英文:
How to insert a bracket between the specified alphabet (i.e. A, D, F, R) and the digit using regular expression?
问题
如何使用正则表达式在指定的字母（即A、D、F、R）和数字之间插入括号？
输入：
A1
F42
D3
R6
预期输出：
(A)1
(F)42
(D)3
(R)6
我尝试过的方法：
let inputString = "f42";
let expPattern = /[ADFR]\d{1,3}/ig;
console.log(inputString.replace(expPattern, "($&)"));
它只返回"$0"。
我如何实现替换？
英文:
How to insert a bracket between the specified alphabet (i.e. A, D, F, R) and the digit using regular expression?
input:
A1
F42
D3
R6
Expected output:
(A)1
(F)42
(D)3
(R)6
What I have tried:
let inputString="f42"
let expPattern=/[ADFR]\d{1,3}/ig;
console.log(inputString.replace(expPattern,"(`$0`)"));
It returns "$0" only.
How can I implement the replacement?
答案1
得分: 3
你应该将字母和数字分别放入不同的捕获组中，这样你可以在替换字符串中引用它们：
let inputString = "F42";
let expPattern = /([ADFR])(\d{1,3})/ig;
console.log(inputString.replace(expPattern, "($1)$2"));
英文:
You should enclose the alphabet and the digits in separate capture groups so you can reference them in the replacement string:
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
let inputString="F42"
let expPattern=/([ADFR])(\d{1,3})/ig;
console.log(inputString.replace(expPattern,"($1)$2"));
<!-- end snippet -->
答案2
得分: 2
Sure, here are the translated parts:
如果您想在替换中使用完全匹配，可以使用正向预查 (?=\d{1,3}\b) 来断言右侧有 1-3 位数字。
为了防止部分匹配，您可以使用单词边界 \b。
在替换中使用完全匹配，表示为 $&。
请参见正则表达式演示中的所有替换。
let inputString = "F42";
let expPattern = /\b[ADFR](?=\d{1,3}\b)/ig;
console.log(inputString.replace(expPattern, "($&)"));
(Note: I've retained the code as requested and only translated the non-code parts.)
英文:
If you want to use the full match in the replacement, you can make use of a positive lookahead (?=\d{1,3}\b) asserting 1-3 digits to the right.
To prevent partial matches, you can use word boundaries \b
In the replacement use the full match denoted as $&
See a regex demo with all the replacements.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
let inputString="F42"
let expPattern=/\b[ADFR](?=\d{1,3}\b)/ig;
console.log(inputString.replace(expPattern,"($&)"));
<!-- end snippet -->
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