发送来自Android的一个JSON HTTP POST请求 [英] Sending a JSON HTTP POST request from Android

问题描述

大家好，我使用下面的code要发送的发送对象为WCF service.This工程确定，但如果我的WCF服务还需要其他参数会发生什么HTTP POST请求？我怎么能发送他们从我的Andr​​oid客户端？ 这是code我写至今：

  StringBuilder的SB =新的StringBuilder（）;

字符串的HTTP =htt​​p://android.schoolportal.gr/Service.svc/SaveValues​​;


HttpURLConnection类的URLConnection = NULL;
尝试 {
    网址URL =新的URL（HTTP）;
    的URLConnection =（HttpURLConnection类）url.openConnection（）;
    urlConnection.setDoOutput（真正的）;
    urlConnection.setRequestMethod（POST）;
    urlConnection.setUseCaches（假）;
    urlConnection.setConnectTimeout（10000）;
    urlConnection.setReadTimeout（10000）;
    urlConnection.setRequestProperty（内容类型，应用/ JSON）;

    urlConnection.setRequestProperty（主机，android.schoolportal.gr）;
    urlConnection.connect（）;

    //创建的JSONObject这里
    JSONObject的jsonParam =新的JSONObject（）;
    jsonParam.put（ID，25）;
    jsonParam.put（说明，实）;
    jsonParam.put（使能，真）;
    OutputStreamWriter OUT =新OutputStreamWriter（urlConnection.getOutputStream（））;
    out.write（jsonParam.toString（））;
    out.close（）;

    INT型Htt presult = urlConnection.getResponse code（）;
    如果（Htt的presult == HttpURLConnection.HTTP_OK）{
        的BufferedReader BR =新的BufferedReader（新的InputStreamReader（
            urlConnection.getInputStream（），UTF-8））;
        串线= NULL;
        而（（行= br.readLine（））！= NULL）{
            sb.append（行+\ N）;
        }
        br.close（）;

        的System.out.println（+ sb.toString（））;

    }其他{
            的System.out.println（urlConnection.getResponseMessage（））;
    }
}赶上（MalformedURLException异常E）{

         e.printStackTrace（）;
}
赶上（IOException异常E）{

    e.printStackTrace（）;
    }赶上（JSONException E）{
    // TODO自动生成的catch块
    e.printStackTrace（）;
}最后{
    如果（URLConnection的！= NULL）
    urlConnection.disconnect（）;
}
 

解决方案

发布使用参数发布： -

  URL网址;
URLConnection的urlConn;
DataOutputStream类打印输出;
的DataInputStream输入;
URL =新的URL（得到codeBase的（）的toString（）+env.tcgi。）;
urlConn = url.openConnection（）;
urlConn.setDoInput（真正的）;
urlConn.setDoOutput（真正的）;
urlConn.setUseCaches（假）;
urlConn.setRequestProperty（内容类型，应用/ JSON）;
urlConn.setRequestProperty（主机，android.schoolportal.gr）;
urlConn.connect（）;
//创建的JSONObject这里
JSONObject的jsonParam =新的JSONObject（）;
jsonParam.put（ID，25）;
jsonParam.put（说明，实）;
jsonParam.put（使能，真）;
 

您错过的部分是在下面......即，如下：

  //发送POST输出。
打印输出=新DataOutputStream类（urlConn.getOutputStream（））;
printout.write（URLEn coder.en code（jsonParam.toString（），UTF-8））;
printout.flush（）;
printout.close（）;
 

的事情剩下你可以做到这一点。

hi guys i'm using the code below to send an http post request which sends an object to a wcf service.This works ok, but what happens if my wcf service needs also other parameters?how can i send them from my android client? this is the code i've written so far:

StringBuilder sb = new StringBuilder();  

String http = "http://android.schoolportal.gr/Service.svc/SaveValues";  


HttpURLConnection urlConnection=null;  
try {  
    URL url = new URL(http);  
    urlConnection = (HttpURLConnection) url.openConnection();
    urlConnection.setDoOutput(true);   
    urlConnection.setRequestMethod("POST");  
    urlConnection.setUseCaches(false);  
    urlConnection.setConnectTimeout(10000);  
    urlConnection.setReadTimeout(10000);  
    urlConnection.setRequestProperty("Content-Type","application/json");   

    urlConnection.setRequestProperty("Host", "android.schoolportal.gr");
    urlConnection.connect();  

    //Create JSONObject here
    JSONObject jsonParam = new JSONObject();
    jsonParam.put("ID", "25");
    jsonParam.put("description", "Real");
    jsonParam.put("enable", "true");
    OutputStreamWriter out = new   OutputStreamWriter(urlConnection.getOutputStream());
    out.write(jsonParam.toString());
    out.close();  

    int HttpResult =urlConnection.getResponseCode();  
    if(HttpResult ==HttpURLConnection.HTTP_OK){  
        BufferedReader br = new BufferedReader(new InputStreamReader(  
            urlConnection.getInputStream(),"utf-8"));  
        String line = null;  
        while ((line = br.readLine()) != null) {  
            sb.append(line + "\n");  
        }  
        br.close();  

        System.out.println(""+sb.toString());  

    }else{  
            System.out.println(urlConnection.getResponseMessage());  
    }  
} catch (MalformedURLException e) {  

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

    e.printStackTrace();  
    } catch (JSONException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}finally{  
    if(urlConnection!=null)  
    urlConnection.disconnect();  
}  

解决方案

Posting parameters Using POST:-

URL url;
URLConnection urlConn;
DataOutputStream printout;
DataInputStream  input;
url = new URL (getCodeBase().toString() + "env.tcgi");
urlConn = url.openConnection();
urlConn.setDoInput (true);
urlConn.setDoOutput (true);
urlConn.setUseCaches (false);
urlConn.setRequestProperty("Content-Type","application/json");   
urlConn.setRequestProperty("Host", "android.schoolportal.gr");
urlConn.connect();  
//Create JSONObject here
JSONObject jsonParam = new JSONObject();
jsonParam.put("ID", "25");
jsonParam.put("description", "Real");
jsonParam.put("enable", "true");

The part which you missed is in the the following... i.e., as follows..

// Send POST output.
printout = new DataOutputStream(urlConn.getOutputStream ());
printout.write(URLEncoder.encode(jsonParam.toString(),"UTF-8"));
printout.flush ();
printout.close ();

The rest of the thing you can do it.

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