I need to match a string against a regex in Java. The string is multiline and therefore contains multiple \n like the followings
String text = "abcde\n"
+ "fghij\n"
+ "klmno\n";
String regex = "\\S*";
System.out.println(text.matches(regex));
I only want to match whether the text contains at least a non-whitespace character. The output is false. I have also tried \\S*(\n)* for the regex, which also returns false.
In the real program, both the text and regex are not hard-coded. What is the right regex to check is a multiline string contains any non-whitespace character?
The problem is not to do with the multi lines, directly. It is that matches matches the whole string, not just a part of it.
If you want to check for at least one non-whitespace character, use:
"\\s*\\S[\\s\\S]*"
Which means
If you just want to check whether there is at least one non white space character in the string, you can just trim the text and check the size without involving regex at all.
String text = "abcde\n"
+ "fghij\n"
+ "klmno\n";
if (!text.trim().isEmpty()){
//your logic here
}
If you really want to use regex, you can use a simple regex like below.
String text = "abcde\n"
+ "fghij\n"
+ "klmno\n";
String regex = ".*\\S+.*";
Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
Matcher matcher = pattern.matcher(string);
if (matcher.find()){
// your logic here
}
Using String.matches()
!text.matches("\\s*")
Check if the input text consist solely of whitespace characters (this includes newlines), invert the match result with !
Using Matcher.find()
Pattern regexp = Pattern.compile("\\S");
regexp.matcher(text).find()
Will search for the first non-whitespace character, which is more efficient as it will stop on the first match and also uses a pre-compiled pattern.