非捕获组匹配 JavaScript 正则表达式中的空白边界 [英] Non-capturing group matching whitespace boundaries in JavaScript regex

问题描述

我有这个功能可以找到整个单词并替换它们.它标识空格但不应替换它们，即不捕获它们.

I have this function that finds whole words and should replace them. It identifies spaces but should not replace them, ie, not capture them.

function asd (sentence, word) {
     str = sentence.replace(new RegExp('(?:^|\s)' + word + '(?:$|\s)'), "*****");
     return str;
};

然后我有以下字符串:

var sentence = "ich mag Äpfel";
var word = "Äpfel";

结果应该是这样的:

"ich mag *****" 

而不是:

"ich mag*****"

我得到后者.

如何才能在替换单词时识别空格但忽略它?

How can I make it so that it identifies the space but ignores it when replacing the word?

起初这似乎是重复的，但我没有找到这个问题的答案，这就是我问它的原因.

At first this may seem like a duplicate but I did not find an answer to this question, that's why I'm asking it.

谢谢

推荐答案

您应该使用 捕获 组(而不是 非捕获一)在替换模式中使用替换反向引用，您还可以利用右空白边界的前瞻，这在连续匹配的情况下很方便:

You should put back the matched whitespaces by using a capturing group (rather than a non-capturing one) with a replacement backreference in the replacement pattern, and you may also leverage a lookahead for the right whitespace boundary, which is handy in case of consecutive matches:

function asd (sentence, word) {
     str = sentence.replace(new RegExp('(^|\s)' + word + '(?=$|\s)'), "$1*****");
     return str;
};
var sentence = "ich mag Äpfel";
var word = "Äpfel";
console.log(asd(sentence, word));

查看正则表达式演示.

详情

• (^|s) - 组 1(稍后在替换模式中的 $1 占位符的帮助下引用):匹配任一开始的捕获组字符串或空格
• Äpfel - 一个搜索词
• (?=$|s) - 一个正向前瞻，要求字符串结尾或空格紧接在当前位置的右侧.

• (^|s) - Group 1 (later referred to with the help of a $1 placeholder in the replacement pattern): a capturing group that matches either start of string or a whitespace
• Äpfel - a search word
• (?=$|s) - a positive lookahead that requires the end of string or whitespace immediately to the right of the current location.

注意:如果 word 可以包含特殊的正则表达式元字符，转义它们:

NOTE: If the word can contain special regex metacharacters, escape them:

function asd (sentence, word) {
     str = sentence.replace(new RegExp('(^|\s)' + word.replace(/[-/\^$*+?.()|[]{}]/g, '\$&') + '(?=$|\s)'), "$1*****");
     return str;
};

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