Javascript Regex expression for matching words with digits ignoring 4 digit numbers

Question

I have a string that is just words with a single whitespace in-between. Assuming no special characters I would like to match all words containing digits while ignoring 4 digit numbers.

IE

hello12345 12345hello 123456789 12 red hello 1234 5678

Would Match:

hello12345 12345hello 123456789 12

The ultimate goal would be replacing hello12345 12345hello 123456789 12 with an empty string resulting in:

red hello 1234 5678

The following \\w*\\d\\w* matches words with digits and \\b\\d{4}\\b matches all 4 digit numbers. However, I am unsure of how to combine them.

Answer 1

Match and capture what you need and just match what you do not need (see The Best Regex Trick Ever ):

 var re = /\\b\\d{4}\\b|(\\w*\\d\\w*)/g; var str = "hello12345 12345hello 123456789 12 red hello 1234 5678"; var m, res = []; while (m = re.exec(str)) { if (m[1]) res.push(m[1]); } console.log(res); 

The \\b\\d{4}\\b alternative is only matched, but the second one, (\\w*\\d\\w*) , is also captured with the help of the capturing group, (...) . This value is kept in Group 1, accessed via m[1] .

Answer 2

This regex

/((\b(\d{1,3}|\d{5,})\b)|([a-z]+\d\w*|\w*\d[a-z]+))\s*/gi

matches:

// Digit-only words with less than or more than 4 digits
\b(\d{1,3}|\d{5,})\b

// Words that contain at least a number and a letter
[a-z]+\d\w*|\w*\d[a-z]+

incl. whitespace between them.

 var string = "hello12345 12345hello 123456789 12 red hello 1234 5678"; var regex = /((\\b(\\d{1,3}|\\d{5,})\\b)|([az]+\\d\\w*|\\w*\\d[az]+))\\s*/gi console.log(string.replace(regex, "")); 

There might be a simpler one. That's just from the top of my head.