CODEWITHSUNDEEP
bashshellgrep
user3568611
user3568611

Reputation: 141

Grep in a directory

I have a subdirectory called Liste and I have texte files inside, I want to find a text (a telephone number inside):

read $number
while ! grep -l $number * Liste; do
    echo"Le fichier n'existe pas"
    read $number
done
grep -l  $number * Liste

But it doesn't work. Thanks.

I made a mistake here's my code, and it shows me 2 time the ./Liste/"name"

Upvotes: 0

Views: 441

Answers (2)

tripleee
tripleee

Reputation: 189487

The problem is that the argument to read is a variable name, not its value. In other words, drop the dollar sign.

read number
while ! grep -iq "$number" Liste/*; do
     echo "Le numero n'existe pas"
     read number
done
echo " Le numero existe"

Note also how to quote properly, and how to refer to all the files in the directory Liste. Finally, the -q option suppresses the normal output from grep (it would print every match -- every line in every file in the directory if you entered something like .!) Finally, notice also that the space after echo is mandatory.

Update: If you want to list the file names with matches without the leading directory name, cd into the directory first.

cd Liste
read number
while ! grep -il "$number" *; do
    echo "Le numéro n'existe pas"
    read number
done
echo "Le numéro existe"

(I added the -l option on the assumption that you only want the file name; but if you don't, just remove it.)

Upvotes: 2

arco444
arco444

Reputation: 22831

You can use:

grep "$number" ./Liste/* >/dev/null && echo "Le numero existe" || echo "Le fichier n'existe pas"

Or if you want to output the name of the file that matches, use the l option:

grep -l "$number" ./Liste/*

Based on your comment, use:

grep -l "$number" Liste/* | xargs basename

Upvotes: 0

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