我基本上是凌空新手,但与apache http旧库不同,我曾经这样处理响应以及状态代码,以显示关于成功和失败响应的aler对话:
@Override
protected String doInBackground(Void... params) {
String mre = null;
stringEntity=new StringEntity(jsonObject.toString());
httpPost.setEntity(stringEntity);
httpPost.setHeader("Content-type", "application/json");
httpResponse=httpClient.execute(httpPost);
int statusCode=httpResponse.getStatusLine().getStatusCode();
switch (statusCode){
case 200:
entity = httpResponse.getEntity();
System.out.println("Entity post is: "
+ EntityUtils.toString(entity));
mre = "200";
Log.d("SUCCESS","YES FINALLY");
Log.d("Ok",entity.toString());
break;
case 412:
mre = "412";
Log.d("412", "412 WE MEET AGAIN)");
break;
default:
mre="unknown";
Log.i("Unknown","Unknown Server Error");
}
} catch (IOException e) {
e.printStackTrace();
}
return mre;
}
@Override
protected void onPostExecute(String mre) {
super.onPostExecute(mre);
hideDialog();
if(mre.equalsIgnoreCase("200")){
//HANDLE HERE
}
else if(mre.equalsIgnoreCase("412")){
//HANDLE HERE
}
else
if(mre.equalsIgnoreCase("unknown")){
//HANDLE HERE
}
但是在排球中我很困惑,我需要处理从服务器收到的每个状态代码,以便我可以向用户显示一条消息。(成功/失败)
有点帮助会深深感激