JsonObjectRequest stringRequest = new JsonObjectRequest(Request.Method.POST, url, jsonBody,
            new Response.Listener<JSONObject>() {
                @Override
                public void onResponse(JSONObject response) {
                    mTextView.setText(response.toString());
                }
            }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            mTextView.setText("That didn't work!");
        }
    }){
        @Override
        public Map<String, String> getHeaders() throws AuthFailureError {
            Map<String,String> params = new HashMap<String, String>();
            params.put("Authorization", "Token EnVo01jqN2XJIAd8vlLkw");
            params.put("Content-Type","application/json");
            return params;
        }
    };
    queue.add(stringRequest);
}
JSONObject  getPerson(int id, int age) throws JSONException {
    JSONObject person = new JSONObject();
    person.put("id", id);
    person.put("age", age);
    return person;
}

我的Json对象看起来像:

    {  
       "origin_id":1,
       "destination_id":17,
       "date":"2015-08-15",
       "passengers":[  
          {  
             "id":1,
             "age":30
          },
          {  
             "id":2,
             "age":30
          },
          {  
             "id":3,
             "age":1
          }
       ]
    }

2 回复 | 直到 6 年前

1

Wasim Abuzaher 6 年前

试着丢掉那个 final jsonBody JSO对象

编辑:

您将int值作为字符串发送,因此另一端的服务器可能对此不太满意: 发送内容:

 {"origin_id":"1",
"destination_id":"17",
"date":"2015-08-15",
"passengers":[{"id":"1","age":"30"},{"id":"2","age":"30"},{"id":"3","age":"1"}]}

{"origin_id":1,
"destination_id":17,
"date":"2015-08-15",
"passengers":[{"id":1,"age":30},{"id":2,"age":30},{"id":3,"age":1}]}

try {
            passengers.put(getPerson(1, 30));
            passengers.put(getPerson(2, 30));
            passengers.put(getPerson(3, 1));
        } catch (JSONException e){
            e.printStackTrace();
        }
        try {
            jsonBody.put("origin_id", 1);
            jsonBody.put("destination_id", 17);
            jsonBody.put("date", "2015-08-15");
            jsonBody.put("passengers", passengers);
        } catch (JSONException e){
            e.printStackTrace();
        }

以及 getPerson 现在看起来应该是用 int String 对于 id age :

JSONObject getPerson(int id, int age) throws JSONException

编辑2-什么有效

2

0

nupadhyaya 6 年前

params.put("Authorization", "Token EnVo01jqN2XJIAd8vlLkw");

拆下 Token 文本:

params.put("Authorization", "EnVo01jqN2XJIAd8vlLkw");