替换字符串中多次出现的缩写

Question

I need to replace multiple abbreviations in a string.

Result should be:

One DUMMY_0 two DUMMY_0 three DUMMY_0 four profession

What am I doing wrong?

 const abbr = ['vs.', 'po']; let str = 'One vs. two vs. three vs. four profession'; abbr.forEach((ab, index) => { const regex = new RegExp('\\b' + ab + '\\b', 'g'); str = str.replace(regex, 'DUMMY_' + index); }); console.log(str); 

Answer 1

Remove '\\b' from regex.

Also you will have to escape special characters.

 const abbr = ['vs.', 'po']; let str = 'One vs. two vs. three vs. four profession'; abbr.forEach(function(ab, index) { var regex = new RegExp(getParsedString(ab), 'g'); str = str.replace(regex, ('DUMMY_' + index)); }); console.log(str); function getParsedString(str) { var specialChars = /(?:\\.|\\!|@|\\#|\\$|\\%|\\^|\\&|\\*|\\(|\\)|\\_|\\+|\\-|\\=)/g; return str.replace(specialChars, function(v) { return "\\\\" + v; }) } 

Edit 1

• Added a generic parser to handle special characters other than . .
• Replaced const with var inside loop. If using ES6 then let is preferred.
• Removed arrow function as context binding is not required

Answer 2

First, you have to remove \\b in your regex (why did you write that there?), second: you have to escape the dots in your regexes, because dot means "any character" in a regular expression.

The modified code:

 const abbr = ['vs.', 'po']; let str = 'One vs. two vs. three vs. four profession'; abbr.forEach((ab, index) => { const regex = new RegExp(ab.replace(/\\./, "\\\\."), 'g'); str = str.replace(regex, 'DUMMY_' + index); }); console.log(str); 

Edit: Added escaping in the loop

Answer 3

An alternative way of solving this issue, with less code and without regex.

 const abbr = ['vs.', 'po']; let str = 'One vs. two vs. three vs. four profession'; abbr.forEach((ab, index) => { str = str.split(ab).join('DUMMY_' + index); }); console.log(str);