[SOLVED] Delete everything before first pattern match with sed/awk

Issue

Let's say I have a line looking like this:

/Users/random/354765478/Tests/StoreTests/Base64Tests.swift

In this example, I would like to get the result:

Tests/StoreTests/Base64Tests.swift

How can I do if I want to get everything before the first pattern match (either Sources or Tests) using sed or awk?

I am using sed 's/^.*\(Tests.*\).*$/\1/' right now but it's falling:

echo '/Users/random/354765478/Tests/StoreTests/Base64Tests.swift' | sed 's/^.*\(Tests\)/\1/'
Tests.swift

Here's another example using Sources (which seems to work):

echo '/Users/random/741672469/Sources/Store/StoreDataSource.swift' | sed 's/^.*\(Sources\)/\1/'
Sources/Store/StoreDataSource.swift

I would like to get everything before the first, and not the last Sources or Tests pattern match.

Any help would be appreciated!

Solution

How can I do if I want to get everything before the first pattern match (either Sources or Tests).

Easier to use a grep -o here:

grep -Eo '(Sources|Tests)/.*' file

Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift

# where input file is
cat file

/Users/random/354765478/Tests/StoreTests/Base64Tests.swift
/Users/random/741672469/Sources/Store/StoreDataSource.swift

Breakdown:

• Regex pattern (Sources|Tests)/.* match any text that starts with Sources/ or Tests/ until end of the line.
• -E: enables extended regex mode
• -o: prints only matched text instead of full line

---

Alternatively you may use this awk as well:

awk 'match($0, /(Sources|Tests)\/.*/) {
   print substr($0, RSTART)
}' file

Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift

---

Or this sed:

sed -E 's~.*/((Sources|Tests)/.*)~\1~' file

Tests/StoreTests/Base64Tests.swift
Sources/Store/StoreDataSource.swift

Answered By - anubhava

Answer Checked By - Cary Denson (WPSolving Admin)