regex: replacing the 1st occurrence of a character in a string

Question

I have this string:-

ABCDE/Something something:XYZ=0, JKLM=0/SOMETHING Something:some_value

What is the regex so that only the first colon (:) is replaced with underscore (_)?

Answer 1

Just match two groups - the first being everything before the first colon; the second, everything after it. Then just rebuild the string with the underscore in place.

s/([^:]*):(.*)/\1_\2/

You will need differing escaping depending on the language/regex-engine you use.

Answer 2

You can do it in regex using a negative lookbehind, but that's relatively inefficient:

(?<!:.*):

Will only match a colon if no other colon has been previously matched.

However, since you're only replacing one character, not a pattern of characters, I would suggest using the language's native "replace" function. You'll get better performance and readability.

Answer 3

In standard systems, you simply write:

s/:/_/

To achieve a global replace (replacing every instance of colon with underscore) you'd add a qualifier (often 'g') after the substitution.

Different languages use different notations for regular expressions, so the detailed answer depends on the target language. However, what I wrote works in 'sed', 'ed', 'vi', 'vim', and Perl.

Answer 4

if you are on *nix and have tools like sed

$ echo "ABCDE/Something something:XYZ=0, JKLM=0/SOMETHING Something:some_value" | sed 's/:/_/'
ABCDE/Something something_XYZ=0, JKLM=0/SOMETHING Something:some_value

also, if you are using bash as your shell

$ var="ABCDE/Something something:XYZ=0, JKLM=0/SOMETHING Something:some_value"
$ echo ${var/:/_}
ABCDE/Something something_XYZ=0, JKLM=0/SOMETHING Something:some_value