正则表达式：在字符串的所有单词中匹配模式

Question

I have a regex pattern that matches leading characters before (a, b, c or i) in only the first word of a string: /^\\s*[^abci]+(?=\\w)/i such that:

"sanity".replace(/^\s*[^abi]+(?=\w)/i, (pattern) => 'v');
  // "vanity"

How do i define the regex newRegex such that it matches leading characters in every word of a string so that:

 "sanity is a rice".replace(newRegex, (pattern) => 'v');

outputs: vanity is a vice

Answer 1

You can also try using split() and map() to remove the first char and get the desired output:

 function replaceChar(str){ var matchChar = ['a', 'b', 'c', 'i']; var changedArr = str.split(/\\s+/).map((item) => { if(matchChar.includes(item.charAt(1))){ return 'v' + item.substr(1, item.length); } return item; }); return changedArr.join(' '); } var str = 'sanity is a rice'; console.log(replaceChar(str)); str = 'This is a mice'; console.log(replaceChar(str)); 

Answer 2

It seems to me you want to replace any word char other than a , b , i at the beginning of a word.

You may use

.replace(/\b[^\Wabi]/gi, 'v')

See the regex demo .

• \\b - a word boundary
• [^\\Wabi] - a negated character class that matches any char other than a non-word char (so, all word chars are matched except the chars that are also present in this class), a , b and i .

The global modifier g is added so as to match all occurrences.

JS demo:

 console.log( "sanity is a rice".replace(/\\b[^\\Wabi]/gi, 'v') );