I was trying to figure out how to write a shell script to be able to print the name of the file and size to a text file I wanted to know if I was on the right track and am understanding this better.
In my script a for loop be able to print files over 4mb (we charge differently for larger images)
Code:
image="/Home/Desktop/images"
(cd "$image"
for f in *.jpg; do
let fsize=$(stat -c %s "$f")
if (( fsize == 4mb )); then
echo "$f"
fi
done) > mnt/desktop/sizes/results.txt
my understanding:
check variable $image that is being set from the directory
f is all images from *
do f's size and make a variable inside $()
stat -c %s prints the file name?
if condition is met for fsize equals 4mb size
print variable $f
to file results.txt
image="/Home/Desktop/images"
{
cd "$image"
for f in *.jpg
do
let fsize=$(stat -c %s "$f")
if (( fsize > 4194304 ))
then
echo "$f"
fi
done
} > mnt/desktop/sizes/results.txt
I extended my code to check the directory but now my issue is if try to echo $f it gives me the full path and the filename of the image. How can I just get the name to also print next to the directory??
Code:
echo 'checking images sizes'
{
for f in $(find $DIRECTORY/$directoryname -name \*.jpg);
do
let fsize=$(stat -c %s "$f");
if (( fsize > 2097152 ))
then
echo "$directoryname $f"
fi
done
} >> results.txt
EDIT:
and how do you do an && in bash so I can check for .jpeg and .jpg
Last edited by graphicsman; 10-30-2012 at 05:05 PM..
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10-30-2012
