Getting highest extensions value in unix script - unix

Question

I need to create new files with extensions like: file.1, file.2, file.3 and then check if files with certain numbers exist and create file.(n+1) where n is number of highest, existing file. I was trying to get extensions using basename but it doesn't want to get couple of files
file=`basename $file.*`
ext=${file##*.}
It only works when I input whole file name like $file.3

Answer 1

If the filenames are guaranteed not to have newline characters in them, you can, for example, use standard unix text processing tools:
printf '%s\n' file.* | #full list
sed 's/.*\.//' | #extensions
grep '^[0-9][0-9]*$' | #numerical extensions
awk '{ if($0>m) m=$0} END{ print m }' #get maximum

Answer 2

Here's my take on this.
You can do this entirely in standard awk.
$ awk '{ext=FILENAME;sub(/.*\./,"",ext)} ext>n&&ext~/^[0-9]+$/{n=ext}{nextfile} END {print n}' *.*
Broken out for easier reading:
$ awk '
{
# Capture the extension...
ext=FILENAME
sub(/.*\./,"",ext)
}
# Then, if we have a numeric extension that is bigger than "n"...
ext > n && ext ~ /^[0-9]+$/ {
# let "n" be that extension.
n=ext
}
{
# We aren't actually interested in the contents of this file, so move on.
nextfile
}
# No more files? Print our result.
END {print n}
' *.*
The idea here is that we'll step through the list of filenames and let awk do ALL the processing to capture and "sort" the extensions. (We're not really sorting, we're just recording the highest number as we pass through the files.)
There are a few provisos with this solution:
This only works if all the files have a non-zero length. Technically awk conditions are being compared on "lines of the file", so if there are no lines, awk will pass right by that file.
You don't really need to use the ext variable, you can modify FILENAME directly. I included it for improved readability.
The nextfile command is fairly standard, but not universal. If you have a very old machine, or are running an esoteric variety of unix, nextfile may not be included. (I don't expect this to be a problem.)
Another alternative, which might be easier for you, would be to implement the same logic directly in POSIX shell:
$ n=0; for f in *.*; do ext=${f##*.}; if expr "$ext" : '[0-9][0-9]*$' >/dev/null && [ "$ext" -gt "$n" ]; then n="$ext"; fi; done; echo "$n"
Or, again broken out for easier reading (or scripting):
n=0
for f in *.*; do
ext=${f##*.}
if expr "$ext" : '[0-9][0-9]*$' >/dev/null && [ "$ext" -gt "$n" ]; then
n="$ext"
fi
done
echo "$n"
This steps through all files using a for loop, captures the extension, makes sure it's numeric, determines whether it's greater than "n" and records if it it is, then prints its result.
It requires no pipes and no external tools except expr, which is a POSIX.1 tool available on every system.
One proviso for this solution is that if you have NO files with extensions (i.e. *.* returns no files), this script will erroneously report that the highest numbered extension is 0. You can of course handle that easily enough, but I thought I should mention it.

Answer 3

Thanks for all answers, I've came up with quite similar and a bit simpler idea which I'd like to present it:
for i in file.*; do
#reading the extensions
ext=${i##*.}
if [ "$ext" -gt "$n" ];
then
#increasing n
n=$((n+1))
fi
done
then if we want to get number exceeding n by one
until [[ $a -gt "$n" ]]; do
a=$((a+1))
done
and finally a is one number bigger then number of file extensions. So if there are three files: file.1 file.2 file.3 the returned value will be 4.