从url获得响应代码，没有其他数据像（html或xml或json）

Question

我在做android应用程序它需要一个服务器的url调用，只需要响应代码来识别调用已成功（如200），但不需要任何其他数据，如json。

换句话说，我不想实现读者对象（InputStreamReader的）

和唧唧歪歪..

这样的代码会像 通话网址 GET响应代码 出口

Answer 1

你可以做这样的事情：

HttpGet httpRequest = new HttpGet(myUri); 
HttpClient httpclient = new DefaultHttpClient(); 
HttpResponse response = httpclient.execute(httpRequest); 
response.getStatusLine().getStatusCode(); // Your status-code 

Answer 2

使用HTTP HEAD请求，例如使用apache http client。

Answer 3

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() 

Answer 4

试试下面的代码：

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(); 

Answer 5

试试这个，请在​​您的文章从execute方法asyntask方法：

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 
    }