date format shows correct when i execute just date but when I store in a variable, loosing a space in date if it has single digit(need that extra space to grep /var/log/messages). please suggest to get the exact format as it is. thanks!
$date -d '-1 day' '+%b %e'
Aug 1
$echo $(date -d '-1 day' '+%b %e')
Aug 1
$var=`date -d '-1 day' '+%b %e'`
$echo $var
Aug 1
Use double-quotes like this:
$ echo $(date -d '+1 day' '+%b %e')
Aug 2
$ echo "$(date -d '+1 day' '+%b %e')"
Aug 2
Or:
$ var="$(date -d '+1 day' '+%b %e')"
$ echo $var
Aug 2
$ echo "$var"
Aug 2
Without double-quotes, the shell, among other things, applies word splitting to the output and that collapses multiple spaces to one blank.
Related
I want to remove the first 4 words from my string variable "DATES".
Does someone have a simple solution for this?
Here my example:
DATES="31 May 2021 10:22:01 30 May 2021 10:23:01 29 May 2021 10:24:01"
WC=$(echo $DATES | wc -w)
DATE_COUNT=$(( $WC / 4 - 1 ))
for i in {0..$DATE_COUNT}
do
YEAR=$(echo $DATES | awk '{print $3}')
MONTH=$(echo $DATES | awk '{print $2}')
MONTH=$( date --date="$(printf "01 %s" $MONTH)" +"%m")
DAY=$(echo $DATES | awk '{print $1}')
TIME=$(echo $DATES | awk '{print $4}' | sed 's/://g')
DATE_ARRAY[$i]="$YEAR$MONTH$DAY$TIME"
#Remove first 4 words from string
done
Use cut.
DATES="31 May 2021 10:22:01 30 May 2021 10:23:01 29 May 2021 10:24:01"
echo $DATES | cut -d' ' -f 5-
Output:
30 May 2021 10:23:01 29 May 2021 10:24:01
You can even use it for a cleaner solution than awk, like this:
YEAR=$(echo $DATES | cut -d' ' -f 3)
General version to remove n first words
remove_n_first_words(){
echo $2 | cut -d' ' -f $(($1+1))-
}
remove_n_first_words 4 "$DATES"
Using bash regex operator =~:
$ [[ $DATES =~ ^(([^ ]+ +){4})(.*) ]] && echo ${BASH_REMATCH[3]}
30 May 2021 10:23:01 29 May 2021 10:24:01
Maybe use read ?
DATES="31 May 2021 10:22:01 30 May 2021 10:23:01 29 May 2021 10:24:01"
read -ra dates <<< "$DATES"; echo "${dates[#]:4}"
Or just store the data in an array directly.
DATES=(31 May 2021 10:22:01 30 May 2021 10:23:01 29 May 2021 10:24:01)
echo "${DATES[#]:4}"
To get the total words/elements like with wc -c
echo "${#DATES[*]}"
how can I print output in double decimal.
below command will print hour in GMT format but i want output as 06 and for double digit hour it should be 10,11,12.
date -u --date="today" +"%I" | awk -F' ' '{print $1-1}'
6
You may use printf "%02d" in awk to achieve it,
$ date -u --date="today" +"%I" | awk -F' ' '{printf "%02d\n",$1-1}'
06
Perhaps you want:
date -u --date="- 1 hour" +"%I"`
If the time adjustment is part of your date string, the format will not be munged.
Alternately, if what you really want is a way to zero-pad a number in bash or awk, you have a variety of alternatives:
date -u --date="- 1 hour" +"%I" | awk '{printf "%02d\n",$1-1}'
Or in bash alone:
read hour < <( date -u --date="- 1 hour" +"%I" )
printf '%02d\n' "$hour"
Get the idea? Output format happens when you print your output, and printf in whatever language formats your output.
awk is superfluous here. You can use a relative time format with date:
date -u --date="-1 hour" +"%I"
06
Hi I need to format the date command output using awk and add zero before the days starting 1 to 9 .
today=`date | awk {'print $1 " " $2 " " $3'}`
So in the above the output is
Wed Mar 2
I need to add 0 to 2 to get to days of the month 1 through 9
Wed Mar 02
Ho can I add this command using the awk command
for i in 0{1..9}; do echo $i; done
So I need to add 0/zero to $3 when it's between 1 or 9
I tried doing it this way , but something is not working I get error
a3=`date|awk '{
if ($3 <=9)
print $1" "$2" " "0"$3;
else
print $1" "$2" " $3;
}'`
echo $a3
Can you please assist?
Regards
If I were you I'd just specify a format directly:
$ date '+%a %b %d'
Wed Mar 02
date takes a format string preceded by a + as its final argument.
if you must do in awk you can use printf for formatted printing
$ echo 1 2 10 20 | awk -v RS=" " '{printf "%s\t-> %02d\n",$1,$1}'
1 -> 01
2 -> 02
10 -> 10
20 -> 20
Wed Jan 21 20:44:20 EST 2015
Wed Jan 21 19:04:20 EST 2015
I have two dates about, need to get the difference in minutes. Please help
"c=date -d #$(( $(date -d "$b" +%s) - $(date -d "$a" +%s) )) -u +'%H:%M'" -> This command is giving in HH:MM but i want in MM
Thank you
this gives you the result: 100 minutes:
echo $((($(date -d "$a" +%s) - $(date -d "$b" +%s))/60 ))
Note that, it will always give an int value, if you need the precision less than 1 min, like 100.25 you may want to use bc or awk to do the calculation instead of $(( .. ))
I have a date in YYYY.MM.DD HH:SS format (e.g. 2014.02.14 13:30). I'd like to convert it in seconds since epoch using the date command.
The command
date -d"2014.02.14 13:30" +%s
won't work, because of the dots separation.
Any Ideas?
Why don't you make the date format acceptable? Just replace dots with dashes:
$ date --date="`echo '2014.02.14 13:30' | sed 's/\./-/g'`" +%s
1392370200
Here I first change the format:
$ echo '2014.02.14 13:30' | sed 's/\./-/g'
2014-02-14 13:30
and then use the result as a parameter for date.
Note that the result depends on your timezone.
You can use:
s='2014.02.14 13:30'
date -d "${s//./}"
Fri Feb 14 13:30:00 EST 2014
To get EPOCH value:
date -d "${s//./}" '+%s'
1392402600
using awk :
s=`echo "2014.02.14 13:30" | awk '{gsub(/\./,"-",$0);print $0}'`
echo -d "$s"
date -d "$s" +%s
output:
Fri Feb 14 13:30:00 IST 2014
1392364800
Perl: does not require you to munge the string
d="2014.02.14 13:30"
epoch_time=$(perl -MTime::Piece -E 'say Time::Piece->strptime(shift, "%Y.%m.%d %H:%M")->epoch' "$d")
echo $epoch_time
1392384600
Timezone: Canada/Eastern
I Finally solved it using
awk 'BEGIN{FS=","}{ gsub(/./," ",$1);gsub(/:/," ",$2); var=sprintf("%s %s 00",$1,$2); print mktime(var), $3,$4,$5,$6,$7 }' myfile | less
so myfile:
2014.09.24,15:15,1.27921,1.27933,1.279,1.27924,234
became
1411582500 1.27921 1.27933 1.279 1.27924 234
:)