String str = "#aaa# #bbb# #ccc# #ddd#"
Can anybody tell me how can i get the substrings “aaa","bbb","ccc","ddd" (substring which is in the pair of "# #" and the number of "# #" is unknown) using regular expression?
Thanks!
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3 Answers
Using regex:
Pattern p = Pattern.compile("#(\\w+)#");
String input = "#aaa# #bbb# #ccc# #ddd#";
Matcher m = p.matcher(input);
List<String> parts = new ArrayList<String>();
while (m.find())
{
parts.add(m.group(1));
}
// parts is [aaa, bbb, ccc, ddd]
4 Comments
yvetterowe
yeah that works=). Would you please explain to me what does "w+" mean in the regex? Thanks =)
Matt Ball
It's standard regex syntax. Read the JavaDocs linked in my answer. If those aren't clear enough, acquaint yourself with regular-expressions.info.
yvetterowe
hmm if i change the input String into "#aaa# #bbb bbb# #ccc# #ddd#", then the parts will be [aaa,null,ccc,ddd]. Can you figure out a more general way that "bbb bbb"(there is a space between the words in the substring) can also be output? Thanks a lot =D
Matt Ball
Change
(\\w+) to ([\\w\\s]+) or ([^#]+)Try this:
String str = "1aaa2 3bbb4 5ccc6 7ddd8";
String[] data = str.split("[\\d ]+");
Each position in the resulting array will contain a substring, except the first one which is empty:
System.out.println(Arrays.toString(data));
> [, aaa, bbb, ccc, ddd]
1 Comment
Stephen C
... or something like that. As with so many questions on regexes, the OP has not CLEARLY specified his/her requirements.
Here is yet another way of doing it using StringTokenizer
String str="#aaa# #bbb# #ccc# #ddd#";
//# and space are the delimiters
StringTokenizer tokenizer = new StringTokenizer(str, "# ");
List<String> parts = new ArrayList<String>();
while(tokenizer.hasMoreTokens())
parts.add(tokenizer.nextToken());
2 Comments
Matt Ball
-1. "
StringTokenizer is a legacy class that is retained for compatibility reasons although its use is discouraged in new code. It is recommended that anyone seeking this functionality use the split method of String or the java.util.regex package instead." docs.oracle.com/javase/7/docs/api/java/util/…
coderplus
ok. I see your point. Though it's available in Java 7(docs.oracle.com/javase/7/docs/api/java/util/…), this class won't be available in future releases? got my answer in your comment update :-)
