CODEWITHSUNDEEP
bashshellgrep
JackNewman
JackNewman

Reputation: 55

grep a pattern and find the last word of the line

Im trying to search through a file that only contains sequences of four words per line ie. the file looks like

walked the dog fast
jumped the moon high
on my nothing sun
on my not grass

and my code looks like

LINE=$(grep "^$PREV_THREE $PREV_TWO $PREV_ONE" $INPUT_FILE)
NEXT_WORD="${LINE##* }"

I use $PREV_THREE $PREV_TWO $PREV_ONE as the three words to indicate which fourth word to get from a specific line. My issue is that when $PREV_THREE $PREV_TWO $PREV_ONE equals on my not it sets NEXT_WORD equal to sun instead of grass. Is there an elegant way to Grep the first three words to find a perfect match and then set next to the fourth word? Instead of what I have above which just finds an occurrence of my pattern.

Upvotes: 0

Views: 962

Answers (1)

anubhava
anubhava

Reputation: 785156

It is easier to use a single awk command for this:

awk -v p='$PREV_THREE $PREV_TWO $PREV_ONE ' '$0 ~ p {print $4}' "$INPUT_FILE"

grass

btw your grep can also have a trailing space after last search word to make sure you don't match partial word. Consider this gnu grep:

grep -oP "^$PREV_THREE $PREV_TWO $PREV_ONE \K.+" $INPUT_FILE

grass

Upvotes: 2

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