Android 正在获取Youtube api的HttpResponse(未找到)
我正在应用程序中使用YouTube api。我得到了HttpSposeException。我得到以下错误: 我的活动:Android 正在获取Youtube api的HttpResponse(未找到),android,youtube-api,httpresponse,Android,Youtube Api,Httpresponse,我正在应用程序中使用YouTube api。我得到了HttpSposeException。我得到以下错误: 我的活动: searching="https://www.googleapis.com/youtube/v3/search?part=snippet&videoSyndicated=true&type=video&q=len&key=AIzaSyDEWZ7vU_zPi1qgaWEREQEJi2FB3IQCE4g&maxResults=25&p
searching="https://www.googleapis.com/youtube/v3/search?part=snippet&videoSyndicated=true&type=video&q=len&key=AIzaSyDEWZ7vU_zPi1qgaWEREQEJi2FB3IQCE4g&maxResults=25&pageToken=";
protected Long doInBackground(URL... params)
{
//TODO Auto-generated method stub
result = UrltoValue.getValuefromUrl(searching);
return null;
}
日志:
org.apache.http.client.HttpResponseException: Not Found
at org.apache.http.impl.client.BasicResponseHandler.handleResponse(BasicResponseHandler.java:71)
at org.apache.http.impl.client.BasicResponseHandler.handleResponse(BasicResponseHandler.java:59)
at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:657)
at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:627)
at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:616)
用于将url转换为响应的类:
public class UrltoValue {
static String page="zero";
public static String getValuefromUrl(String url) throws TimeoutException
{
try
{
final int TIMEOUT_MILLISEC = 20000;
HttpParams httpParams = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(httpParams, TIMEOUT_MILLISEC);
HttpConnectionParams.setSoTimeout(httpParams, TIMEOUT_MILLISEC);
HttpConnectionParams.setTcpNoDelay(httpParams, true);
HttpClient httpClient = new DefaultHttpClient(httpParams);
//DefaultHttpClient httpClient = new DefaultHttpClient();
URLEncoder.encode(url, "UTF-8");
HttpPost httpPost = new HttpPost(url);
ResponseHandler<String> resHandler = new BasicResponseHandler();
page = httpClient.execute(httpPost, resHandler);
Log.v("PAGE",page);
return page;
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
String zero="zero";
return zero;
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
String zero="zero";
return zero;
} catch(IllegalArgumentException e) {
e.printStackTrace();
String zero="zero";
return zero;
//Toast.makeText(context, "Please remove % symbol", Toast.LENGTH_SHORT);
} catch(Exception e) {
e.printStackTrace();
String zero="zero";
return zero;
}
}
}
public类UrltoValue{
静态字符串page=“零”;
公共静态字符串getValuefromUrl(字符串url)引发TimeoutException
{
尝试
{
最终整数超时_毫秒=20000;
HttpParams HttpParams=新的BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(httpParams,超时0毫秒);
HttpConnectionParams.setSoTimeout(httpParams,超时0毫秒);
HttpConnectionParams.setTcpNoDelay(httpParams,true);
HttpClient HttpClient=新的默认HttpClient(httpParams);
//DefaultHttpClient httpClient=新的DefaultHttpClient();
编码(url,“UTF-8”);
HttpPost HttpPost=新的HttpPost(url);
ResponseHandler resHandler=new BasicResponseHandler();
page=httpClient.execute(httpPost,resHandler);
Log.v(“第页”,第页);
返回页面;
}捕获(客户端协议例外e){
//TODO自动生成的捕捉块
e、 printStackTrace();
字符串zero=“zero”;
返回零;
}捕获(IOE异常){
//TODO自动生成的捕捉块
e、 printStackTrace();
字符串zero=“zero”;
返回零;
}捕获(IllegalArgumentException e){
e、 printStackTrace();
字符串zero=“zero”;
返回零;
//Toast.makeText(上下文“请删除%symbol”,Toast.LENGTH\u SHORT);
}捕获(例外e){
e、 printStackTrace();
字符串zero=“zero”;
返回零;
}
}
}
您指向的是一个不存在的url。这就是stacktrace所说的:
org.apache.http.client.HttpResponseException: Not Found
原因是youtube url不接受POST请求
更改此项:
HttpPost httpPost = new HttpPost(url);
page = httpClient.execute(httpPost, resHandler);
为此:
HttpGet httpGet = new HttpGet(url);
page = httpClient.execute(httpGet, resHandler);
一切都会好起来的