I'm trying to write a script that will display the name of oldest file within the directory that the script is executed from.
This is what I have so far:
#!/bin/bash
for arg in $*
do
oldest=$1
if [[ $arg -ot $oldest ]]
then
oldest=$arg
fi
done
echo "Oldest file: $oldest"
I'm not sure how to increment to the next file to check if it is older
for example:
oldest=$2
oldest=$3
etc..
trying to run this script in the bash shell given the following args:
myScript `ls -a`
I get a result of:
Oldest File: .
The ls
program has an option to sort on time and you can just grab the last file from that output::
# These are both "wun", not "ell".
# v v
oldest="$(ls -1t | tail -1)"
If you want to avoid directories, you can strip them out beforehand:
# This one's an "ell", this is still a "wun".
v v
oldest="$(ls -lt | grep -v '^d' | tail -1 | awk '{print $NF}')"
I wouldn't normally advocate parsing ls
output but it's fine for quick and dirty jobs, and if you understand its limitations.
If you want a script that will work even for crazies who insist on putting control characters in their file names :-) then this page has some better options, including:
unset -v oldest
for file in "$dir"/*; do
[[ -z $oldest || $file -ot $oldest ]] && oldest=$file
done
Though I'd suggest following that link to understand why ls
parsing is considered a bad idea generally (and hence why it can be useful in limited circumstances such as when you can guarantee all your files are of the YYYY-MM-DD.log
variety for example). There's a font of useful information over there.
You can use this function to find oldest file/directory in any directory:
oldest() {
oldf=
for f in *; do
# not a file, ignore
[[ ! -f "$f" ]] && continue
# find oldest entry
[[ -z "$oldf" ]] && oldf="$f" || [[ "$f" -ot "$oldf" ]] && oldf="$f"
done
printf '%s\n' "$oldf"
}
And call it in any directory as:
oldest
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