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用于在字符串中查找模式的 Javascript 正则表达式,排除可能的前面单词匹配

[英]Javascript regex to find a pattern in a string, excluding a possible preceding word match

There is a string of writing/text in a variable:变量中有一个文字/文本字符串:

var copypaste, replaced;
var verbdb = ["keep","know","enjoy"];
var arrayLength = verbdb.length;

for (var i = 0; i < arrayLength; i++) {
    copypaste = "One of the reasons first_name keep waking up at night, is blah blah blah. Try this, let first_name know and see blah blah blah. Remember that first_name enjoy being challenged and blah blah blah.";

    replaced = copypaste.replace(RegExp("[^let]? first_name " + verbdb[i] + "(\s|\.)","g")," first_name " + verbdb[i] + "_object_s\$1");
}

What I am seeking to get from the replaced variable is to exclude ONLY when first_name is preceded by the word let .我试图从replaced变量中得到的是仅当 first_name 前面有let一词时才排除。

"One of the reasons first_name keep_object_s waking up at night, is blah blah blah. Try this, let first_name know and see blah blah blah. Remember that first_name enjoy_object_s being challenged and blah blah blah."; “first_name keep_object_s 晚上醒来的原因之一,是等等等等。试试这个,让 first_name 知道,看看等等等等。记住 first_name enjoy_object_s 受到挑战,等等等等。”;


So in this example:所以在这个例子中:

  1. the first pattern MATCHES (keep) and is replaced with _object_s added in,第一个模式匹配(保持)并替换为添加的_object_s
  2. the second pattern does NOT match (know) because of the word "let", so no replacement由于“让”这个词,第二个模式不匹配(知道),所以没有替换
  3. the third pattern MATCHES (enjoy) and is replaced with _object_s added in.第三个模式 MATCHES (enjoy) 并替换为添加的_object_s

You can try this :你可以试试这个:

(?:\b)(?!let)(?=\w+\sfirst_name\s+(know|keep|enjoy))(\w+ \w+)\s+(\w+)

Explanation解释

 const regex = /(?:\\b)(?!let)(?=\\w+\\sfirst_name\\s+(know|keep|enjoy))(\\w+ \\w+)\\s+(\\w+)/g; const str = `One of the reasons first_name keep waking up at night, is blah blah blah. Try this, let first_name know and see blah blah blah. Remember that first_name enjoy being challenged and blah blah blah`; const subst = `$2 _object_s`; const result = str.replace(regex, subst); console.log(result);

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