Select directories with date in their names from a date range - bash

Question

I'm creating a list of directories with the requested date range in their name.
Directories are all labeled other_2019-07-18T00-00-00. The T is messing me up!
Copied this loop from somewhere.
#!/bin/bash
curdate=$(date +%Y-%m-%d%H-%M-%S)
#
for o in other_*; do
tmp=${o##other_}
tmp=$(echo "$tmp" | sed 's/T//') # clean up prefixes
fdate=$(date -d "${tmp}")
(( curdate < fdate )) && echo "$o"
done
I expect the final echo to include the path of all dir that match.

Answer 1

Unlike AWK, Bash comparison operator < works only on numerical values.
Please try instead:
#!/bin/bash
curdate=$(date +%Y%m%d%H%M%S)
for o in other_*; do
tmp=${o##other_}
fdate=$(echo "$tmp" | sed 's/[-T]//g') # numeralization
(( curdate < fdate )) && echo "$o"
done
As an alternative, you can compare epoch times:
#!/bin/bash
curdate=$(date +%s)
for o in other_*; do
tmp=${o##other_}
tmp=$(echo "$tmp" | sed 's/T/ /' | sed 's/\([0-9][0-9]\)-\([0-9][0-9]\)-\([0-9][0-9]\)$/\1:\2:\3/')
fdate=$(date -d "$tmp" +%s)
(( curdate < fdate )) && echo "$o"
done
Hope this helps.

Answer 2

Instead of dropping T...
date -d 2019-03-23T00-06-28
date: invalid date '2019-03-23T00-06-28'
ok, but:
date -d 2019-03-23T00:06:28
Sat Mar 23 00:06:28 UTC 2019
So we have to replace last two dashes by ::
As your question is tagged bash:
file="somepath/other_2019-07-18T00-00-00.extension"
time=${file#*other_} # suppress from left until 'other_'
time=${time%.*} # suppress extension
time=${time//-/:} # replace all dash by a `:`
time=${time/:/-} # replace 1nd `:` by a dash
time=${time/:/-} # replace 1nd `:` by a dash (again)
date -d $time
Thu Jul 18 00:00:00 UTC 2019
This could by written:
printf -v now "%(%s)T" -1 # bashism for current date to variable $now
for file in somepath/other_*.ext ;do
time=${file#*other_} time=${time%.*} time=${time//-/:}
time=${time/:/-} time=${time/:/-}
read fdate < <(date +%s -d $time)
((fdate > now)) && { echo $file: $((fdate - now)) ; }
done
Reducing forks (to date) improve quickness:
for matching your sample, you could replace for file in somepath/other_*.ext ;do by for file in other_*; do. This must work quite same.
fifo=/tmp/fifo-date-$$
mkfifo $fifo
exec 5> >(exec stdbuf -o0 date -f - +%s >$fifo 2>&1)
echo now 1>&5
exec 6< $fifo
read -t 1 -u 6 now
rm $fifo
for file in otherdates/*.ext ; do
time=${file#*other_} time=${time%.*} time=${time//-/:}
time=${time/:/-} time=${time/:/-}
echo $time 1>&5 && read -t 1 -u 6 fdate
((fdate > now)) && {
echo $file: $((fdate - now))
}
done
exec 6>&-
exec 5>&-
In this, I run date +%s in background, with -f argument, date will interpret each incoming line, then answer UNIX_TIME. So $now is first read from date process, by:
echo now >&5 && # now, the string
read -u 6 now # read will populate `$now` variable
Nota, once fifo opened by both input and output, they could be deleted. It will remain for process until process close them.

Answer 3

there is no space between day and hour, which causes date not to be able to read the date. Try:
sed 's/T/ /'