在 Java 中匹配零或一次出现的单词的正则表达式

Question

I have written a regex to match following pattern:

Any characters followed by hyphen followed by number followed by space followed by an optional case insensitive keyword followed by space followed by any char .

Eg,

1. TXT-234 #comment anychars
2. TXT-234 anychars

The regular expression I have written is as follows:

(?<issueKey>^((\\s*[a-zA-Z]+-\\d+)\\s+)+)((?i)?<keyWord>#comment)?\\s+(?<comment>.*)

But the above doesn't capture the zero occurrence of '#comment', even though I have specified the '?' for the regular expression. The case 2 in the above example always fails and the case 1 succeeds.

What am I doing wrong?

Answer 1

#comment won't match #keyword. That is why you don't have a match try. This one it should work:

 ([a-zA-Z]*-\\d*\\s(((?i)#comment|#transition|#keyword)+\\s)?[a-zA-Z]*)

Answer 2

This may help;

String str = "1. TXT-234 #comment anychars";
String str2 = "2. TXT-234 anychars";
String str3 = "3. TXT-2a34 anychars";
String str4 = "4. TXT.234 anychars";
Pattern pattern = Pattern.compile("([a-zA-Z]*-\\d*\\s(#[a-zA-Z]+\\s)?[a-zA-Z]*)");
Matcher m = pattern.matcher(str);
if (m.find()) {
    System.out.println("Found value: " + m.group(0));
    System.out.println("Found value: " + m.group(1));
    System.out.println("Found value: " + m.group(2));
}
m = pattern.matcher(str2);
if (m.find()) {
    System.out.println("Found value: " + m.group(0));
}
m = pattern.matcher(str3);
if (m.find()) {
    System.out.println("Found value: " + m.group(0));
} else {
    System.out.println("str3 not match");
}
m = pattern.matcher(str4);
if (m.find()) {
    System.out.println("Found value: " + m.group(0));
} else {
    System.out.println("str4 not match");
}