如何在android中执行GET方法?
我想得到这些数据-->“#1$Egypt$1605;!\@” 从这个链接-->“http://184.173.7.132/mobile.aspx?action=4” 我使用了此代码,但响应为如何在android中执行GET方法?,android,get,Android,Get,我想得到这些数据-->“#1$Egypt$1605;!\@” 从这个链接-->“http://184.173.7.132/mobile.aspx?action=4” 我使用了此代码,但响应为NULL public HttpResponse doGet(String url) throws Exception { HttpResponse response = null; try { HttpClient httpclient = new Default
NULL
public HttpResponse doGet(String url) throws Exception {
HttpResponse response = null;
try {
HttpClient httpclient = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
response = httpclient.execute(httpget);
}
catch (IOException e) {
e.printStackTrace();
}
return response;
}
使用此代码,它是为您的url工作。在清单中添加internet权限
String page;
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpGet httpGet = new HttpGet("http://184.173.7.132/mobile.aspx?action=4");
// httpGet.addHeader("Content-type","application/json");
ResponseHandler<String> resHandler = new BasicResponseHandler();
try
{
page = httpClient.execute(httpGet, resHandler);
}
catch (ClientProtocolException e)
{
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
}
Log.e("response",page);
字符串页;
DefaultHttpClient httpClient=新的DefaultHttpClient();
HttpGet HttpGet=新的HttpGet(“http://184.173.7.132/mobile.aspx?action=4");
//addHeader(“内容类型”、“应用程序/json”);
ResponseHandler resHandler=new BasicResponseHandler();
尝试
{
page=httpClient.execute(httpGet,resHandler);
}
捕获(客户端协议例外e)
{
e、 printStackTrace();
}捕获(IOE异常){
//TODO自动生成的捕捉块
}
Log.e(“答复”,第页);
您可能会收到IOException吗?-您是否已将INTERNET权限添加到清单中?这一行是什么“Log.e”(“response”,page);“do”?仅为了在logcat中查看响应,您可以使用Toast而不是Head。解析结果字符串时还有另一个问题。。。。这一行代码“String[]splitc=page.split($”);”中的数组“splitc”在解析后仍然具有值#1$Egypt$م@。。它出了什么问题:S