shell script for copying log files into a single compressed file

Question

We have a folder in our embedded board "statuslogs", this folder contains logs which are of the format : daily_status_date_time.log.

We need to get all the files of a particular year into a single file, for fetching from the server.

We did the following in our script

gzip -c statuslogs/daily_status_2017*.log > status_2017.gz
gzip -c statuslogs/daily_status_2018*.log > status_2018.gz
gzip -c statuslogs/daily_status_2019*.log > status_2019.gz
gzip -c statuslogs/daily_status_2020*.log > status_2020.gz
gzip -c statuslogs/daily_status_2021*.log > status_2021.gz

The problem with this logic is that it will still create status_*.gz file for the years 2019,2020,2021.

I tried writing the following logic

if [ - f statuslogs/daily_status_2017*.log ] but it fails due to regex may be. And I am not using bash, the interpreter is ash.

Can you please help me to optimize the script

Thanks for your time

Answer 1

You have a syntax error. It's -f, not - f. Example:

if [ -f statuslogs/daily_status_2017*.log ]; then
    gzip -c statuslogs/daily_status_2017*.log > status_2017.gz
fi

However, with this you will probably run into a "too many arguments" error, which will happen if you have more than one matching file. So this would work better:

if find statuslogs/daily_status_2017*.log -mindepth 0 -maxdepth 0|head -n1; then 
    gzip -c statuslogs/daily_status_2017*.log > status_2017.gz
fi

It would be better to instead stop the loop when you reach the current year. For example,

for year in $(seq 2017 $(date +%Y)); do