Question

我是android新手。我正在学习获取，发布，删除的http请求。从此我学会了获取和删除，并发布请求。但是在post请求中发送数组时出现了问题。

这是我的帖子数据结构..

{
 "customerId": "CUST01",
 "orderId": "101010",
 "orderTotal": 99.99,
 "orderDetailList": [
  {
    "lineId": "1",
    "itemNumber": "ABC",
     "quantity": 9,
     "price": 10.0
   },
   {
     "lineId": "2",
     "itemNumber": "XYZ",
     "quantity": 1,
     "price": 9.99
   }
 ]
}

如何在帖子中发送数组？

Answer 1

这里我发布一些代码来将值发布到服务器..

       public void postData() {
      // Create a new HttpClient and Post Header
       HttpClient httpclient = new DefaultHttpClient();
     HttpPost httppost = new HttpPost("http://www.yoursite.com/script.php");

      try {
// Add your data
//you can add all the parameters your php needs in the BasicNameValuePair. 
//The first parameter refers to the name in the php field for example
// $id=$_POST['customerId']; the second parameter is the value.
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("customerId", "CUST01"));
nameValuePairs.add(new BasicNameValuePair("orderId", "101010"));
  nameValuePairs.add(new BasicNameValuePair("orderTotal", "99.99"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

// Execute HTTP Post Request
 HttpResponse response = httpclient.execute(httppost);

      } catch (ClientProtocolException e) {
     // TODO Auto-generated catch block
      } catch (IOException e) {
       // TODO Auto-generated catch block
  }}

Answer 2

在这里，我发布了一些如何从网址获取内容的代码。

现在你必须在Array中传递String。

try{


            HttpClient httpclient = getNewHttpClient();

            HttpGet httpget = new HttpGet("url");


            HttpResponse response = httpclient.execute(httpget);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

        }catch(Exception e){
            Log.e("log_tag", "Error in http connection "+e.toString());
        }

        //convert response to string
        try{
            BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            result=sb.toString();
        }catch(Exception e){
            Log.e("log_tag", "Error converting result "+e.toString());
        }

Answer 3

我假设您的请求结构是JSONObject，上面给出了示例数据。

现在，只需在JSONObject类的帮助下创建一个JSONObject，在这里您可以浏览我的Web API & Its Integration in Android.

演示文稿之一

示例：

    JSONObject myJSONRequest = new JSONObject();
    myJSONRequest.put("customerId", "CUST01");
    myJSONRequest.put("orderId","101010");
    .........
    .........
    JSONArray arrayOrder = new JSONArray();
    for(int i=cntLine; i<n; i++)
    {
       JSONObject objSub = new JSONObject();
       objSub .put("lineId", String.valueOf(i));
       objSub .put("itemNumber", String.valueOf(i));
       .............
       .............

       arrayOrder.put(objSub);
    }
    myJSONRequest.put("orderDetailList", arrayOrder.toString());



    // create complete request object by placing all the values inside it.


    // Below code is for posting request data to web api

    DefaultHttpClient client = new DefaultHttpClient();
    HttpPost post = new HttpPost(url);
    post.setEntity(new StringEntity(myJSONRequest.toString(), "utf-8"));
    HttpResponse response = client.execute(post);

Answer 4

在android中它建议开发人员使用HttpUrlConnection ...

你需要上面的json作为字符串。

步骤1）创建新的URL对象。

URL url = new URL("www.url.com/demoservice");

步骤2）创建HttpURLConnection对象。

HttpUrlConnection connection = url.openConnection();

步骤3）将request属性设置为HttpPost ..

connection.setRequestProperty("request-Method","POST");
connection.setRequestProperty("content-type","application/json");

步骤4）获取输出流参考。

OutputStream stream = connection.getOutputStream();

步骤5）将json字符串写入输出流。

stream.write(jsonString.toBytes());

步骤6）关闭流。

stream.close();

希望这会有所帮助..

如何制作httppost请求

4 个答案: