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大家好,我正在做一个远程控制我的遥控车的项目,安装了 adruino 板和 wifly shied。wifly shied 有自己的网络服务器设置,并设置了上下左右移动的配置。但是,当我单击前进按钮时,我在我的 android 应用程序中遇到了关于如何访问网络服务器的问题。下面是示例代码,我从那里被困住了。

public class GetUrl extends Activity implements OnClickListener {

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

        View forward_Button = findViewById(R.id.forwardButton);
        forward_Button.setOnClickListener(this);
}
@Override
    public void onClick(View v) {

        switch (v.getId()) {
        case R.id.forwardButton:
            HttpClient httpclient = new DefaultHttpClient();
            HttpGet httpget = new HttpGet("http://192.168.1.3/?LED=Fowd");
            HttpResponse response = httpclient.execute(httpget);

            break;
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1 回答 1

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JAVA 101:httpclient.execute(httpget);引发您必须捕获的异常:

try {
  httpclient.execute(httpget);
} catch (ClientProtocolException e) {
  e.printStackStrace();
}
于 2011-01-06T13:31:30.843 回答