Bash replace string where specific condition is true

Question

I am trying to replace a string where a specific condition is true in the same line

I have a file with some lines. I to replace a word (word1) with another (word2) in every line that starts with another word (word3)

For example:

foo moo see
kaa haa qee
foo dee see
uuu ooo rrr
foo dee laa

I want to replace the word "see" with "raa" in every line which begins with "foo". So doing that will result in the lines being like:

foo moo raa
kaa haa qee
foo dee raa
uuu ooo rrr
foo dee laa

I was told that I can use sed to do that but I didn't figure out how to do that.

Please help! (with sed only)

Answer 1

This is very basic sed:

$ sed '/^foo/s/see/raa/g' file
foo moo raa
kaa haa qee
foo dee raa
uuu ooo rrr
foo dee laa

Explanation

• sed 's/something/replacement/g' replaces something strings with replacement. The g indicates that it is to be done every time it can.
• Adding /^foo/ makes it just take into consideration the lines starting with foo.
• If you want the file to be edited in place, that is, to have the file content updated with the substitution, add -i: sed -i.bak '/^foo/s/see/raa/g' file. file.bak will contain the previous file.

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Update

How to use variables instead of foo, see, raa ? – Khaleal

Like this and with double quotes:

$ d="foo"
$ e="see"
$ sed "/^$d/s/$e/raa/g" file
foo moo raa
kaa haa qee
foo dee raa
uuu ooo rrr
foo dee laa

and so on.

Answer 2

I know its asked for sed, but posting an awk for other to read, if they like.

awk '/^foo/ {sub(/see/,"raa")}1' file