1, [abc] a, b or c
[^abc] Any character except a, b or c
[a-zA-Z] a to z or A to Z, both inclusive
[ad[mp]] a to d or m to p: [a-dm-p] (union)
[az&&[def]] d, e or f (intersection)
2, * any character
\d Digits: [0-9]
\D Not a number: [^0-9]
\s whitespace character: [\t\n\x0B\f\r]
\S non-whitespace characters: [^\s]
\w word character: [a-zA-Z_0-9]
\W non-word character: [^\w]
3. Greedy Quantifier
X? once or not at all
X* zero or more times
X+ one or more times
X{n} exactly n times
X{n,} at least n times
X{n, m} at least n times, but not more than m times string
4. Replacement function
String s = "Hello123World";
String regex = "\\d";
String s2 = s.replaceAll(regex, "");
5. Grouping function
Capture groups can be numbered by counting their opening brackets from left to right, ((A)(B(C))) there are 4 such groups
1 ((A)(B(C)))
2 (A)
3 (B(C))
4 (C)
Application: Redundant word cutting: "sdqqfgkkkhjppppkl";
String regex = "(.)\\1+"; String[] arr = s.split(regex);
Application: put the string "I I....I...I.I...want...to learn....learn to learn..learn.edit.edit.program.program... Chengcheng...cheng" output is "I want to learn programming"
String s = "我我....我...我.我...要要...要学....学学..学.编.编编.编.程...程程...程";
String s2 = s.replaceAll("\\.+", "");
String s3 = s2.replaceAll("(.)\\1+", "$1");
6， Pattern和Matcher
Pattern p = Pattern.compile(a*b); //获取到正则表达式
Matcher m = p.matcher("aaaaaab"); //获取匹配器
boolean b = m.matches(); //看是否能匹配，匹配就返回true
等价于System.out.println("aaaaaab".matches("a*b"));
应用：获取字符串中的电话号码
String s = "我的电话号码是18988888888,曾经用过18987654321,还用过18812345678";
String regex = "1[3578]\\d{9}";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(s);
while(m.find()) System.out.println(m.group()); //必须先find再group