Question

I have log file in which I’m trying to delete all entries older than a specified date. Though I haven’t succeeded with this yet. What I’ve tested so far is having an input for what the entries must be older than to be deleted and then loop like this:

#!/bin/bash

COUNTER=7
DATE=$(date -d "-${COUNTER} days" +%s)
DATE=$(date -d -@${DATE} "+%Y-%m-%d")

while [ -n "$(grep $DATE test.txt)" ]; do

    sed -i "/$DATE/d" test.txt

    COUNTER=$((${COUNTER}+1))
    DATE=$(date -d "-${COUNTER} days" +%s)
    DATE=$(date -d @${DATE} +"%Y-%m-%d")
done

This kind of works except when a log entry doesn’t exist for date. When it doesn’t find a match it aborts the loop and the even older entries are kept.

Update

This was how I solved it:

#!/bin/bash

COUNTER=$((7+1))
DATE=$(date -d "-${COUNTER} days" +%s)
DATE=$(date -d -@${DATE} "+%Y-%m-%d")

if [ -z "$(grep $DATE test.txt)" ]; then
   exit 1
fi

sed -i "1,/$DATE/d" test.txt

Answer 1

Sorry for answering my own question but I went with Martin Frost’s suggestion in the comments. It was much easier than the other suggestions.

This was my implementation:

#!/bin/bash

# requirements for script script

COUNTER=$((7+1))
DATE=$(date -d "-${COUNTER} days" +%s)
DATE=$(date -d -@${DATE} "+%Y-%m-%d")

sed -i "1,/$DATE/d" test.txt

Thanks for all the help!