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

问题描述

我具有此功能，可以查找整个单词并应替换它们.它标识了空格，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";

结果应该类似于:

"ich mag *****" 

而非:

"ich mag*****"

我得到了后者.

如何使它识别空格，但在替换单词时忽略空格?

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，但是我没有找到这个问题的答案，这就是为什么我要问这个问题.

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 captureing 组(而不是 non-captureing )放回匹配的空格一个)，并在替换模式中使用替换后向引用，并且您还可以利用超前查询来找到正确的空白边界，这在连续匹配的情况下非常方便:

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));

请参见 regex演示.

详细信息

• (^ | \ s)-第1组(后来在替换模式中借助 $ 1 占位符进行引用):与任一开始匹配的捕获组字符串或空格
• Äpfel-搜索词
• (?= $ | \ s)-正向超前，需要在当前位置的右侧紧跟着字符串或空格的结尾.

• (^|\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，请转义它们:

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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