Bash script to find and display oldest file

Question

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: . ```

Accepted Answer

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.

Bash script to find and display oldest file

Solution

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