关于如何从我的Android应用程序向数据库发送json对象，我有点不知所措
由于我是新手，我不太确定我哪里出错了，我已经从XML中提取数据，我不知道如何将对象发布到我们的服务器上。
任何建议都会非常感激
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://52.88.194.67:8080/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;
}
}
答案 0 :(得分：15)
您需要使用AsyncTask类与服务器进行通信。像这样：
这是在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();
}
}
});
这是您活动课程中的新课程。
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");使其成为HTTP POST请求，应作为服务器上的POST请求处理。
然后在您的服务器上，您需要从＆＃34; PostData＆＃34;创建一个新的JSON对象。已在HTTP POST请求中发送。如果您告诉我们您在服务器上使用的语言，那么我们可以为您编写一些代码。
答案 1 :(得分：1)
根据您当前的代码实现，您使用onPostExecute方法，但没有onPreExecute和doInBackgound方法。从Android 3.0开始，所有网络操作都需要在后台线程上完成。因此，您需要使用Asynctask在后台执行请求的实际发送，并onPostExecute处理doInbackground方法返回的结果。
这是你需要做的。
sendDeviceDetails方法最终会进入doInBackgound方法。 onPostExecute将处理返回的结果。就发送JSON对象而言，您可以按照以下步骤进行操作，
从here
借来的代码段 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();
}
这只是其中一种方式。您也可以进行Asynctask实施。
答案 2 :(得分：0)
您应该使用网络服务将数据从您的应用发送到您的服务器，因为它可以让您的工作轻松顺畅。为此，您必须使用任何服务器端语言（如php，.net）创建Web服务，甚至可以使用jsp（java服务器页面）。
您必须将所有项目从Edittexts传递到Web服务。将数据添加到服务器的工作将由Web服务处理
答案 3 :(得分：0)
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());
} catch (JSONException e) {
e.printStackTrace();
}
}
});