RegEx for the string at the beginning of which there may be a version

Question

So, my program can accept strings like:

• v00240004 Lorem ipsum dolor sit amet, consectetur adipiscing elit.
• v0345 Duis nec eleifend nulla..

and also a string without a version number:

• Vivamus blandit et nibh nec placerat.
• In hac habitasse platea dictumst. Suspendisse potenti.

I tried to write a regular expression to remove the version from the string. And it came out something like this: /^.*?\s/ . It finds a match for the version and successfully replaces it with an empty string. But the problem is that it also works on strings without a version. That is, it removes the first word before the space.

So, my question is, how do I write a regular expression that works only on sentences starting with the letter "v" followed by numbers and replaces the version before the first space?

Example: A program receives 2 strings. The first v0255 Example1 string1 and the second Example2 string2 . The result after the regular expression should be Example1 string1 and Example2 string2 .

Answer 1

Use the following regex:

^v.*?\s

See regex proof .

Answer 2

This regex is going to help you match only those strings which have first word before space as 'v' then version number.

/^v\d.*?\s/gm