how to handle spaces in url android

Question

i have an error while i tried to run some spaces in url android, this is the code :

              String strUrlLoginFullpath = "";


              strUrlLoginFullpath = strUrlMain+"exl?u="+strUser+"&p="+strPass+"&t=1";



              URL url = new URL(strUrlLoginFullpath);
              publishProgress(40);

              //membuka koneksi
              URLConnection urlConn = url.openConnection();
              urlConn.setConnectTimeout(50000);
              urlConn.setReadTimeout(50000);

              //digunakan untuk menangkap nilai dari server
              BufferedReader in = new BufferedReader(
                      new InputStreamReader(
                              urlConn.getInputStream()));
              publishProgress(60);
              String inputLine;
              int i=0;
              while ((inputLine = in.readLine()) != null)
              {     
                  Log.v("Login",i+", return:" +inputLine+"; url: "+strUrlLoginFullpath);
                  if(i == 5)
                      {
                        Log.v("Login","RESULT >"+i+", return:" +inputLine+"; url: "+strUrlLoginFullpath);
                        str2 = inputLine;
                      }
                  i++;
              }
              in.close();


              publishProgress(80);

              publishProgress(100);
          } 
          catch (MalformedURLException e) 
          {
                Log.v("KxL", "MalformedURLException=" + e.toString());
          } 
          catch (IOException e) 
          {
                Log.v("KxL", "IOException=" + e.toString());
          } 
          return null;
    }

as you see.. i have the command for take login validation in strUrlLoginFullpath = strUrlMain+"exl?u="+strUser+"&p="+strPass+"&t=1"; but there's condition that strUser sometimes containing blank spaces, that can make my program doesn't run. so how the way to solve this problem.?

http://stackoverflow.com/questions/3286067/url-encoding-in-android — waqaslam, Apr 09 '13 at 07:43

laalto · Accepted Answer · 2013-04-09T07:48:24.330

3

URLEncoder.encode() your URL parameter values so that spaces and other special characters get correctly encoded.

Example:

strUrlLoginFullpath = strUrlMain + "exl?u=" + URLEncoder.encode(strUser, "UTF-8") +
    "&p=" + URLEncoder.encode(strPass, "UTF-8") + "&t=1";

edited Apr 09 '13 at 07:48

answered Apr 09 '13 at 07:42

laalto

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Added an example on how to use it. – laalto Apr 09 '13 at 07:49
@Pragnani true, better be explicit. Been bitten by that once. Edited. – laalto Apr 09 '13 at 07:52
still doesn't work.. url will be like this : http://10.10.2/exl?u=yu+kitai&p=MTIzNDU2%0A&t=1 – Aoyama Nanami Apr 09 '13 at 07:53
`%0A` is the line-feed character, you probably want to `trim()` your password first. – laalto Apr 09 '13 at 07:55
i've trim my password... the user name still yu+kita all that i want is user name as yu%20kita – Aoyama Nanami Apr 09 '13 at 08:00
`+` is a perfectly legal encoding for space `0x20`. – laalto Apr 09 '13 at 08:02
i'e fixed it the code must be strUrlLoginFullpath = strUrlMain + "exl?u=" + URLEncoder.encode(strUser, "UTF-8") + "&p=" +strPass+ "&t=1"; thank u very much @laalto – Aoyama Nanami Apr 09 '13 at 08:03

score 1 · Answer 2 · answered Apr 09 '13 at 07:43

Spaces aren't allowed in URLs. Instead, you should URL encode them - replace all the spaces with %20 and things should come out right in the other end.

If there is any chance any other special character (as considered special in URLs) might be in any of those strings, which for passwords at least seems quite likely, you should really URL encode the whole string using a suitable URL encoding function to avoid any problems or security issues.

Ignoring, that is, the security issue of passing a password in a URL's query string.

Rakeeb Rajbhandari · Answer 3 · 2013-04-09T12:48:41.223

0

try using the URLEncoder.encode()

strUrlLoginFullpath = strUrlMain+"exl?u="+ URLEncoder.encode(strUser)+"&p="+URLEncoder.encode(strPass)+"&t=1";

edited Apr 09 '13 at 12:48

answered Apr 09 '13 at 08:42

Rakeeb Rajbhandari

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score 0 · Answer 4 · answered Jul 18 '20 at 17:00

0

After creating the url String just do this:

strUrlLoginFullpath = strUrlMain+"exl?u="+strUser+"&p="+strPass+"&t=1"; strUrlLoginFullpath.replace(" ", "20%")

answered Jul 18 '20 at 17:00

K P Ramanath

1

how to handle spaces in url android

4 Answers4