Grep for each element of array in a loop and write to newfile with BASH

Question

I am using grep and a for loop to try and search for each element of an array (ids) in a file (4700921_combined_subs.txt) and then write each line to a new file which is named as the search term. There are multiple occurances of each search term in the file being searched.

Can someone help me decipher what is wrong with my code? It does everything i want except the resultant files are all empty. any ideas?

    for i in "${ids[@]}"
        do 
        grep '^$i' 4700921_combined_subs.txt > $i.txt
        done

Many thanks in advance

anubhava · Accepted Answer · 2013-11-19 21:31:40Z

3

Use double quotes otherwise shell doesn't expand variables:

grep "$i" 4700921_combined_subs.txt > "$i.txt"

answered Nov 19, 2013 at 21:31

anubhava

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2 Comments

Tom Smith Over a year ago

Excellent, worked wonders. Thanks a lot - its always the simple things.

anubhava Over a year ago

You're welcome. If it woks for you please consider marking it as "accepted", so users facing a similar problem in the future will be able to see it easily.

chepner · Accepted Answer · 2013-11-19 21:52:56Z

1

You can do this much more efficiently with awk. Depending on how you obtain $ids, you can probably simplify this. Only two calls, one to grep, one to awk, are needed. And there's probably a way to avoid the call to grep, if I could think of it.

grep -f <( printf '%s\n' "${ids[@]}" ) 4700921_combined_subs.txt |
   awk '{f=$1.txt; print >> f }'

answered Nov 19, 2013 at 21:52

chepner

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