What is a good way to parse parameters in a Url with a "&" char as a value?

Question

In my Java app, I can get name , value pairs from url, but a url with "&" as part of a value is causing problem, it looks like this:

http://localhost:6600/Resume_App?Id=Edit&File_Name=AT&T.txt

I know I can replace the string "AT&T" with "ATT" to avoid it, but I don't want to avoid it, there might be other cases, that I can't predict where this char might occur, so I want to solve it, I'm trying to come up with an intelligent way to tell if "&" is part of a value or not. So for values with "&" char, what's a good way to get name value pairs from it ?

You should be escaping the `&` to start with: `http://localhost:6600/Resume_App?Id=Edit&File_Name=AT%26T.txt` — Jon Skeet, Aug 26 '15 at 16:35
possible duplicate of [escaping ampersand in url](http://stackoverflow.com/questions/16622504/escaping-ampersand-in-url) — lc., Aug 26 '15 at 16:35
Once you have tidied up the URL, you can try using `String.split()`, and then split again on the `=` sign. — Tim Biegeleisen, Aug 26 '15 at 16:37
need to encode those characters http://www.w3schools.com/tags/ref_urlencode.asp — chrismillah, Aug 26 '15 at 16:37
Need more info before I can know what to suggest... are you getting the URLs over HTTP (i.e., is your Java application web-enabled and processing incoming URLs), or are you processing a list of stored URLs someplace (in a text file, in a database, etc.) using your app? The context of how you are obtaining the URLs to deal with is crucial to knowing what to do about the problem. — Matt Campbell, Aug 26 '15 at 17:46
@MattCampbell : thanks for the suggestion. my java web app is processing incoming URLs, trying to get the name and values. — Frank, Aug 26 '15 at 23:57

score 1 · Answer 1 · answered Aug 26 '15 at 16:38

1

I think you're going at it the wrong way. Instead of trying to figure out if a certain & is part of a value or an argument delimiter, encode the values before sending them to your app (e.g., by using URLEncoder), and decode it on the other side.

answered Aug 26 '15 at 16:38

Mureinik

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score 1 · Accepted Answer · answered Aug 27 '15 at 14:30

After some trial and error, I finally figured it out :

[1] Use this to encode the URL :

  public static String encode(String s)
  {
    try { return URLEncoder.encode(s,"UTF-8"); }
    catch (UnsupportedEncodingException e)
    {
      Resume_App.Save_To_Log("UTF-8 is not a recognized encoding\n"+Tool_Lib_Simple.Get_Stack_Trace(e));
      throw new RuntimeException(e);
    }
  }

[2] Use "exchange.getRequestURI().getRawQuery()" to get the raw query, I was using exchange.getRequestURI().getQuery() eariler, and it decodes the query for me, and made things confusing, what I need is the raw query.

[3] Use the following to get the name value map :

  public LinkedHashMap<String,String> queryToMap(String query)
  {
    LinkedHashMap<String,String> result=new LinkedHashMap();
    for (String param : query.split("&"))
    {
      String pair[]=param.split("=");
      if (pair.length>1) result.put(pair[0],pair[1]);
      else result.put(pair[0],"");
    }
    return result;
  }

[4] Use the following to get decoded name value pairs :

    for (Map.Entry<String,String> entry : params.entrySet())
    {
      paramName=entry.getKey().replace("+"," ");
      try { paramValue=URLDecoder.decode(entry.getValue(),"utf-8"); }
      catch (Exception e) { e.printStackTrace(); }
    }

I have vote up, nice way to solve it, I have just for fun build also a method and it and works for me too, my solution is further work and based on previous solution I made http://stackoverflow.com/questions/29381446/how-to-convert-url-parameteres-to-json-format/29388163#29388163. — Maytham Fahmi, Aug 27 '15 at 15:45

What is a good way to parse parameters in a Url with a "&" char as a value?

2 Answers2