Reputation: 2410
What is the best way to extract the time from this filename and converting it in seconds since 00:00 of current day (no timestamp) and using pure bash?
/a/very/long/path/to/my/files/2019-07-02_06_12_55_SOME_FOO_BAR.log
Output excpected :
extractTimeinSecond() {
// code
}
filetime=$(extractTimeinSecond $file)
echo $filetime # 22375
I'd have done something like this, but this isn't very sexy as I'm sure there is a better way to get it.
extractTimeinSecond() {
file=$1 # /a/very/long/path/to/my/files/2019-07-02_06_12_55_SOME_FOO_BAR.log
shortfile=$(basename $file) # 2019-07-02_06_12_55_SOME_FOO_BAR.log
time=$(echo $shortfile | cut -d '_' -f 2,3,4) # 06_12_55
h=$(( $(echo $time | cut -d '_' -f1) * 60 * 60 )) # 06 * 60 * 60 = 21600
m=$(( $(echo $time | cut -d '_' -f2) * 60 )) # 12 * 60 = 720
s=$(echo $time | cut -d '_' -f3) # 55
echo $(( $h + $m + $s )) # 21600 + 720 + 55 = 22375
}
Upvotes: 1
Views: 39
Reputation: 15438
To cleanly collect time components:
Assuming log=/a/very/long/path/to/my/files/2019-07-02_06_12_55_SOME_FOO_BAR.log
,
$: f="${log##*/}" # strip path
$: IFS=_ read date h m s trash <<< "$f" # parse time elements
$: echo "$date $h:$m:$s" # check values
2019-07-02 06:12:55
$: secstamp=$( date +%s -d "$date $h:$m:$s" )
echo $secstamp
1562065975
I assume you want the seconds into the day of the log's timestamp, even if checking on a different day?
$: midnight=$( date +%s -d "$date 00:00:00" )
$: echo $midnight
1562043600
Subtract for the seconds.
$: echo $(( secstamp - midnight ))
22375
An awk
solution will be much faster, but this is a great trick to add to the toolbox.
Upvotes: 2
Reputation: 67929
Using awk:
extractTimeinSecond() {
basename "$1" | awk -F'[-_]' '{print ($4*60+$5)*60+$6}'
}
Using bash only:
extractTimeinSecond() {
a=${1##*/}
b=${a#*_}
hour=${b/_*}
c=${b#*_}
minute=${c/_*}
d=${c#*_}
second=${d/_*}
echo $(((hour*60+minute)*60+second));
}
Upvotes: 2