Linux – Read a file and pipe to grep

greplinux

Wondering if anyone can help me, I'm very rusty with bash and seem to hit a bit of an impasse.

I'm storing a list of strings in a file and would like to read the file and pipe each line returned to grep which in turn searches a directory for files containing the string.

Initial attempt:

cat filename | grep -lr *

However this is not returning any output.

Can anyone give me some directions on the best approach?

Best Answer

Avoid that useless use of cat. You can of course solve this with xargs and the like. But that's over-complex compared to a simple while loop.

while read i 
do
    grep -r -- "$i" directory/
done < filename

Related Solutions

Linux – grep a file and pipe the output to sed and store sed output in a file

To select any line containing gold from source.txt and replace the first occurrence of green with red:

$ sed  -n '/gold/{s/green/red/; p}' source.txt 
gold  red white black blue

To save that in a file:

sed  -n '/gold/{s/green/red/; p}' source.txt  >pol.txt

How it works

-n tells sed not to print lines unless we explicitly ask it to.
/gold/ selects lines that match the regex gold.
s/green/red/ performs the substitution
p prints.

Using awk

With the same logic:

$ awk '/gold/{gsub(/green/, "red"); print}' source.txt 
gold  red white black blue

Using grep

If we are forced, for reasons not yet explained, to use a grep pipeline, then try:

$ grep -l --null "gold" source.txt | xargs -0 sed  -n '/gold/s/green/red/p'
gold  red white black blue