Replace words with sed

Goal

The file testdata/app.log mentions timeout in several places. Print the whole file with every occurrence of timeout replaced by TIMEOUT, and everything else left exactly as it is.

Example

If testdata/app.log held:

info all fine
warn timeout on db
error timeout, then another timeout

your script prints:

info all fine
warn TIMEOUT on db
error TIMEOUT, then another TIMEOUT

Notice that the first line had no match and still gets printed, and that the last line had two matches and both changed.

Rules

• Edit answer.sh. Read testdata/app.log and print the result to the screen.
• Print every line of the file, in the original order, including lines with no match.
• Replace every match on a line, not just the first one.
• Matching is case-sensitive: only the lower-case word timeout changes. A word like Timeout or TIMEOUT stays as it is.
• Change only the matching text. Any other characters on the line stay untouched, and a longer word that contains timeout (such as timeouts) still has its timeout part replaced.
• Do not edit testdata/app.log itself. Your script only prints the changed text.

What you'll practice

sed 's/old/new/g' and the difference between replacing the first match and replacing every match.

Run it

gonomicon fetch week-1/03-03-sed-substitute-words
cd 03-03-sed-substitute-words
gonomicon test
gonomicon submit

Hints

1. sed rewrites text as it flows through. The s command means substitute: s/old/new/.
2. Without the trailing g, sed only changes the first match on each line, and some lines here have two.
3. The shape is sed 's/OLD/NEW/g' FILE. Put the file path last.