Android系统中如何发送Get请求

东方尚智沈大海Android技术开发系列讲座：www.3gdci.com
/*URL可以随意改*/
        String uriAPI = "http://192.168.1.100:8080/test/test.jsp?u=wangyi&p=456";
        /*建立HTTP Get对象*/
        HttpGet httpRequest = new HttpGet(uriAPI);
        try
        {
          /*发送请求并等待响应*/
          HttpResponse httpResponse = new DefaultHttpClient().execute(httpRequest);
          /*若状态码为200 ok*/
          if(httpResponse.getStatusLine().getStatusCode() == 200) 
          {
            /*读*/
            String strResult = EntityUtils.toString(httpResponse.getEntity());
            /*去没有用的字符*/
            strResult = eregi_replace("(\r\n|\r|\n|\n\r)","",strResult);
            mTextView1.setText(strResult);
          }
          else
          {
            mTextView1.setText("Error Response: "+httpResponse.getStatusLine().toString());
          }
        }
        catch (ClientProtocolException e)
        { 
          mTextView1.setText(e.getMessage().toString());
          e.printStackTrace();
        }
        catch (IOException e)
        { 
          mTextView1.setText(e.getMessage().toString());
          e.printStackTrace();
        }
        catch (Exception e)
        { 
          mTextView1.setText(e.getMessage().toString());
          e.printStackTrace(); 
        }