Question

请教大神，安卓开发 如何能检测是否能连接到指定网址，比如百度？

RT

Answer 1

是否能连接到指定网址？不太清楚是什么意思，
用HttpClient，如果返回相应正常就是连接成功了。若超时就是没有成功。
HttpClient分get请求和post请求。
给你个网上的demo
get方式：
//先将参数放入List，再对参数进行URL编码
List<BasicNameValuePair> params = new LinkedList<BasicNameValuePair>();
params.add(new BasicNameValuePair("param1", "中国"));
params.add(new BasicNameValuePair("param2", "value2"));

//对参数编码
String param = URLEncodedUtils.format(params, "UTF-8");

//baseUrl
String baseUrl = "http://ubs.free4lab.com/php/method.php";

//将URL与参数拼接
HttpGet getMethod = new HttpGet(baseUrl + "?" + param);

HttpClient httpClient = new DefaultHttpClient();

try {
HttpResponse response = httpClient.execute(getMethod); //发起GET请求

Log.i(TAG, "resCode = " + response.getStatusLine().getStatusCode()); //获取响应码
Log.i(TAG, "result = " + EntityUtils.toString(response.getEntity(), "utf-8"));//获取服务器响应内容
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}

post 方式:

//和GET方式一样，先将参数放入List
params = new LinkedList<BasicNameValuePair>();
params.add(new BasicNameValuePair("param1", "Post方法"));
params.add(new BasicNameValuePair("param2", "第二个参数"));

try {
HttpPost postMethod = new HttpPost(baseUrl);
postMethod.setEntity(new UrlEncodedFormEntity(params, "utf-8")); //将参数填入POST Entity中

HttpResponse response = httpClient.execute(postMethod); //执行POST方法
Log.i(TAG, "resCode = " + response.getStatusLine().getStatusCode()); //获取响应码
Log.i(TAG, "result = " + EntityUtils.toString(response.getEntity(), "utf-8")); //获取响应内容

} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}

Answer 2

/** 是否连接网络 **/
	public static final boolean IsHaveInternet(Context context) {
		try {
			ConnectivityManager manger = (ConnectivityManager) context
					.getSystemService(Context.CONNECTIVITY_SERVICE);
			NetworkInfo info = manger.getActiveNetworkInfo();
			return (info != null && info.isConnected());
		} catch (Exception e) {
			return false;
		}
	}

Answer 3

手机试试