I'm not really sure how to phrase this question, but here goes:
I have a command which returns the date of last connection to an app in the format of 'adDateLastConnected: Thursday, August 23, 2018 at 12:00:07 PM'. I managed to strip out the first part, and only display the date, but I'd like to view this in a more standard date format (like 08-23-2018 12:00:07 PM).
Is this a trivial thing, or is it far more complicated than I suspect? I'm not sure of how deep my ignorance goes :-)
GNU date is pretty good about interpreting human style date formats:
#!/bin/bash
if [[ $(date --version) != *GNU* ]]
then
echo "Sorry, this script only works with GNU date." >&2
exit 1
fi
# Make locale independent
export LC_ALL=C
# Strip the " at " part
x="${1// at / }"
echo "ISO time: $(date --iso-8601=seconds -d "$x")"
echo "Unix time: $(date -d "$x" +%s)"
Example:
$ ./myscript "Thursday, August 23, 2018 at 12:00:07 PM"
ISO time: 2018-08-23T12:00:07-07:00
Unix time: 1535050807
Its not difficult. You can format with sed or gawk (awk), any one of theses will do the job for you.
An example with AWK:
awk '
BEGIN { FS = OFS = "," }
{ split($3, date, /\//)
$3 = date[3] "-" date[2] "-" date[1]
print $0
}
' infile
Thursday, August 23, 2018 at 12:00:07 PM as the input. You're trying to split on /, but the input doesn't contain any slashes.