我正在尝试从URL获取getInputStream,连接响应代码为200
,但是当我尝试getInputStream时,我收到异常FileNotFoundException
,这是我的代码:
url = new URL("http://...");
connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestMethod("POST");
int status = connection.getResponseCode();
if(status >= 400){
request = new OutputStreamWriter(connection.getOutputStream());
request.write(params);
request.flush();
request.close();
String line;
InputStreamReader isr = new InputStreamReader(connection.getInputStream());
BufferedReader reader = new BufferedReader(isr);
StringBuilder sb = new StringBuilder();
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
}
堆栈跟踪:
W/System.err: java.io.FileNotFoundException: http://...
W/System.err: at com.android.okhttp.internal.huc.HttpURLConnectionImpl.getInputStream(HttpURLConnectionImpl.java:238)
W/System.err: at com.apps.topinformers.sharedreading.AddGroupMembersFragment.postText(AddGroupMembersFragment.java:112)
W/System.err: at com.apps.topinformers.sharedreading.AddGroupMembersFragment.access$000(AddGroupMembersFragment.java:26)
W/System.err: at com.apps.topinformers.sharedreading.AddGroupMembersFragment$PostDataAsyncTask.doInBackground(AddGroupMembersFragment.java:65)
W/System.err: at com.apps.topinformers.sharedreading.AddGroupMembersFragment$PostDataAsyncTask.doInBackground(AddGroupMembersFragment.java:55)
W/System.err: at android.os.AsyncTask$2.call(AsyncTask.java:295)
W/System.err: at java.util.concurrent.FutureTask.run(FutureTask.java:237)
W/System.err: at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:234)
W/System.err: at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1113)
W/System.err: at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:588)
W/System.err: at java.lang.Thread.run(Thread.java:818)
问题是什么,以及我如何调试它?
答案 0 :(得分:3)
连接响应代码为200
不,不是。此代码中没有任何内容可以检查响应代码。 FileNotFoundException
表示响应代码为404.
NB setDoOutput(true)
将方法设置为POST。你不需要自己设置,
答案 1 :(得分:-1)
获得@EJP的学分我已经检查了回复如下:
url = new URL("http://...");
connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestMethod("POST");
request = new OutputStreamWriter(connection.getOutputStream());
request.write(params);
request.flush();
request.close();
String line;
int status = connection.getResponseCode();
if(status < 400) {
InputStreamReader isr = new InputStreamReader(connection.getInputStream());
...
} else{
InputStreamReader isr = new InputStreamReader(connection.getErrorStream());
...
}
错误流向我显示我请求的URL的实现代码存在一些问题。