CODEWITHSUNDEEP
regexshellsed
tbasien
tbasien

Reputation: 91

Extracting a string with sed

I've checked a few articles and found sed with regex. I mapped the tips to my problem, but without success.

This is not working for me:

echo "uri=https://myserver.domain.de:1234" | sed 's|//\(.+\):|\1|'

I expected

myserver.domain.de

But got the whole string

uri=https://myserver.domain.de:1234

Upvotes: 0

Views: 52

Answers (3)

Mateusz Piotrowski
Mateusz Piotrowski

Reputation: 9117

There's no need for sed here. Here's a way to do it using a portable POSIX feature called parameter expansion:

full="uri=https://myserver.domain.de:1234"
withoutport="${full%:[[:digit:]]*}" # Strip the trailing port number (":1234")
desired="${withoutport#uri=https://}" # Strip the undesired prefix
printf '%s\n' "$desired"

You can read more about it here in Open Group Standard Vol. 3: Shell and Utilities, Issue 7: 2.6.2 Parameter Expansion at The Open Group Publications Server.

If you insist on using sed, however, then here is pretty readable solution:

sed -e 's,^uri=https://,,' -e 's,:[0-9]\+$,,'

Upvotes: 0

tbasien
tbasien

Reputation: 91

I found this solution now:

echo "uri=https://myserver.domain.de:1234" | sed -r 's|(.+//)([^:]+)(:.+)|\2|'

answer

myserver.domin.de

Upvotes: 0

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626689

You need to match the part before the match and after it, and replace + with * (or escape the +, which will make it work in GNU sed with the BRE POSIX pattern):

echo "uri=https://myserver.domain.de:1234" | sed 's|.*//\(.*\):.*|\1|'

Result: myserver.domain.de.

See an online demo.

Here is an alternative pattern:

sed 's|.*//\([^:]*\).*|\1|'

where .* inside the capturing group is replaced with [^:]* (any 0+ chars other than :, see below).

Details

  • .* - any 0+ chars, as many as possible, up to the last occurrence of subsequent subpatterns
  • // - a // substring
  • \(.*\) - Group 1: any 0+ chars as many as possible (or, to constrain the engine a bit, you may use [^:]* here instead of .* (to match any 0+ chars other than :)
  • : - a colon
  • .* - the rest of the line

The \1 backreference will keep the captured value only.

Upvotes: 1

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