如何将json对象从我的android应用程序发送到服务器
如何将json对象从我的android应用程序发送到服务器
提问于 2016-02-14 10:54:58
对于如何将json对象从我的安卓应用程序发送到数据库，我有点不知所措
由于我是新手，所以我不太确定自己哪里出了问题，我已经从XML中提取了数据，我不知道如何将对象发布到服务器上。
任何建议都将不胜感激。
package mmu.tom.linkedviewproject;
import android.content.Intent;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.ImageButton;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
import java.io.IOException;
/**
* Created by Tom on 12/02/2016.
*/
public class DeviceDetailsActivity extends AppCompatActivity {
private EditText address;
private EditText name;
private EditText manufacturer;
private EditText location;
private EditText type;
private EditText deviceID;
@Override
protected void onCreate(Bundle savedInstanceState){
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_device_details);
ImageButton button1 = (ImageButton) findViewById(R.id.image_button_back);
button1.setOnClickListener(new View.OnClickListener() {
Class ourClass;
public void onClick(View v) {
Intent intent = new Intent(DeviceDetailsActivity.this, MainActivity.class);
startActivity(intent);
}
});
Button submitButton = (Button) findViewById(R.id.submit_button);
submitButton.setOnClickListener(new View.OnClickListener() {
Class ourClass;
public void onClick(View v) {
sendDeviceDetails();
}
});
setContentView(R.layout.activity_device_details);
this.address = (EditText) this.findViewById(R.id.edit_address);
this.name = (EditText) this.findViewById(R.id.edit_name);
this.manufacturer = (EditText) this.findViewById(R.id.edit_manufacturer);
this.location = (EditText) this.findViewById(R.id.edit_location);
this.type = (EditText) this.findViewById(R.id.edit_type);
this.deviceID = (EditText) this.findViewById(R.id.edit_device_id);
}
protected void onPostExecute(JSONArray jsonArray) {
try
{
JSONObject device = jsonArray.getJSONObject(0);
name.setText(device.getString("name"));
address.setText(device.getString("address"));
location.setText(device.getString("location"));
manufacturer.setText(device.getString("manufacturer"));
type.setText(device.getString("type"));
}
catch(Exception e){
e.printStackTrace();
}
}
public JSONArray sendDeviceDetails() {
// URL for getting all customers
String url = "http://IP-ADDRESS:8080/IOTProjectServer/registerDevice?";
// Get HttpResponse Object from url.
// Get HttpEntity from Http Response Object
HttpEntity httpEntity = null;
try {
DefaultHttpClient httpClient = new DefaultHttpClient(); // Default HttpClient
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
httpEntity = httpResponse.getEntity();
} catch (ClientProtocolException e) {
// Signals error in http protocol
e.printStackTrace();
//Log Errors Here
} catch (IOException e) {
e.printStackTrace();
}
// Convert HttpEntity into JSON Array
JSONArray jsonArray = null;
if (httpEntity != null) {
try {
String entityResponse = EntityUtils.toString(httpEntity);
Log.e("Entity Response : ", entityResponse);
jsonArray = new JSONArray(entityResponse);
} catch (JSONException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
return jsonArray;
}
}回答 4
Stack Overflow用户
回答已采纳
发布于 2016-02-14 11:19:14
您需要使用AsyncTask类与服务器进行通信。就像这样：
这在您的onCreate方法中。
Button submitButton = (Button) findViewById(R.id.submit_button);
submitButton.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
JSONObject postData = new JSONObject();
try {
postData.put("name", name.getText().toString());
postData.put("address", address.getText().toString());
postData.put("manufacturer", manufacturer.getText().toString());
postData.put("location", location.getText().toString());
postData.put("type", type.getText().toString());
postData.put("deviceID", deviceID.getText().toString());
new SendDeviceDetails().execute("http://52.88.194.67:8080/IOTProjectServer/registerDevice", postData.toString());
} catch (JSONException e) {
e.printStackTrace();
}
}
});这是您活动类中的一个新类。
private class SendDeviceDetails extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
String data = "";
HttpURLConnection httpURLConnection = null;
try {
httpURLConnection = (HttpURLConnection) new URL(params[0]).openConnection();
httpURLConnection.setRequestMethod("POST");
httpURLConnection.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(httpURLConnection.getOutputStream());
wr.writeBytes("PostData=" + params[1]);
wr.flush();
wr.close();
InputStream in = httpURLConnection.getInputStream();
InputStreamReader inputStreamReader = new InputStreamReader(in);
int inputStreamData = inputStreamReader.read();
while (inputStreamData != -1) {
char current = (char) inputStreamData;
inputStreamData = inputStreamReader.read();
data += current;
}
} catch (Exception e) {
e.printStackTrace();
} finally {
if (httpURLConnection != null) {
httpURLConnection.disconnect();
}
}
return data;
}
@Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
Log.e("TAG", result); // this is expecting a response code to be sent from your server upon receiving the POST data
}
}行：httpURLConnection.setRequestMethod("POST");使这是一个HTTP请求，应该作为您服务器上的POST请求来处理。
然后，在您的服务器上，您需要从HTTP请求中发送的"PostData“创建一个新的JSON对象。如果您让我们知道您在服务器上使用的是什么语言，那么我们可以为您编写一些代码。
Stack Overflow用户
发布于 2016-02-14 11:15:06
根据您当前的代码实现，您有onPostExecute方法，但是没有onPreExecute和doInBackgound方法。从Android3.0开始，所有的网络操作都需要在后台线程上完成。因此，您需要使用Asynctask，它将在后台执行实际发送请求，并在onPostExecute中处理doInbackground方法返回的结果。
这是你需要做的。
- 创建Asynctask类并覆盖所有必要的方法。
sendDeviceDetails方法最终将进入doInBackgound方法中。onPostExecute将处理返回的结果。
对于发送JSON对象，您可以这样做，
从here借用的代码片段
protected void sendJson(final String email, final String pwd) {
Thread t = new Thread() {
public void run() {
Looper.prepare(); //For Preparing Message Pool for the child Thread
HttpClient client = new DefaultHttpClient();
HttpConnectionParams.setConnectionTimeout(client.getParams(), 10000); //Timeout Limit
HttpResponse response;
JSONObject json = new JSONObject();
try {
HttpPost post = new HttpPost(URL);
json.put("email", email);
json.put("password", pwd);
StringEntity se = new StringEntity( json.toString());
se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
post.setEntity(se);
response = client.execute(post);
/*Checking response */
if(response!=null){
InputStream in = response.getEntity().getContent(); //Get the data in the entity
}
} catch(Exception e) {
e.printStackTrace();
createDialog("Error", "Cannot Estabilish Connection");
}
Looper.loop(); //Loop in the message queue
}
};
t.start();
}这只是方法之一。您也可以使用Asynctask实现。
Stack Overflow用户
发布于 2016-02-14 11:16:11
您应该使用web服务将数据从您的应用程序发送到您的服务器，因为它将使您的工作轻松和顺利。为此，您必须使用任何服务器端语言创建web服务，如php、.net，甚至可以使用jsp(java服务器页)。
您必须将编辑文本中的所有项传递给web service.Work，将数据添加到服务器将由web服务处理。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/35390928
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