Without replaceall Output: [3, 2, 1]
With replaceall Output: 3,2,1
Is there anyway to do this using single replaceAll method, something like
set.toString().replaceAll("\\[\\]\\s+","");
Now Code
Set<String> set = new HashSet<String>();
set.add("1");
set.add("2");
set.add("3");
System.out.println(set.toString().replaceAll("\\[", "").replaceAll("\\]", "").replaceAll("\\s+", ""));
How about using Guava's Joiner:
String joined = Joiner.on(",").join(set);
System.out.println(joined); // 1,2,3
Or, if you can't use any 3rd party library, then following replaceAll would work:
System.out.println(set.toString().replaceAll("[\\[\\]]|(?<=,)\\s+", "")); // 1,2,3
Well, you won't get always the same output, as HashSet doesn't preserve insertion order. If you want that, use LinkedHashSet.
|.How about using this regex, [\\[\\]].
System.out.println(set.toString().replaceAll("[\\[\\]]", "")); // Output is 3,2,1
If you want to remove the white space also, then use this regex, [\\[\\]\\s] (but the comma will be there).
replaceAll("[^\\d|,]", ""); will replace everything that is not a digit (\\d) or (|) a comma (,). The hat symbol ^ means not in Java regex and the square brackets [] denote a set. So our set is "everything that is not a digit or a comma".
Set<String> set = new HashSet<String>();
set.add("1");
set.add("2");
set.add("3");
System.out.println(set.toString().replaceAll("[^\\d|,]", ""));
Output:
3,2,1
