正则的扩展详解

// 第一种情况
let regex = new RegExp('abc', 'i')
// 第二种情况
let regex2 = /abc/i

let s = 'abc'
regex.test(s) == regex2.test(s); // true

let regex4 = new RegExp(/abc/ig, 'i')

/^\uD83D/u.test('\uD83D\uDC2A');  // false

/^\uD83D/.test('\uD83D\uDC2A'); // true 

let z = '正'
/^.$/.test(z);   // true
/^.$/u.test(z);  // true

/\u{61}/.test('a'); // false
/\u{61}/u.test('a'); // true

/[a-z]/i.test('\u212A'); // false 
/[a-z]/iu.test('\u212A'); // true
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let a = 'aaa-aa-a'
let r = /a+/y
let r2 = /a+/g
// 第一次匹配
r.exec(a); // aaa
r2.exec(a); // aaa

// 第二次匹配
r.exec(a); // null  
r2.exec(a); // aa  

let a1 = /heelo\d/y
console.log(a1.sticky); // true

let rul = /abc/ig
console.log(rul.source); // abc
console.log(rul.flags); // ig

let f = /foo.bar/.test('foo\nbar')
console.log(f); // false // 因为.不匹配 \n ,所以正则表达式返回false
// 解决方法 : s 修饰符
/* 使用s修饰符，可以使 . 匹配任意单个字符 */
let f2 = /foo.bar/s.test('foo\nbar')
console.log(f2); // true

let f3 = /foo.bar/s
console.log(f3.dotAll); // true  // 开始了dotAll模式

let look = '100% 东方不败'
/\d+(?=%)/.exec(look); // 100  // 数字在百分号前面才匹配
let look2 = '100! 东方不败'
console.log(/\d+(?=%)/.exec(look2));  // null 没有匹配到%前的数字

let look3 = '东方不败 $100 西方求败'
/(?<=\$)\d+/.exec(look3); // 100  表示在\$后的字符才会被匹配

let rul = /(\d{4})-(\d{2})-(\d{2})/
let r = rul.exec('2022-12-13')
console.log(r[1]);  // 2022
console.log(r[2]);  // 12
console.log(r[3]);  // 13

let rul2 = /(?<year>\d{4})-(?<month>\d{2})-(?<day>\d{2})/
let r2 = rul2.exec('2022-12-13')
console.log(r2.groups.year);  // 2022
console.log(r2.groups.month);  // 12
console.log(r2.groups.day);  // 13
console.log(r2.groups.minute);  // undefined  如果没有匹配，那么对象属性为undefined

let {groups : {hours,minute}} = /^(?<hours>.*):(?<minute>.*)/.exec('12:49')
console.log(hours,minute);  // 12  49  // 通过解构赋值提取

let rul4 = /(?<year>\d{4})-(?<month>\d{2})-(?<day>\d{2})/u
console.log('2022-12-13'.replace(rul4,'$<day>/$<month>/$<year>')); // 13-12-2022
console.log(rul4);  // /(?<year>\d{4})-(?<month>\d{2})-(?<day>\d{2})/u

let data = '2022-12-13'.replace(rul4,(matched,params1,param2,params3,position,sources,groups)=>{
        let {day,month,year} = groups;
        return `${day}/${month}/${year}`
       })
console.log(data);   // 13-12-2022

let str = /^(?<word>[a-z]+)!\k<word>$/
str.test('abc!abc');  // true  引用了具名匹配，前后都是abc,前后字符一致
str.test('abc!ab');   // false 引用了具名匹配，前后字符不匹配，但是第二个少一个c

let str2 = /^(\d{2})([a-z]+)\1$/
console.log(str2.test('11abcd'));  // false 
console.log(str2.test('11aa11'));  // true

let str3 = /^(?<word>[a-z]+)-(\d{2})-\k<word>-\2$/
str3.test('a-11-a-11'); // true
str3.test('a-11-a-aa'); // false

let text = 'abcdefg'
let te = /bc/d
let result = te.exec(text)
result.index;  // 1   拿到bc匹配的起始位置，下标为1
result.indices;  // [1,3]  拿到bc匹配的开始和结束位置的下一个字符的位置

let text2 = 'abccdef'
let te2 = /bc+(de)/d
let result2 = te2.exec(text2)
console.log(result2.indices);  // [1,6][4,6] 

let text3 = 'abccdefgh'
let te3 = /bc+(de(fg))/d
let result3 = te3.exec(text3)
result3.indices;  // [1,8][4,8][6,8] 

let text4 = 'abcdefgh'
let te4 = /bc+(?<word>de)/d  // 具名组<word>
let result4 = te4.exec(text4)
result4.indices; // [1,5][3,5]
result4.indices.groups; // {word : [3,5]}

let strings = 'test1test2test3';
let regex = /t(e)(st(\d?))/g;
for(let match of strings.matchAll(regex)){
   console.log(match);  // 返回所有匹配
  }

[...strings.matchAll(regex)]; // 返回所有匹配，结果为数组

Array.from(strings.matchAll(regex)); // 返回所有匹配，结果为数组