发送一个字符串从android的
我想从一个android应用程序发送一个字符串到一个servlet,然后检索到我的android应用程序的字符串,但是当我尝试调用该servlet它强制关闭我 和我不知道为什么(IM很新的Android,这是一个实践锻炼对我来说) 这里是我的android简单的应用程序:发送一个字符串从android的
ANDROID
package com.theopentutorials.android;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLConnection;
import android.app.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.TextView;
public class HttpGetServletActivity extends Activity implements OnClickListener {
Button button;
TextView outputText;
public static String request = "kjo ishte e gjitha";
public static final String URL = ("http://10.0.2.2:8080/HttpGetServlet/HelloWorldServlet?param1=" + request);
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
findViewsById();
button.setOnClickListener(this);
}
private void findViewsById() {
button = (Button) findViewById(R.id.button);
outputText = (TextView) findViewById(R.id.outputTxt);
}
public void onClick(View view) {
GetXMLTask task = new GetXMLTask();
task.execute(new String[] { URL });
}
private class GetXMLTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... urls) {
String output = null;
for (String url : urls) {
output = getOutputFromUrl(url);
}
return output;
}
private String getOutputFromUrl(String url) {
StringBuffer output = new StringBuffer("");
try {
InputStream stream = getHttpConnection(url);
BufferedReader buffer = new BufferedReader(
new InputStreamReader(stream));
String s = "";
while ((s = buffer.readLine()) != null)
output.append(s);
} catch (IOException e1) {
e1.printStackTrace();
}
return output.toString();
}
private InputStream getHttpConnection(String urlString)
throws IOException {
InputStream stream = null;
URL url = new URL(urlString);
URLConnection connection = url.openConnection();
try {
HttpURLConnection httpConnection = (HttpURLConnection) connection;
httpConnection.setRequestMethod("GET");
httpConnection.connect();
if (httpConnection.getResponseCode() == HttpURLConnection.HTTP_OK) {
stream = httpConnection.getInputStream();
}
} catch (Exception ex) {
ex.printStackTrace();
}
return stream;
}
@Override
protected void onPostExecute(String output) {
outputText.setText(output);
}
}
}
,这里是我的简单的servlet
SERVLET
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
@WebServlet("/HelloWorldServlet")
public class HelloWorldServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public HelloWorldServlet() {
super();
}
protected void doGet(HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException {
String par1 = request.getParameter("param1");
PrintWriter out = response.getWriter();
out.println(par1);
}
}
而且logcat的日志错误说
01-10 13:36:50.014: E/AndroidRuntime(1187): FATAL EXCEPTION: AsyncTask #1
01-10 13:36:50.014: E/AndroidRuntime(1187): java.lang.RuntimeException: An error occured while executing doInBackground()
01-10 13:36:50.014: E/AndroidRuntime(1187): at android.os.AsyncTask$3.done(AsyncTask.java:299)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:352)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.util.concurrent.FutureTask.setException(FutureTask.java:219)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.util.concurrent.FutureTask.run(FutureTask.java:239)
01-10 13:36:50.014: E/AndroidRuntime(1187): at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1080)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:573)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.lang.Thread.run(Thread.java:856)
01-10 13:36:50.014: E/AndroidRuntime(1187): Caused by: java.lang.NullPointerException: lock == null
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.io.Reader.<init>(Reader.java:64)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.io.InputStreamReader.<init>(InputStreamReader.java:122)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.io.InputStreamReader.<init>(InputStreamReader.java:59)
01-10 13:36:50.014: E/AndroidRuntime(1187): at com.theopentutorials.android.HttpGetServletActivity$GetXMLTask.getOutputFromUrl(HttpGetServletActivity.java:64)
01-10 13:36:50.014: E/AndroidRuntime(1187): at com.theopentutorials.android.HttpGetServletActivity$GetXMLTask.doInBackground(HttpGetServletActivity.java:54)
01-10 13:36:50.014: E/AndroidRuntime(1187): at com.theopentutorials.android.HttpGetServletActivity$GetXMLTask.doInBackground(HttpGetServletActivity.java:1)
01-10 13:36:50.014: E/AndroidRuntime(1187): at android.os.AsyncTask$2.call(AsyncTask.java:287)
01-10 13:36:50.014: E/AndroidRuntime(1187): at java.util.concurrent.FutureTask.run(FutureTask.java:234)
01-10 13:36:50.014: E/AndroidRuntime(1187): ... 4 more
行的也有人有什么想法? 非常感谢您的帮助! 祝您有美好的一天!
您需要添加的互联网访问权限
<uses-permission android:name="android.permission.INTERNET"/>
我发现,你应该使用类URLEncoder来编码URL,因为您的网址包含空格。请检查http://developer.android.com/reference/java/net/URLEncoder.html
我已经有了,我试图直接从servlet发送servlet响应像一个字符串,它的工作.. –
我想通了它是由param1值的空间造成的。尝试转义它像kjoi%20shtegjitha – Kai
我曾尝试发送它就像一个单词像'公共静态最终字符串URL = (“http://10.0.2.2:8080/HttpGetServlet/HelloWorldServlet?param1=CkemiBota”); '所以不需要urlEncoder,但没有什么,再次同样的错误:/ –
你忘了从android应用程序中设置param1。
connection.setRequestProperty("param1", "Your String Value");
然后你会从Servlet中得到回应的值。
sure.go上.. @ p3rand0r –
谢谢您的答复,但我还有2问题,我不知道我必须把这行代码,在这之后我也想编辑'public static final String URL =(“http://10.0.2.2:8080/HttpGetServlet/HelloWorldServlet?param1=” + request);'to'public static final String URL =(“http://10.0.2.2:8080/HttpGetServlet/HelloWorldServlet”);'对吗?另外我发布了logcat,如果这是为了任何帮助。 –
作为你的setRequestProperty,那么你会得到param1值到你的servlet,因此你也会得到一个response.just设置requestproperty线,你完成了。保持所有其他线路。 –
发布您的logcat可以是有用 – ben75
谢谢您的回复,我刚刚发布的logcat的,但我不能充分理解它 –