[問題] Regular Expressions Pattern的Capturing Groups問題

看板java作者d0068267 (Wei-Tse,Hsu)時間2年前 (2021/08/04 20:18)推噓3(3推 0噓 6→)

留言9則, 3人參與討論串1/1

大家好，我是Java的菜雞，目前在學到Regular Pattern的地方想請問正規表達式的問題，程式碼如下： import java.util.regex.Matcher; import java.util.regex.Pattern; public class RegexMatches { public static void main( String args[] ) { // String to be scanned to find the pattern. String line = "This order was placed for QT3000! OK?"; String pattern = "(.*)(\\d+)(.*)"; // Create a Pattern object Pattern r = Pattern.compile(pattern); // Now create matcher object. Matcher m = r.matcher(line); 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) ); }else { System.out.println("NO MATCH"); } } } 我想請問的是Output的部分 Found value: This order was placed for QT3000! OK? Found value: This order was placed for QT300 Found value: 0 為什麼"(.*)(\\d+)(.*)"; 的m.group(1)是 This order was placed for QT300 就我目前的理解在API裡面查到的內容為 . Any character (may or may not match line terminators) X* X, zero or more times X+ X, one or more times \d A digit: [0-9] .是任何字元、\\d是數字那麼digit不是從3開始到300，也就是 m.group(1):This order was placed for QT m.group(2):3000 m.group(3):! OK? 為什麼跟我想的不一樣，是我誤會(.*)(\\d+)(.*)的意思嗎？ -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 1.165.119.107 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/java/M.1628079531.A.30C.html