2013-06-25 86 views

回答

2

創建一個Uri對象超出你的網址/字符串。

String myUrl = "http://www.sample.com/json/just_in.asp?variable1=value1&variable2=value2"; 

Uri uri = Uri.parse(myUrl); 
String variable1 = uri.getQueryParameter("variable1"); 
String variable2 = uri.getQueryParameter("variable2"); 

有關該主題的更多信息,請參閱http://developer.android.com/reference/android/net/Uri.html

3

像這樣:

HashMap<String, String> map = new HashMap<String, String>(); 
try { 
    String url = "http://www.sample.com/json/just_in.asp?variable1=value1&variable2=value2"; 
    List<NameValuePair> parameters = URLEncodedUtils.parse(new URI(url), "UTF-8"); 
    for (NameValuePair p : parameters) { 
     map.put(p.getName(), p.getValue()); 
    } 
} catch (URISyntaxException e) { 
    e.printStackTrace(); 
} 

map將包含參數的鍵/值對。

1

使用這個簡單的代碼

final String[] params = url.split("&"); 
     final Map<String, String> map = new HashMap<String, String>(); 
     try { 
      for (final String param : params) { 
       final String name = param.split("=")[0]; 
       final String value = param.split("=")[1]; 
       map.put(name, value); 
      } 
     } catch (final Exception e) { 
      e.printStackTrace(); 
     }