BASH - file test within a variable

Question

I'm trying to find out whether a file is an image or not and then do something according to the result.This works fine for file in the present working directory and in other directories:

file -ib "$i" | grep "image/"

But this only works with files in the present directory (where the script was called from)

TEST=$(file -ib "$i" | grep "image/")

if [ "$TEST" == "" ]
then
    echo "Image"
else
    echo "Not an image"
fi

So something like ../../my.jpg works in the first case, but always gives "Not an image" in the second case. I would like to use my script on files anywhere on the system.

Thanks for your help.

What does bash -x second-fragment.sh show as the value in $TEST? That'll tell you what you need to do to fix the problem. — Jonathan Leffler
– Jonathan Leffler, Commented Aug 18, 2015 at 17:45
I assume you are setting i=$1 or similar in the script case? — Etan Reisner
– Etan Reisner, Commented Aug 18, 2015 at 17:45
You seem to have your condition wrong. TEST will be empty if grep doesn't find image. — Qualia
– Qualia, Commented Aug 18, 2015 at 17:53
Do you need the value of TEST later? If not you can do the command directly in the if — Eric Renouf
– Eric Renouf, Commented Aug 18, 2015 at 18:03

NarūnasK · Accepted Answer · 2015-08-18 18:55:10Z

1

Please try:

path="path/to/images/folder"
for i in $(find $path -type f); do
  file -ib "$i" | grep -q "image/" && echo "$i: Image" || echo "$i: Not an image";
done

answered Aug 18, 2015 at 18:55

NarūnasK

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Comments

Jonathan Leffler · Accepted Answer · 2015-08-18 18:17:42Z

0

Try:

TEST=$(file -i "$i" | grep ': *image/')

if [ "$TEST" = "" ];then 
  echo "not an image"
else 
  echo "it is an image"
fi

edited Aug 18, 2015 at 18:17

Jonathan Leffler

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answered Aug 18, 2015 at 17:49

michael501

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