Java 从url获取响应代码,无需其他数据,如(html、xml或json)
我在做android应用程序,它需要一个服务器url调用,只需要识别调用成功的响应代码(如200),但不需要任何其他数据,如json 换句话说,我不想实现reader对象(inputStreamReader)和bla-bla-bla 所以代码是这样的 呼叫url 获取响应代码Java 从url获取响应代码,无需其他数据,如(html、xml或json),java,android,Java,Android,我在做android应用程序,它需要一个服务器url调用,只需要识别调用成功的响应代码(如200),但不需要任何其他数据,如json 换句话说,我不想实现reader对象(inputStreamReader)和bla-bla-bla 所以代码是这样的 呼叫url 获取响应代码 退出使用HTTP HEAD请求,例如使用。您可以执行以下操作: HttpGet httpRequest = new HttpGet(myUri); HttpClient httpclient = new DefaultHt
退出使用HTTP HEAD请求,例如使用。您可以执行以下操作:
HttpGet httpRequest = new HttpGet(myUri);
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response = httpclient.execute(httpRequest);
response.getStatusLine().getStatusCode(); // Your status-code
使用状态代码:
HttpGet get = new HttpGet(url);
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response = httpclient.execute(get);
int code = response.getStatusLine().getStatusCode()
请尝试以下代码:
URL obj = new URL("http://www.google.com/");
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
// you can set the required parameter for your connection object
con.setRequestMethod("GET");
int responseCode = con.getResponseCode();
请在asyntask方法的post execute方法上尝试以下操作:
StringBuilder builder = new StringBuilder();
// Set up HTTP post
// HttpClient is more then less deprecated. Need to change to URLConnection
HttpClient client = new DefaultHttpClient( );
HttpGet httpGet = new HttpGet(params[0]);
HttpResponse response = client.execute(httpGet);
StatusLine statusLine = response.getStatusLine();
int statusCode = statusLine.getStatusCode();
if (statusCode == 200) {
//do some work
}
你的飞机要做什么?这不是基础设施的工作方式。你不只是得到一个int响应,它包含了许多其他信息,你需要捕捉这些信息,然后可以忽略它们。嗨@sky你能看看下面我的答案吗