awk system date with -d option

Shell Programming and Scripting
1

Hi

I get problems when using the following command :

cat logs | awk -F";" '{ system("date -d "1970-01-01 UTC+0100 $1 seconds""); }'
date: date invalide `1968641199401200'
date: date invalide `1968641199381709'

this is what i have in my log file :

cat logs
1199401200;a
1199381709;b

I don't know where this 196864 comes form

Any idea ?

Thx

2 
while IFS=";" read a b
do
    date -d "1970-01-01 UTC+0100 $a seconds"
done  < file
3

thx for this idea but it does not work for me :frowning:
I made something that gives the same output your code should give but this is not what i am searching for :

for mytimestamp in $(cat logs| awk -F";" '{ print $1; }'); do echo $(date -d �1970-01-01 UTC+0100 $mytimestamp seconds�); done

What i want is to replace all the timestamps (first field) in a log file with the date in UTC+0100 format but i want to keep the other fields (file scheme)...

4

If you have GAWK you can use the strftime function or you can fit the script of gostdog74 like:

while IFS=";" read a b
do
    echo $(date -d "1970-01-01 UTC+0100 $a seconds")";"$b
done  < file

Regards