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java - android doInBackground() 错误

转载 作者:行者123 更新时间:2023-12-01 06:51:59 25 4
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我正在尝试从 android Activity (一个简单的登录表单)执行 http 请求/响应,但是在我完成代码后它给了我一个错误,因为我必须将它放在一个线程中,我不熟悉异步任务,但我尝试实现它,但是,现在当我运行程序并点击提交时,logcat 给了我类似的东西

FATAL EXCEPTION: AsyncTask #1
Process: com.example.httpgetandroidexample, PID: 1131
java.lang.RuntimeException: An error occured while executing doInBackground()
at android.os.AsyncTask$3.done(AsyncTask.java:300)
at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:355)
at java.util.concurrent.FutureTask.setException(FutureTask.java:222)
at java.util.concurrent.FutureTask.run(FutureTask.java:242)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
at java.lang.Thread.run(Thread.java:841)
Caused by: android.view.ViewRootImpl$CalledFromWrongThreadException: Only the original thread that created a view hierarchy can touch its views.
at android.view.ViewRootImpl.checkThread(ViewRootImpl.java:6094)
at android.view.ViewRootImpl.requestLayout(ViewRootImpl.java:824)
at android.view.View.requestLayout(View.java:16431)
at android.view.View.requestLayout(View.java:16431)
at android.view.View.requestLayout(View.java:16431)
at android.view.View.requestLayout(View.java:16431)
at android.widget.RelativeLayout.requestLayout(RelativeLayout.java:352)
at android.view.View.requestLayout(View.java:16431)
at android.widget.TextView.checkForRelayout(TextView.java:6600)
at android.widget.TextView.setText(TextView.java:3813)
at android.widget.TextView.setText(TextView.java:3671)
at android.widget.TextView.setText(TextView.java:3646)
at com.example.httpgetandroidexample.MainActivity$AsyncTaskRunner.doInBackground(MainActivity.java:87)
at com.example.httpgetandroidexample.MainActivity$AsyncTaskRunner.doInBackground(MainActivity.java:1)
at android.os.AsyncTask$2.call(AsyncTask.java:288)
at java.util.concurrent.FutureTask.run(FutureTask.java:237)
... 4 more

我不明白为什么!这是我的 Android Activity :

package com.example.httpgetandroidexample;
import java.io.UnsupportedEncodingException;
import java.net.URLEncoder;

import org.apache.http.client.HttpClient;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;

import android.app.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

public class MainActivity extends Activity {

public TextView content;
EditText name,pass;

@Override
protected void onCreate(Bundle savedInstanceState) {

super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

name = (EditText)findViewById(R.id.name);
pass = (EditText)findViewById(R.id.pass);
content = (TextView)findViewById(R.id.text);

Button button=(Button)findViewById(R.id.but);

button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
AsyncTaskRunner runner = new AsyncTaskRunner();
runner.execute();
}
});


}
class AsyncTaskRunner extends AsyncTask<String, String, String> {

private String resp;

@Override
protected String doInBackground(String... params) {
publishProgress("Working..."); // Calls onProgressUpdate()
try{

// URLEncode user defined data

String nameValue = URLEncoder.encode(name.getText().toString(), "UTF-8");
String passValue = URLEncoder.encode(pass.getText().toString(), "UTF-8");

// Create http cliient object to send request to server

HttpClient Client = new DefaultHttpClient();

// Create URL string

String URL = "http://localhost:8080/login/web?user="+nameValue+"&pass="+passValue;

//Log.i("httpget", URL);

try
{
String SetServerString = "";

// Create Request to server and get response

HttpGet httpget = new HttpGet(URL);
ResponseHandler<String> responseHandler = new BasicResponseHandler();
SetServerString = Client.execute(httpget, responseHandler);

// Show response on activity

content.setText(SetServerString);
}
catch(Exception ex)
{
content.setText("fail");
System.out.println(ex);
resp="fail";
}
}
catch(UnsupportedEncodingException ex)
{
content.setText("FailAndroid!");
resp="fail";
}
return resp;
}
@Override
protected void onPostExecute(String result) {
content.setText(result);
}
protected void onPreExecute() {
}

@Override
protected void onProgressUpdate(String... text) {
content.setText(text[0]);

}
}
}

我的代码没有显示错误,甚至没有警告,但是当我按下提交按钮时,它在出现“工作”消息后强制关闭,有人知道为什么吗?预先感谢您

在你们的帮助下(谢谢大家!)我已经修改了我的 Activity 代码,如下所示:

package com.example.httpgetandroidexample;
import java.io.UnsupportedEncodingException;
import java.net.URLEncoder;

import org.apache.http.client.HttpClient;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;

import android.app.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

public class MainActivity extends Activity {

public TextView content;
EditText name,pass;

@Override
protected void onCreate(Bundle savedInstanceState) {

super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

name = (EditText)findViewById(R.id.name);
pass = (EditText)findViewById(R.id.pass);
content = (TextView)findViewById(R.id.text);

Button button=(Button)findViewById(R.id.but);

button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
AsyncTaskRunner runner = new AsyncTaskRunner();
runner.execute();
}
});


}
class AsyncTaskRunner extends AsyncTask<String, String, String> {

private String resp;

@Override
protected String doInBackground(String... params) {
publishProgress("Working..."); // Calls onProgressUpdate()
try{

// URLEncode user defined data

String nameValue = URLEncoder.encode(name.getText().toString(), "UTF-8");
String passValue = URLEncoder.encode(pass.getText().toString(), "UTF-8");

// Create http cliient object to send request to server

HttpClient Client = new DefaultHttpClient();

// Create URL string

String URL = "http://localhost:8080/login/web?user="+nameValue+"&pass="+passValue;

//Log.i("httpget", URL);

try
{
String SetServerString = "";

// Create Request to server and get response

HttpGet httpget = new HttpGet(URL);
ResponseHandler<String> responseHandler = new BasicResponseHandler();
SetServerString = Client.execute(httpget, responseHandler);

// Show response on activity

return SetServerString;
}
catch(Exception ex)
{
Log.i("MyActivity","Exception");
}
}
catch(UnsupportedEncodingException ex)
{

Log.i("MyActivity","UnsupportedEncodingException");
}
return resp;
}
@Override
protected void onPostExecute(String result) {
content.setText(result);
}
protected void onPreExecute() {
}

@Override
protected void onProgressUpdate(String... text) {
content.setText(text[0]);

}
}
}

但现在我收到了 Log.i("MyActivity","Exception"); 我放在上面的消息,我仍然做错了什么?

更新

在我最后一次尝试中,我按照一些在线教程再次修改了代码,这就是我所做的

package com.example.httpgetandroidexample;
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.net.URLEncoder;

import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;

import android.app.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

public class MainActivity extends Activity {

public TextView content;
EditText name,pass;
String URL;
@Override
protected void onCreate(Bundle savedInstanceState) {

super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

name = (EditText)findViewById(R.id.name);
pass = (EditText)findViewById(R.id.pass);
content = (TextView)findViewById(R.id.text);

Button button=(Button)findViewById(R.id.but);
try {
String nameValue = URLEncoder.encode(name.getText().toString(), "UTF-8");
String passValue = URLEncoder.encode(pass.getText().toString(), "UTF-8");
URL = "http://localhost:8080/login/web?user="+nameValue+"&pass="+passValue;
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}

button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
AsyncTaskRunner runner = new AsyncTaskRunner();
runner.execute(new String[ ] { URL });
}
});


}
private class AsyncTaskRunner extends AsyncTask<String, Void, String> {


@Override
protected String doInBackground(String... urls) {
String response = "";
for (String url : urls) {
DefaultHttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet(url);
try {
HttpResponse execute = client.execute(httpGet);
InputStream content = execute.getEntity().getContent();

BufferedReader buffer = new BufferedReader(new InputStreamReader(content));
String s = "";
while ((s = buffer.readLine()) != null) {
response += s;
}

} catch (Exception e) {
e.printStackTrace();
}
}
return response;
}

@Override
protected void onPostExecute(String result) {
content.setText(result);
}
}
}

这是详细模式下的 logcat:

Permission denied (missing INTERNET permission?)
at java.net.InetAddress.lookupHostByName(InetAddress.java:418)
at java.net.InetAddress.getAllByNameImpl(InetAddress.java:236)
at java.net.InetAddress.getAllByName(InetAddress.java:214)
at org.apache.http.impl.conn.DefaultClientConnectionOperator.openConnection(DefaultClientConnectionOperator.java:137)
at org.apache.http.impl.conn.AbstractPoolEntry.open(AbstractPoolEntry.java:164)
at org.apache.http.impl.conn.AbstractPooledConnAdapter.open(AbstractPooledConnAdapter.java:119)
at org.apache.http.impl.client.DefaultRequestDirector.execute(DefaultRequestDirector.java:360)
at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:555)
at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:487)
at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:465)
at com.example.httpgetandroidexample.MainActivity$AsyncTaskRunner.doInBackground(MainActivity.java:69)
at com.example.httpgetandroidexample.MainActivity$AsyncTaskRunner.doInBackground(MainActivity.java:1)
at android.os.AsyncTask$2.call(AsyncTask.java:288)
at java.util.concurrent.FutureTask.run(FutureTask.java:237)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
at java.lang.Thread.run(Thread.java:841)
Caused by: libcore.io.GaiException: getaddrinfo failed: EAI_NODATA (No address associated with hostname)
at libcore.io.Posix.getaddrinfo(Native Method)
at libcore.io.ForwardingOs.getaddrinfo(ForwardingOs.java:61)
at java.net.InetAddress.lookupHostByName(InetAddress.java:405)
... 17 more
Caused by: libcore.io.ErrnoException: getaddrinfo failed: EACCES (Permission denied)
... 20 more

据我所知,它缺少权限,但问题是我的 list 文件中有它:

<application
android:allowBackup="true"
android:icon="@drawable/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme" android:permission="android.permission.INTERNET">

最佳答案

Logcat 错误:

原因:android.view.ViewRootImpl$CalledFromWrongThreadException:只有创建 View 层次结构的原始线程才能触摸其 View 。

您尝试在 doInBackground 中使用 UI 线程,

content.setText("fail");

您无法从后台线程使用 onPostExecute 触摸 ui 线程中的任何内容用于使用 UI

这是您想要的示例。

 private class DownloadWebPageTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... urls) {
String response = "";
for (String url : urls) {
DefaultHttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet(url);
try {
HttpResponse execute = client.execute(httpGet);
InputStream content = execute.getEntity().getContent();

BufferedReader buffer = new BufferedReader(new InputStreamReader(content));
String s = "";
while ((s = buffer.readLine()) != null) {
response += s;
}

} catch (Exception e) {
e.printStackTrace();
}
}
return response;
}

@Override
protected void onPostExecute(String result) {
textView.setText(result);
}
}

您需要从doInBackground返回您的字符串

示例中:return response;

退回时您需要找零:

在示例中:

1- AsyncTask<String, Void, String> <--- 最后一个值是您想要从 doInBackground 获得的值更多信息This site

2- protected String doInBackground <--- 将返回类型更改为 String

3 - 返回您的值 return response;

4- protected void onPostExecute(String result) <--- 这也必须是字符串

然后使用 onPostExecute 中的 UI

在示例中onPostExecute(String result) <---结果是你的字符串

Vogella Article 复制代码6.5. Example: AsyncTask

更新

您再次使用 UI:

String nameValue    = URLEncoder.encode(name.getText().toString(), "UTF-8");
String passValue = URLEncoder.encode(pass.getText().toString(), "UTF-8");

如果你想要这个值,你可以将它们发送到 asynctask 的构造函数或作为参数发送到 doInBackground ,我推荐第二种方式

由于您在 try catch 语句中的代码,您的应用程序未关闭,并且由于您没有在 catch 上打印任何有用的内容,因此您无法收到错误,如果您使用 catch ,那么打印错误也更合适,例如:

catch(UnsupportedEncodingException ex)
{
ex.printStackTrace();
}

更新2

您在代码中声明您的 Android 权限,如下所示:

android:permission="android.permission.INTERNET">

但您必须添加您的权限,例如:

<manifest xlmns:android...>
...
<uses-permission android:name="android.permission.INTERNET"></uses-permission>
</manifest>

您的 list 必须类似于:

<?xml version="1.0" encoding="utf-8"?>
<manifest
......
......
.....>

<uses-sdk
...... />

<uses-permission android:name="android.permission.INTERNET" />

<application
.......
.......
</application>

</manifest>

关于java - android doInBackground() 错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22935106/

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