Android应用在启动时崩溃

时间:2011-12-03 22:02:45

标签: java android

我已经写了一个快速的android添加来显示连接到的wifi网络的SSID以及设备的IP地址。代码是:

public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        Context context = getApplicationContext();
        WifiManager wifi_man = (WifiManager)context.getSystemService(Context.WIFI_SERVICE);

        if(wifi_man.isWifiEnabled()==true)
        {
            System.out.println("inwifi");
            WifiInfo wifi_info = wifi_man.getConnectionInfo();
            String ssid_name = wifi_info.getSSID();

            TextView temp = (TextView)findViewById(R.id.widget40);
            CharSequence sentence = "SSID NAME : "+ssid_name;
            temp.setText(sentence);

            int temp2 = wifi_info.getIpAddress();
            String ip_addr = Formatter.formatIpAddress(temp2);
            temp = (TextView)findViewById(R.id.widget41);
            CharSequence sent = "IP ADDRESS : "+ip_addr;
            temp.setText(sent);

        }
    }    

然而,应用程序进程一旦启动就会崩溃...任何想法为什么。我也很困惑我的代码的入口点。我只是将我的应用代码放在onCreate流程的末尾吗?

1 个答案:

答案 0 :(得分:1)

所以,正如我在评论中猜到的那样,你在清单中错过了这一行:

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