I have a below log files who have date mention in filename. I want to delete the files have 30 days old date in filename. someone please provide the solution.

log files:-

fortinetPerformanceMet_01_11_2022.log
fortinetTokenGeneration_05_11_2022.log
fortinetTopDestination_10_12_2022.log
fortinetTopSource_11_12_2022.log

I want to delete a logfiles who have 30 days old date in filename. after deleting 30 days old files the expected output from above file will be

fortinetTopDestination_10_12_2022.log
fortinetTopSource_11_12_2022.log

I tried below

find /path/to/files/ -type f -name "*_*_*_$2022.log" -mtime 30 -delete, but it will delete the files according to the file timestamp.

CodePudding user response：

Assuming:

• Your date command supports -d option.
• The timestamp in the filenames represents dd_mm_yyyy.

Then would you please try:

#!/bin/bash

oldday=$(LC_TIME=C date -d "-30 days" " %Y_%m_%d")
while IFS= read -r f; do
    if [[ $f =~ _([0-9]{2})_([0-9]{2})_([0-9]{4})\.log ]]; then
        timestamp="${BASH_REMATCH[3]}_${BASH_REMATCH[2]}_${BASH_REMATCH[1]}"
        if [[ $oldday > $timestamp ]]; then
            echo rm -- "$f"
        fi
    fi
done < <(find /path/to/files/ -type f -name "*_*_*_*.log")

• oldday is assigned to the date 30 days ago in the format yyyy_mm_dd.
• timestamp is re-arranged as yyyy_mm_dd.
• Now we can compare them with the > operator.

If the output looks good, drop echo.

CodePudding user response：

If I understand the timestamp is given in the filename only.

• We can use regex to extract the dd_mm_yyyy components from the filename
• We can rearrange this to yyyymmdd format
• Then we can use yyyymmdd to convert that to a second relative to an epoch date in seconds using date %s
• We can also get today's date relative to the same epoch date in seconds using date %s
• Then we can simply subtract the two dates and change the units from seconds to days
• If the days is less than 30 we don't delete

Here's the code:

now=$(date  %s)
for f in $(find . -type f -name "*_*_*.log")
do
    if [[ $f =~ _([0-9]{2})_([0-9]{2})_([0-9]{4})\.log ]]; then
        yyyymmdd=${BASH_REMATCH[3]}${BASH_REMATCH[2]}${BASH_REMATCH[1]}
        old=$(date  %s -d $yyyymmdd)
        ((age = (now -old) / 60 / 60 / 24))
        echo "file: $f age: $age"
        if (( age < 30 )); then continue; fi
        echo delete $f
        # rm $f # UNCOMMENT WHEN YOU'RE HAPPY
    fi
done

For your sample data, the above script outputs:

file: ./fortinetPerformanceMet_01_11_2022.log age: 41
delete ./fortinetPerformanceMet_01_11_2022.log
file: ./fortinetTopDestination_10_12_2022.log age: 2
file: ./fortinetTopSource_11_12_2022.log age: 1
file: ./fortinetTokenGeneration_05_11_2022.log age: 37
delete ./fortinetTokenGeneration_05_11_2022.log