Question

目前，我正在使用HttpClient，HttpPost从PHP server向我的Android app发送数据，但所有这些方法都在API 22中弃用，并在API 23，那么它有哪些备选方案呢？

我到处搜索，但我找不到任何东西。

Answer 1

我也遇到过这个问题要解决我自己上课的问题。其中基于java.net，并支持Android的API 24 请检查一下： HttpRequest.java

您可以轻松地使用此课程：

发送Http GET请求
发送Http POST请求
发送Http PUT请求
发送Http DELETE
发送没有额外数据的请求参数＆amp;检查回复HTTP status code
添加自定义HTTP Headers以请求（使用varargs）
将数据参数添加为String查询以请求
将数据参数添加为HashMap {key = value}
接受回复为String
接受回复为JSONObject
接受响应为byte []字节数组（对文件有用）

以及它们的任意组合 - 只需一行代码）

以下是一些例子：

//Consider next request: 
HttpRequest req=new HttpRequest("http://host:port/path");

示例1 ：

//prepare Http Post request and send to "http://host:port/path" with data params name=Bubu and age=29, return true - if worked
req.prepare(HttpRequest.Method.POST).withData("name=Bubu&age=29").send();

示例2 ：

// prepare http get request,  send to "http://host:port/path" and read server's response as String 
req.prepare().sendAndReadString();

示例3 ：

// prepare Http Post request and send to "http://host:port/path" with data params name=Bubu and age=29 and read server's response as JSONObject 
HashMap<String, String>params=new HashMap<>();
params.put("name", "Groot"); 
params.put("age", "29");
req.prepare(HttpRequest.Method.POST).withData(params).sendAndReadJSON();

示例4 ：

//send Http Post request to "http://url.com/b.c" in background  using AsyncTask
new AsyncTask<Void, Void, String>(){
        protected String doInBackground(Void[] params) {
            String response="";
            try {
                response=new HttpRequest("http://url.com/b.c").prepare(HttpRequest.Method.POST).sendAndReadString();
            } catch (Exception e) {
                response=e.getMessage();
            }
            return response;
        }
        protected void onPostExecute(String result) {
            //do something with response
        }
    }.execute();

示例5 ：

//Send Http PUT request to: "http://some.url" with request header:
String json="{\"name\":\"Deadpool\",\"age\":40}";//JSON that we need to send
String url="http://some.url";//URL address where we need to send it 
HttpRequest req=new HttpRequest(url);//HttpRequest to url: "http://some.url"
req.withHeaders("Content-Type: application/json");//add request header: "Content-Type: application/json"
req.prepare(HttpRequest.Method.PUT);//Set HttpRequest method as PUT
req.withData(json);//Add json data to request body
JSONObject res=req.sendAndReadJSON();//Accept response as JSONObject

示例6 ：

//Equivalent to previous example, but in a shorter way (using methods chaining):
String json="{\"name\":\"Deadpool\",\"age\":40}";//JSON that we need to send
String url="http://some.url";//URL address where we need to send it 
//Shortcut for example 5 complex request sending & reading response in one (chained) line
JSONObject res=new HttpRequest(url).withHeaders("Content-Type: application/json").prepare(HttpRequest.Method.PUT).withData(json).sendAndReadJSON();

示例7 ：

//Downloading file
byte [] file = new HttpRequest("http://some.file.url").prepare().sendAndReadBytes();
FileOutputStream fos = new FileOutputStream("smile.png");
fos.write(file);
fos.close();

Answer 2

HttpClient文档指出了正确的方向：

org.apache.http.client.HttpClient：

此级别在API级别22中已弃用。请改用openConnection（）。请访问此网页了解更多详情。

表示您应切换到java.net.URL.openConnection()。

以下是你如何做到的：

URL url = new URL("http://some-server");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("POST");

// read the response
System.out.println("Response Code: " + conn.getResponseCode());
InputStream in = new BufferedInputStream(conn.getInputStream());
String response = org.apache.commons.io.IOUtils.toString(in, "UTF-8");
System.out.println(response);

IOUtils文档：Apache Commons IO
IOUtils Maven依赖：http://search.maven.org/#artifactdetails|org.apache.commons|commons-io|1.3.2|jar

Answer 3

以下代码位于AsyncTask中：

在我的后台流程中：

String POST_PARAMS = "param1=" + params[0] + "&param2=" + params[1];
URL obj = null;
HttpURLConnection con = null;
try {
    obj = new URL(Config.YOUR_SERVER_URL);
    con = (HttpURLConnection) obj.openConnection();
    con.setRequestMethod("POST");

    // For POST only - BEGIN
    con.setDoOutput(true);
    OutputStream os = con.getOutputStream();
    os.write(POST_PARAMS.getBytes()); 
    os.flush();
    os.close();
    // For POST only - END

    int responseCode = con.getResponseCode();
    Log.i(TAG, "POST Response Code :: " + responseCode);

    if (responseCode == HttpURLConnection.HTTP_OK) { //success
         BufferedReader in = new BufferedReader(new InputStreamReader(con.getInputStream()));
         String inputLine;
         StringBuffer response = new StringBuffer();

         while ((inputLine = in.readLine()) != null) {
              response.append(inputLine);
         }
         in.close();

         // print result
            Log.i(TAG, response.toString());
            } else {
            Log.i(TAG, "POST request did not work.");
            }
        } catch (IOException e) {
            e.printStackTrace();
        }

参考： http://www.journaldev.com/7148/java-httpurlconnection-example-to-send-http-getpost-requests

Answer 4

这是我已经应用于httpclient在此版本的android 22中弃用的问题的解决方案。

 public static final String USER_AGENT = "Mozilla/5.0";



public static String sendPost(String _url,Map<String,String> parameter)  {
    StringBuilder params=new StringBuilder("");
    String result="";
    try {
    for(String s:parameter.keySet()){
        params.append("&"+s+"=");

            params.append(URLEncoder.encode(parameter.get(s),"UTF-8"));
    }


    String url =_url;
    URL obj = new URL(_url);
    HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

    con.setRequestMethod("POST");
    con.setRequestProperty("User-Agent", USER_AGENT);
    con.setRequestProperty("Accept-Language", "UTF-8");

    con.setDoOutput(true);
    OutputStreamWriter outputStreamWriter = new OutputStreamWriter(con.getOutputStream());
    outputStreamWriter.write(params.toString());
    outputStreamWriter.flush();

    int responseCode = con.getResponseCode();
    System.out.println("\nSending 'POST' request to URL : " + url);
    System.out.println("Post parameters : " + params);
    System.out.println("Response Code : " + responseCode);

    BufferedReader in = new BufferedReader(new InputStreamReader(con.getInputStream()));
    String inputLine;
    StringBuffer response = new StringBuffer();

    while ((inputLine = in.readLine()) != null) {
        response.append(inputLine + "\n");
    }
    in.close();

        result = response.toString();
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (ProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }catch (Exception e) {
        e.printStackTrace();
    }finally {
    return  result;
    }

}

Answer 5

您可以继续使用HttpClient。谷歌仅弃用他们自己的Apache组件版本。你可以像我在这篇文章中描述的那样安装新的，功能强大且不被弃用的Apache HttpClient版本：https://stackoverflow.com/a/37623038/1727132

Answer 6

哪个客户最好？

Apache HTTP客户端在Eclair和Froyo上的错误更少。这是最好的   这些版本的选择。

对于Gingerbread而言，更好的是，HttpURLConnection是最佳选择。它的   简单的API和小巧的尺寸使其非常适合Android ...

参考here了解更多信息（Android开发人员博客）

Answer 7

如果针对API 22及更早版本，则应将以下行添加到build.gradle

dependencies {
    compile group: 'org.apache.httpcomponents' , name: 'httpclient-android' , version: '4.3.5.1'
}

如果针对API 23及更高版本，则应将以下行添加到build.gradle

dependencies {
    compile group: 'cz.msebera.android' , name: 'httpclient', version: '4.4.1.1'
}

如果仍想使用httpclient库，在Android Marshmallow（sdk 23）中，您可以添加：

useLibrary 'org.apache.http.legacy'

以android {}部分中的build.gradle作为解决方法。这似乎是Google自己的一些gms库所必需的！

Answer 8

您可以使用我易于使用的自定义类。只需创建抽象类的对象（Anonymous）并定义onsuccess（）和onfail（）方法。 https://github.com/creativo123/POSTConnection

Answer 9

我在使用 HttpClent 和 HttpPost 方法时遇到类似问题，因为我不想更改我的代码，所以我在build.gradle（模块）中找到了备用选项删除＆＃39; rc3＆＃39;来自buildToolsVersion＆＃34; 23.0.1 rc3＆＃34;它对我有用。希望有助于。

我需要Android中的HttpClient替代选项，以便将数据发送到PHP，因为它不再受支持

9 个答案: