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我正在尝试做一个必须登录到站点的应用程序，但是当我启动管理连接的活动时，程序关闭并弹出一个窗口，提示应用程序意外中断。

这是活动代码（它不完整且尚未完成）：

package com.example.myfirstapp;

import android.os.Bundle;
import android.app.Activity;
import android.view.MenuItem;
import android.support.v4.app.NavUtils;
import android.annotation.TargetApi;
import android.os.Build;
import android.content.Context;
import android.content.Intent;
import android.widget.TextView;
import android.net.ConnectivityManager;
import android.net.NetworkInfo;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.net.URL;
import java.net.URLConnection;
import java.net.URLEncoder;

public class DisplayMessageActivity extends Activity {

    private TextView textView;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        Intent intent = getIntent();
        String username = intent.getStringExtra(MainActivity.EXTRA_USERNAME);
        String password = intent.getStringExtra(MainActivity.EXTRA_PASSWORD);

        if(connect()){  
            try {

            URL url = new URL("http://www.example.com/"); //obviously non the real url

            // Construct data
            String data = URLEncoder.encode("username", "UTF-8") + "=" + URLEncoder.encode(username, "UTF-8");
            data += "&" + URLEncoder.encode("pass", "UTF-8") + "=" + URLEncoder.encode(password, "UTF-8");

            // Send data
            URLConnection conn = url.openConnection();
            conn.setDoOutput(true);
            OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
            wr.write(data);
            wr.flush();

            // Get the response
            BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            String line;
            while ((line = rd.readLine()) != null) {
                System.out.println(line);
            }
            wr.close();
            rd.close();
            } catch (Exception e) {
            }
        }else   {
            textView.setText("Errore di connessione.");
        }

        setContentView(textView);
        // Show the Up button in the action bar.
        setupActionBar();
    }

    /**
     * Set up the {@link android.app.ActionBar}, if the API is available.
     */
    @TargetApi(Build.VERSION_CODES.HONEYCOMB)
    private void setupActionBar() {
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.HONEYCOMB) {
            getActionBar().setDisplayHomeAsUpEnabled(true);
        }
    }

    private boolean connect(){
        ConnectivityManager connMgr = (ConnectivityManager) 
        getSystemService(Context.CONNECTIVITY_SERVICE);
        NetworkInfo networkInfo = connMgr.getActiveNetworkInfo();
        if(networkInfo != null && networkInfo.isConnected())    {
            return true;
        }
        else{
            return false;
        }
    }


    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        switch (item.getItemId()) {
        case android.R.id.home:
            // This ID represents the Home or Up button. In the case of this
            // activity, the Up button is shown. Use NavUtils to allow users
            // to navigate up one level in the application structure. For
            // more details, see the Navigation pattern on Android Design:
            //
            // http://developer.android.com/design/patterns/navigation.html#up-vs-back
            //
            NavUtils.navigateUpFromSameTask(this);
            return true;
        }
        return super.onOptionsItemSelected(item);
    }

}

和清单：

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.example.myfirstapp"
    android:versionCode="1"
    android:versionName="1.0" >

    <uses-sdk
        android:minSdkVersion="8"
        android:targetSdkVersion="17" />

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

    <application
        android:allowBackup="true"
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name"
        android:theme="@style/AppTheme" >
        <activity
            android:name="com.example.myfirstapp.MainActivity"
            android:label="@string/app_name" >
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
        <activity
            android:name="com.example.myfirstapp.DisplayMessageActivity"
            android:label="@string/title_activity_display_message"
            android:parentActivityName="com.example.myfirstapp.MainActivity" >
            <meta-data
                android:name="android.support.PARENT_ACTIVITY"
                android:value="com.example.myfirstapp.MainActivity" />
        </activity>
    </application>

</manifest>

谢谢你的帮助。

4

3 回答 3

2

您没有初始化 textView。直到你没有初始化 textview。

它保持为空，你会NullPointerException。

首先通过调用该方法设置活动布局setContentView并移除setContentView(textView);

private TextView textView;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    textView = (TextView)findViewById(R.id.yourTextView);
    .......

}

除此之外，不要在主线程上进行网络操作，这在 android 版本 >= 3.0 中是不允许的。

使用异步任务

于 2013-03-19T16:59:26.727 回答
1

那行不通：

}else{
  textView.setText("Errore di connessione.");
}

TextView未连接到您的布局，也未以其他方式启动。在使用 textView 之前试试这个：

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    Intent intent = getIntent();
    textView = (TextView) findViewById(R.id.YOUR_ID_FROM_LAYOUT);

我相信你有一个NPE...

于 2013-03-19T16:58:34.530 回答
0

很明显，这是行不通的。您可能有 NullPointerException 因为您的 TextView 没有界限。做这样的事情，

在您的 onCreate 方法中 -

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(your_layout);
    textView = (TextView)findViewById(R.id.textViewId);

    ....

    }else   {
            textView.setText("Errore di connessione.");
        } 
}
于 2013-03-19T17:08:37.120 回答