This is my code:
String httpURL = "http://codespeed_server:8000/result/add/";
URL myurl = new URL(httpURL);
HttpURLConnection con = (HttpURLConnection) myurl.openConnection();
con.setDoOutput(true);
DataOutputStream output = new DataOutputStream(con.getOutputStream());
output.writeBytes(query);
Can anybody tell me why this only works when I trail to it:
con.getResponseCode();
? The server only gets new entries when I call getResponseCode(). Is it a normal behavior ? Or is it a server-side issue ?
The URLConnection does not make a connection until one of the getResponse*, getInputStream methods (or other method which requires response data) is called (or connect()).
Related
I have a weird issue with my code:
URL url = new URL(searchUrlPOST.replace("%accessToken", accessToken));
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setConnectTimeout(5000);
conn.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setRequestMethod("POST");
OutputStream os = conn.getOutputStream();
os.write(json.getBytes("UTF-8"));
os.close();
// read the response
InputStream in = new BufferedInputStream(conn.getInputStream());
That works perfectly fine as long as the server is responding with a proper status code. If he however responds with something like a 400 and i replace the conn.getInputStream() with a conn.getErrorStream() I get a
Exception in thread "main" java.io.IOException: Stream closed
How come?
I'm not sure but:
Please make sure that
searchUrlPOST.replace("%accessToken", accessToken)
returns a valid url, this error can been thrown by using an invalid url at "new Url(String url)"
I am newbie developer in Java. STEP 1 I have already done the the following:
Logged in to REST API server (with login&password)
Received a token in XML format which i parsed with SAX parser so now i
am in a position of a token. Below is the sample code for Login:
Java code:
String url1 = "https://api4.liverail.com/login";
URL obj = new URL(url1);
HttpsURLConnection con1 = (HttpsURLConnection) obj.openConnection();
String urlParameters ="username=paania#gmail.com&password=d372a15b714bd250e";
con1.setDoOutput(true);
con1.setRequestMethod("POST");
DataOutputStream wr = new DataOutputStream(con1.getOutputStream());
wr.writeBytes(urlParameters);
STEP 2: I want to pass the token to REST API to obtain some information e.g a list from category but when i send the request via GET method , i get a response in XML saying [CDATA[You need to be logged in]] This is the code in Java:
String url = "http://api4.liverail.com/advertising/category/list/?token="72938howdwoi";
URL obj = new URL(url);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
con.setRequestMethod("GET");
BufferedReader in = new BufferedReader(newInputStreamReader(con.getInputStream()));
in.close();
con.disconnect();
I am not sure what i am missing here.
Any suggestions?
Just changed your url for request of data :
String url = "http://api4.liverail.com/advertising/category/list/?token=72938howdwoi";
So I have a problem where if I type this link on the browser and hit enter, an activation happens. I just want to do the same through Java. I don't need any kind of response from the URL. It should just do the same as entering the URL on a browser. Currently my code doesn't throw an error, but I don't think its working because the activation is not happening. My code:
public static void enableMachine(String dns){
try {
String req= "http://"+dns+"/username?username=sputtasw";
URL url = new URL(req);
URLConnection connection = url.openConnection();
connection.connect();
/*BufferedReader br = new BufferedReader(new InputStreamReader(url.openStream()));
String strTemp = "";
while (null != (strTemp = br.readLine())) {
System.out.println(strTemp);
}*/
} catch (Exception ex) {
ex.printStackTrace();
}
}
What's the problem?
If you want to do that with an URLConnection, it isn't sufficient to just open the connection with connect, you also have to send e.g. an HTTP request etc.
That said, i think it would be easier, if you use an HTTP client like the one from Apache HttpComponents (http://hc.apache.org/). Just do a GET request with the HTTP client, this would be the same as visiting the page with a browser (those clients usually also supports redirection etc.).
You may use HttpUrlConnectionClass to do the job:
URL url = new URL("http://my.url.com");
HttpURLConnection httpCon = (HttpURLConnection) url.openConnection();
httpCon.setRequestProperty("Content-Type", "application/json");
httpCon.setDoOutput(true);
httpCon.setRequestMethod("POST");
String params = "foo=42&bar=buzz";
DataOutputStream wr = new DataOutputStream(httpCon.getOutputStream());
wr.writeBytes(params);
wr.flush();
wr.close();
httpCon.connect();
int responseCode = httpCon.getResponseCode();
You may as well use "GET" request method and just append parameters to the url.
I am currently developing an application that needs to interact with the server but i'm having a problem with receiving the data via POST. I'm using Django and then what i'm receiving from the simple view is:
<QueryDict: {u'c\r\nlogin': [u'woo']}>
It should be {'login': 'woooow'}.
The view is just:
def getDataByPost(request):
print '\n\n\n'
print request.POST
return HttpResponse('')
and what i did at the src file on sdk:
URL url = new URL("http://192.168.0.148:8000/data_by_post");
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
urlConnection.setDoInput(true);
urlConnection.setDoOutput(true);
urlConnection.setChunkedStreamingMode(0);
String parametros = "login=woooow";
urlConnection.setRequestMethod("POST");
urlConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
urlConnection.setRequestProperty("charset","utf-8");
urlConnection.setRequestProperty("Content-Length", "" + Integer.toString(parametros.getBytes().length));
OutputStream os = urlConnection.getOutputStream();
BufferedWriter writer = new BufferedWriter( new OutputStreamWriter(os, "UTF-8"));
writer.write(parametros);
writer.close();
os.close();
I changed the Content-Length to see if that was a problem and then the problem concerning thw value of login was fixed but it was by hard coding (which is not cool).
ps.: everything except the QueryDict is working well.
What could i do to solve this? Am i encoding something wrong at my java code?
thanks!
Just got my problem solved with a few modifications concearning the parameters and also changed some other things.
Having parameters set as:
String parameters = "parameter1=" + URLEncoder.encode("SOMETHING","UTF-8");
then, under an AsyncTask:
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setRequestMethod("POST");
//not using the .setRequestProperty to the length, but this, solves the problem that i've mentioned
conn.setFixedLengthStreamingMode(params.getBytes().length);
conn.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
PrintWriter out = new PrintWriter(conn.getOutputStream());
out.print(params);
out.close();
String response = "";
Scanner inStream = new Scanner(conn.getInputStream());
while (inStream.hasNextLine()) {
response += (inStream.nextLine());
}
Then, with this, i got the result from django server:
<QueryDict: {u'parameter1': [u'SOMETHING']}>
which is what i was wanting.
String album = "http://picasaweb.google.com/data/feed/api/user/"+email;
HttpURLConnection con = (HttpURLConnection) new URL(albumUrl).openConnection();
// request method, timeout and headers
con.setRequestMethod("GET") ;
con.setReadTimeout(15000);
con.setRequestProperty("Authorization", "GoogleLogin auth="+auth);
con.setRequestProperty("GData-Version", "2");
// set timeout and that we will process output
con.setReadTimeout(15000);
con.setDoOutput(true);
// connnect to url
con.connect();
// read output returned for url
BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
Problem : Everytime i call con.getInputStream() it gives me file not found exception.
But when i load the same url in the desktop browser then it is displaying correct data.
I am confused why on android it is throwing exception.
Thanks in advance.
Did you get this? Maybe you just missed the https
below example uses default for authenticated user and the experimental fields list.
url = "https://picasaweb.google.com/data/feed/api/user/default?kind=album&access=public&fields="
+ URLEncoder
.encode("entry(title,id,gphoto:numphotosremaining,gphoto:numphotos,media:group/media:thumbnail)",
"UTF-8");
https://developers.google.com/picasa-web/docs/2.0/developers_guide_protocol#ListAlbums