Bash script condition always passing even when grep should return nothing

Question

I am having issues when trying to return the one image file that fits the parameters. $1 is the search parameter, in this incident it is "real" which is a tag on one of the images (not two) in the given folder. What happens when I call it "./test.sh real" is it prints off both images, rather than just the one. I imagine it has to deal with me setting up the function return as a variable and/or my condition statement, but I am not quite sure.

#!/bin/bash
for f in specim/*.jpg
do
    image=$(exiftool -EXIF:XPKeywords $f | grep "$1")
    if [[ !  -z  "$image// }"  ]]; then
    echo $f
fi

what's returned:

../../test.sh real
specim/image2.jpg
specim/image.jpg

This bash script prints off what I want, only the one image rather than both (as the exiftool stuff which I don't want, but this was just a test):

#!/bin/bash
for f in *.jpg
do
    exiftool -EXIF:XPKeywords $f | grep $1
done

result:

./test2.sh real
XP Keywords                     : name;real

any help would be appreciated and it's probably super simple... Thanks

Answer 1

It is probably as you suspected, just a minor change to fix, try instead:

if [[ !  -z  "$image"  ]]; then

Explanation

Let's say when there is a match by exiftool and grep, then your $image variable contains this:

abcabcabc

But when there is no output, $image contains: (nothing)

In your test condition, you had:

if [[ !  -z  "$image// }"  ]]; then

So, in the first case bash sees this:

if [[ !  -z  "abcabcabc// }"  ]]; then

But in the second case, bash sees this:

if [[ !  -z  "// }"  ]]; then

The test is saying if "// }" is NOT zero-value, then... however "// }" is always going to be not zero value, it is something, it is a string consisting of two slashes a space and a curly brace. So since there is something there, something not zero-value, that is why the then part is triggered even when you have no matches in $image. So by removing this // } it should work.

Answer 2

Another option might be:

if [[ -n "$image" ]]; then

to check whether only the $image variable contains something, rather than ! -z ...

Answer 3

You are running exiftool and grep (+ a test and echo) for each .jpg in that directory... not very efficient...
exiftool can do all that by itself via the -if EXPR option, e.g.

exiftool -q -s3 -if '$tag=~/pattern/' -filename ./*.jpg

(here combined with -q - quiet and -s[NUM] - short output format: print values only, see the manual for more details)
So, in your particular case the whole script could be replaced by:

exiftool -q -s3 -if \$Exif:XPKeywords=~/"$1"/ -filename specim/*.jpg

e.g. running the test.sh script with four test files that have the following XPKeywords values:

exiftool -Exif:XPKeywords specim/*.jpg
======== specim/1.jpg
XP Keywords                     : xname;real
======== specim/2.jpg
XP Keywords                     : yname;different
======== specim/3.jpg
XP Keywords                     : wname;true
======== specim/4.jpg
XP Keywords                     : wname;unreal
    4 image files read

running test.sh real returns:

1.jpg
4.jpg

and test.sh unreal returns:

4.jpg

Answer 4

grep's exit status is whether or not it matched something, so why capture grep's output and then test it, as a string, for emptiness? You could do

if exiftool -EXIF:XPKeywords $f | grep -q "$1" ; then
    echo $f
fi

if you wanted.