问题
When I cat the file an example of the output format is:
ok: servername Mon May 23 00:00:00 EDT 2018
ok: servername Thu Jul 16 00:00:00 EDT 2019
I would like the format to be something like
ok: servername 05/23/2018
ok: servername 07/16/2019
I need to use the Linux bash shell to do it. If any one could help me I be very grateful.
回答1:
When performance matters. Put this in script.awk:
BEGIN{
m["Jan"]="01"; m["Feb"]="02"; m["Mar"]="03"; m["Apr"]="04";
m["May"]="05"; m["Jun"]="06"; m["Jul"]="07" # to be completed
}
{
print $1, $2, m[$4] "/" $5 "/" $8
}
Usage: awk -f script.awk logfile
Output:
ok: servername 05/23/2018 ok: servername 07/16/2019
回答2:
With GNU date, you can specify an input file containing all your date strings; combined with cut and paste:
paste -d ' ' \
<(cut -d ' ' -f-2 infile) \
<(date -f <(cut -d ' ' -f3- infile) '+%m/%d/%Y')
Output:
ok: servername 05/23/2018
ok: servername 07/16/2019
This uses process substitution to build a temporary input file for date -f, and for paste to build a new output file.
回答3:
According man date, the command is able to
display time described by STRING, not 'now'
via --date or -d. So if you store your values in a variable
DATE="Thu Jul 16 00:00:00 EDT 2019"
you could use something like
date --date="${DATE}" +%m/%d/%Y
to reformat the timestamp.
For more information you may have a look into Convert date formats in bash.
来源:https://stackoverflow.com/questions/61015078/formatting-date-strings-in-a-file-with-linux-bash-shell