Java finding substring

Question

I have the following String:

oauth_token=safcanhpyuqu96vfhn4w6p9x&**oauth_token_secret=hVhzHVVMHySB**&application_name=Application_Name&login_url=https%3A%2F%2Fapi-user.netflix.com%2Foauth%2Flogin%3Foauth_token%3Dsafcanhpyuqu96vfhn4w6p9x

I am trying to parse out the value for oauth_token_secret. I need everything from the equals sign (=) to the next ampersand sign (&). So I need to parse out: hVhzHVVMHySB

Currently, I have the following code:

Const.OAUTH_TOKEN_SECRET = "oauth_token_secret";

Const.tokenSecret = 
  content.substring(content.indexOf((Const.OAUTH_TOKEN_SECRET + "="))
    + (Const.OAUTH_TOKEN_SECRET + "=").length(), 
      content.length());

This will start at the beginning of the oauth_token_string, but it will not stop at the next ampersand. I am unsure how to specify to stop at the end of the following ampersand. Can anyone help me?

cletus · Accepted Answer · 2010-02-20 09:47:50Z

8

The indexOf() methods allow you to specify an optional fromIndex. This allows you to find the next ampersand:

int oauth = content.indexOf(Const.OAUTH_TOKEN_SECRET);
if (oauth != -1) {
  int start = oath + Const.OATH_TOKEN_SECRET.length(); // or
  //int start = content.indexOf('=', oath) + 1;
  int end = content.indexOf('&', start);
  String tokenSecret = end == -1 ? content.substring(start) : content.substring(start, end);
}

edited Feb 20, 2010 at 9:47

answered Feb 19, 2010 at 4:05

cletus

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1 Comment

littleK Over a year ago

That did the trick, thanks for showing me how this can be done!

marklai · Accepted Answer · 2010-02-19 04:20:14Z

2

public static Map<String, String> buildQueryMap(String query)  
{  
  String[] params = query.split("&");  
  Map<String, String> map = new HashMap<String, String>();  
  for (String param : params)  
  {
    String[] pair = param.split("=");
    String name = pair[0];  
    String value = pair[1];  
    map.put(name, value);  
  }  
  return map;  
}

// in your code
Map<String, String> queryMap = buildQueryMap("a=1&b=2&c=3....");
String tokenSecret = queryMap.get(Const.OAUTH_TOKEN_SECRET);

answered Feb 19, 2010 at 4:20

marklai

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1 Comment

Stephen C Over a year ago

+1 this is the most robust, though a malformed input string could give null pointer or array index exceptions.

missingfaktor · Accepted Answer · 2010-02-19 04:18:58Z

1

Using String.split gives a much cleaner solution.

static String getValue(String key, String content) {
  String[] tokens = content.split("[=&]");
  for(int i = 0; i < tokens.length - 1; ++i) {
    if(tokens[i].equals(key)) {
      return tokens[i+1];
    }
  }
  return null;
}

Click here for a test drive! ;-)

edited Feb 19, 2010 at 4:18

answered Feb 19, 2010 at 4:06

missingfaktor

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Comments

helpermethod · Accepted Answer · 2010-02-19 10:58:31Z

0

A much better solution is using the Pattern and corresponding Matcher class.

By using a capturing group you can check and "cut out" the the appropriate substring in one step.

answered Feb 19, 2010 at 10:58

helpermethod

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