CODEWITHSUNDEEP
regexbashfilter
Kevin Goslar
Kevin Goslar

Reputation: 445

Filter lines of a file through partial matches in Bash

I have a file that contain redundant Git statements, like this:

git fetch
git checkout foo
git checkout bar
git checkout baz
git merge origin/baz

The checkout statements are redundant, and I would like to keep only the last one, so that the file ends up looking like this:

git fetch
git checkout baz
git merge origin/baz

I can only use Bash script. How could I go about this?

Upvotes: 1

Views: 197

Answers (2)

buydadip
buydadip

Reputation: 9417

sed has a convenient way of printing the last occurrence of a pattern in a line:

sed '/checkout/h; $!d; x' file

Which, in your case would extract the string git checkout baz.

You can then use this in a script, save the sed output to a variable, and use it to print out all the lines which you specify:

file=$1
line=$(sed '/checkout/h; $!d; x' $file)
sed -n "/checkout/!p;/$line/p" $file

The last sed states, print all lines that do not contain checkout, as well as the one that contains the output from the first sed command.

Upvotes: 0

kev
kev

Reputation: 161674

Try this awk command:

tac input.txt | awk '{cmd = $1$2}; cmd != last; {last = cmd}' | tac

result:

git fetch
git checkout baz
git merge origin/baz

Upvotes: 1

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