如何使用Android中的HTTPClient在JSON中发送POST请求？

Question

我试图弄清楚如何使用HTTPClient从Android发布JSON.我一直试图解决这个问题,我已经在网上找到了很多例子,但我无法让它们中的任何一个工作.我相信这是因为我缺乏一般的JSON /网络知识.我知道那里有很多例子,但有人可以指点我一个实际的教程吗？我正在寻找一步一步的过程,包括代码和解释为什么你要做每一步,或者说那个步骤做了什么.它不需要是复杂的,简单的就足够了.

再一次,我知道那里有很多例子,我只是在寻找一个例子来解释究竟发生了什么以及为什么这样做.

如果有人知道关于这个的好Android手册,请告诉我.

再次感谢@terrance的帮助,这是我在下面描述的代码

public void shNameVerParams() throws Exception{
     String path = //removed
     HashMap  params = new HashMap();

     params.put(new String("Name"), "Value"); 
     params.put(new String("Name"), "Value");

     try {
        HttpClient.SendHttpPost(path, params);
    } catch (Exception e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
 }

Answer 1

在这个答案中,我使用的是Justin Grammens发布的一个例子.

关于JSON

JSON代表JavaScript Object Notation.在JavaScript中,可以像这样引用属性object1.name,就像这样 object['name'];.本文中的示例使用了这一点JSON.

部件
一个粉丝对象,电子邮件为密钥,foo @ bar.com为值

{
  fan:
    {
      email : 'foo@bar.com'
    }
}

所以等价的对象是fan.email;或fan['email'];.两者都具有相同的值'foo@bar.com'.

关于HttpClient请求

以下是我们的作者用于制作HttpClient请求的内容.我并不认为自己是这方面的专家,所以如果有人有更好的方式来解释一些术语,那就免费了.

public static HttpResponse makeRequest(String path, Map params) throws Exception 
{
    //instantiates httpclient to make request
    DefaultHttpClient httpclient = new DefaultHttpClient();

    //url with the post data
    HttpPost httpost = new HttpPost(path);

    //convert parameters into JSON object
    JSONObject holder = getJsonObjectFromMap(params);

    //passes the results to a string builder/entity
    StringEntity se = new StringEntity(holder.toString());

    //sets the post request as the resulting string
    httpost.setEntity(se);
    //sets a request header so the page receving the request
    //will know what to do with it
    httpost.setHeader("Accept", "application/json");
    httpost.setHeader("Content-type", "application/json");

    //Handles what is returned from the page 
    ResponseHandler responseHandler = new BasicResponseHandler();
    return httpclient.execute(httpost, responseHandler);
}

地图

如果您不熟悉Map数据结构,请查看Java Map参考.简而言之,地图类似于字典或散列.

private static JSONObject getJsonObjectFromMap(Map params) throws JSONException {

    //all the passed parameters from the post request
    //iterator used to loop through all the parameters
    //passed in the post request
    Iterator iter = params.entrySet().iterator();

    //Stores JSON
    JSONObject holder = new JSONObject();

    //using the earlier example your first entry would get email
    //and the inner while would get the value which would be 'foo@bar.com' 
    //{ fan: { email : 'foo@bar.com' } }

    //While there is another entry
    while (iter.hasNext()) 
    {
        //gets an entry in the params
        Map.Entry pairs = (Map.Entry)iter.next();

        //creates a key for Map
        String key = (String)pairs.getKey();

        //Create a new map
        Map m = (Map)pairs.getValue();   

        //object for storing Json
        JSONObject data = new JSONObject();

        //gets the value
        Iterator iter2 = m.entrySet().iterator();
        while (iter2.hasNext()) 
        {
            Map.Entry pairs2 = (Map.Entry)iter2.next();
            data.put((String)pairs2.getKey(), (String)pairs2.getValue());
        }

        //puts email and 'foo@bar.com'  together in map
        holder.put(key, data);
    }
    return holder;
}

请随时评论有关这篇文章的任何问题,或者如果我没有做出明确的事情,或者我没有触及你仍然感到困惑的事情......等等.

(如果Justin Grammens不同意,我会退出.但如果没有,那么感谢Justin对此感到很酷.)

更新

我刚刚发表了关于如何使用代码的评论,并意识到返回类型中存在错误.方法签名设置为返回一个字符串,但在这种情况下它没有返回任何东西.我将签名更改为HttpResponse并将引用此链接获取响应主体的HttpResponse 路径变量是url并更新以修复代码中的错误.

Answer 2

这是@Ttrance答案的另一种解决方案.您可以轻松地外包转换.该GSON图书馆做一朵奇葩把各种数据结构到JSON和周围的其他方式.

public static void execute() {
    Map<String, String> comment = new HashMap<String, String>();
    comment.put("subject", "Using the GSON library");
    comment.put("message", "Using libraries is convenient.");
    String json = new GsonBuilder().create().toJson(comment, Map.class);
    makeRequest("http://192.168.0.1:3000/post/77/comments", json);
}

public static HttpResponse makeRequest(String uri, String json) {
    try {
        HttpPost httpPost = new HttpPost(uri);
        httpPost.setEntity(new StringEntity(json));
        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");
        return new DefaultHttpClient().execute(httpPost);
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    return null;
}

类似的可以通过使用杰克逊而不是Gson 来完成.我还建议你看看Retrofit,它为你隐藏了很多样板代码.对于更有经验的开发人员,我建议您尝试使用RxAndroid.

Answer 3

我建议使用这个HttpURLConnection代替HttpGet.正如HttpGetAndroid API级别22中已弃用的那样.

HttpURLConnection httpcon;  
String url = null;
String data = null;
String result = null;
try {
  //Connect
  httpcon = (HttpURLConnection) ((new URL (url).openConnection()));
  httpcon.setDoOutput(true);
  httpcon.setRequestProperty("Content-Type", "application/json");
  httpcon.setRequestProperty("Accept", "application/json");
  httpcon.setRequestMethod("POST");
  httpcon.connect();

  //Write       
  OutputStream os = httpcon.getOutputStream();
  BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
  writer.write(data);
  writer.close();
  os.close();

  //Read        
  BufferedReader br = new BufferedReader(new InputStreamReader(httpcon.getInputStream(),"UTF-8"));

  String line = null; 
  StringBuilder sb = new StringBuilder();         

  while ((line = br.readLine()) != null) {  
    sb.append(line); 
  }         

  br.close();  
  result = sb.toString();

} catch (UnsupportedEncodingException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
} 

Answer 4

此任务的代码太多,请检查此库https://github.com/kodart/Httpzoid 内部使用GSON并提供适用于对象的API.隐藏所有JSON详细信息.

Http http = HttpFactory.create(context);
http.get("http://example.com/users")
    .handler(new ResponseHandler<User[]>() {
        @Override
        public void success(User[] users, HttpResponse response) {
        }
    }).execute();