I am completely new to shell scripting.
I need to change the format of given date to customized format like i have date in a variable with format as MM/DD/YY HH:MM:SS but i want date in a format as MM/DD/YYY HH:MM:SS which is having four digits of the year.
We can change the sys date format but i need the same in a variable.
My code as below
START_DATE="12/20/14 05:59:01"
yr=`echo $START_DATE | cut -d ' ' -f1 | cut -d '/' -f3`
yr_len=`echo $yr | wc -c`
if [ $yr_len -lt 4 ]
then
tmp_yr="20${yr}";
else
1=1;
fi
ln=`echo $tmp_yr|wc -c`
After this i strucked in reframe the same date in wanted format.
Can some one please help me
Regards,
Sai.
Using GNU date:
date -d'02/16/15 09:16:04' "+%m/%d/%Y %T"
produces
02/16/2015 09:16:04
which is what you want. See man date for details about the formatting, or this question for a number of great examples.
One option may be using the date/time functions inside awk. Here is a oneliner:
echo '02/16/15 09:16:04' | sed 's\[/:]\ \g' | awk '{d0=$3+2000FS$1FS$2FS$4FS$5FS$6; d1=mktime(d0);print strftime("%m/%d/%Y %T", d1) }'
output is:
02/16/2015 09:16:04
You can find more strftime formats in https://www.gnu.org/software/gawk/manual/html_node/Time-Functions.html
Related
I have a file that consists of the following...
false|aaa|user|aaa001|2014-12-11|
false|bbb|user|bbb||
false|ccc|user|ccc|2021-10-19|
false|ddd|user|ddd|2018-11-16|
false|eee|user|eee|2020-06-02|
I want to use the date in the 5th column to calculate the number of days from the current date and append it to each line in the file.
The end result would be a file that looks like the following, assuming the current date is 1/13/2022...
false|aaa|user|aaa001|2014-12-11|2590
false|bbb|user|bbb||
false|ccc|user|ccc|2021-10-19|86
false|ddd|user|ddd|2018-11-16|1154
false|eee|user|eee|2020-06-02|590
Some lines in the file will not contain a date value (which is expected). I need a solution for a Bash script on Linux.
I am able to submit a command using echo for a single line and then calculate the number of days from the current date by using cut on the 5th field (see below)...
echo "false|aaa|user|aaa001|2014-12-11" | echo $(( ($(date --date=date +"%Y-%m-%d" +%s) - $(date --date=cut -d'|' -f5 +%s) )/(60*60*24) ))
2590
I don't know how to do this one line at a time, capture the 'number of days' value and then append it to each line in the file.
Here's an approach using
paste to append the outputs
sed to arrange the empty lines and
awk to calculate the desired days.
This works with GNU date. BSD date has to use something like date -jf x +%s.
EDIT: Updated the date to compare with to current day.
% current=$(date +%m/%d/%Y)
% paste -d"\0" file <(cut -d"|" -f5 file |
sed 's/^$/#/' |
xargs -Ix date -d x +%s 2>&1 |
awk -v cur="$(date -d "$current" +%s)" '/invalid/{print 0; next}
{print int((cur-$1)/3600/24)}')
false|aaa|user|aaa001|2014-12-11|2590
false|bbb|user|bbb||0
false|ccc|user|ccc|2021-10-19|86
false|ddd|user|ddd|2018-11-16|1154
false|eee|user|eee|2020-06-02|590
Also date returns date: invalid date ‘#’ in the empty case. If any other implementation behaves differently the awk regex has to be adjusted accordingly.
Data
% cat file
false|aaa|user|aaa001|2014-12-11|
false|bbb|user|bbb||
false|ccc|user|ccc|2021-10-19|
false|ddd|user|ddd|2018-11-16|
false|eee|user|eee|2020-06-02|
i have time logs in timestamp (epoch unix time) format :
1515365117236
1515365123162
1515365139963
i would like to convert it to a regular date like
2017-01-07 23:48:01
2017-01-07 23:48:02
2017-01-07 23:48:03
any ideas what approach would be the fastest?
cat ff1.csv | while read line ; do echo $line\;$(date -d +"%Y-%m-%d %H:%M:%S") ; done > somefile.csv
this takes awful lot of time and just appends the current time
Another approach that must be much faster , using printf of bash version >4.2 :
$ printf '%(datefmt)T\n' epoch
For datefmt you need a string accepted by strftime(3) - see man 3 strftime
Testing:
$ cat file10
1515365117236
1515365123162
1515365139963
$ printf '%(%F %H:%M:%S)T\n' $(cat file10)
49990-01-04 04:47:16
49990-01-04 06:26:02
49990-01-04 11:06:03
In this case , printf format string is:
%F Equivalent to %Y-%m-%d (the ISO 8601 date format). (C99)
%H The hour as a decimal number using a 24-hour clock (range 00 to 23).(Calculated from tm_hour.)
%M The minute as a decimal number (range 00 to 59). (Calculated from tm_min.)
%S The second as a decimal number (range 00 to 60). (The range is up to 60 to allow for occasional leap seconds.- Calculated from tm_sec.)
Update to remove milliseconds:
$ printf '%(%F %T)T\n' $(printf '%s/1000\n' $(<file10) |bc)
2018-01-08 00:45:17
2018-01-08 00:45:23
2018-01-08 00:45:39
The way to transform epoch to date is date -d #epochtime +format
An alternative way is to use date --file switch to read dates from a file directly.
$ cat file10
1515365117236
1515365123162
1515365139963
In order date to understand that these lines are epoch time you need to add # in the beginning of each line.
This can be done like bellow:
$ sed -i 's/^/#/g' file10 #caution - this will make changes in your file
$ date --file file10 +"%Y-%m-%d %H:%M:%S"
Alternativelly, you can do it on the fly without affecting the original file:
$ sed 's/^/#/g' file10 |date --file - +"%Y-%m-%d %H:%M:%S"
PS: in this case --file reads from - == stdin == pipe
In both cases, the result is
49990-01-04 04:47:16
49990-01-04 06:26:02
49990-01-04 11:06:03
PS: by the way, the timestamps you provide seems invalid, since it seems to refer at year 49990
Your input data aren't epoch unix time, it has miliseconds. If you wish to use any method on bash first you must convert to timestamp:
cat ff1.csv | while read LINE; do echo "#$(expr $LINE \/ 1000)" | date +"%Y-%m-%d %H:%M:%S" --file - ; done
First divide by 1000 to delete miliseconds parts, the rest is the same that explain George Vasiliou
I have a number of files in the form foo_[SECONDS.MILLISECONDS]_bar.tar.gz and for each file I would like to be to get a datetime value (YYYYMMDDHHMMSS) for each file.
So far I have
ls -1 /filestore/*.tar.gz | cut -d _ -f 2 | date -f -
But this errors along the lines of
date: invalid date '1467535262.712041352'
How should a bash pipeline of epoch values be converted into a datetime string?
MWE
mkdir tmpBLAH
touch tmpBLAH/foo_1467483118.640314986_bar.tar.gz
touch tmpBLAH/foo_1467535262.712041352_bar.tar.gz
ls -1 tmpBLAH/*.tar.gz | cut -d _ -f 2 | date -f -
To convert epoch time to datetimem, please try the following command:
date -d #1346338800 +'%Y%m%d%H%M%S'
1346338800 is a epoch time.
About your case, for comand line as following:
echo 1467535262.712041352 | cut -d '.' -f 1 | xargs -I{} date -d #{} +'%Y%m%d%H%M%S'
you will get:
20160703174102
Something like this?
for f in /filestore/*.tar.gz; do
epoch=${f#*_}
date -d #${epoch%%.*} +%Y%m%d%H%M%S
done
The syntax of the date command differs between platforms; I have assumed GNU date, as commonly found on Linux. (You could probably use date -f if you add the # before each timestamp, but I am not in a place where I can test this right now.) Running a loop makes some things easier, such as printing both the input file name and the converted date, while otherwise a pipeline would be the most efficient and idiomatic solution.
As an aside, basically never use ls in scripts.
First, the -1 option to ls is useless, because ls prints its output one file per line by default, it's just that when the output is a terminal (not a pipe), it pretty-prints in columns. You can check that fact by just running ls | cat.
Then, date converts epoch timestamps safely only if prefixed with an #.
% date -d 0
Sun Jul 3 00:00:00 CEST 2016
% LANG=C date -d #0
Thu Jan 1 01:00:00 CET 1970
% date -d 12345
date: invalid date '12345'
% date -d #12345
Thu Jan 1 04:25:45 CET 1970
Which gives:
printf "%s\n" tmpBLAH/foo_*_bar.tar.gz | sed 's/.*foo_/#/; s/_bar.*//' | date -f -
You can do:
for i in foo_*_bar.tar.gz; do date -d "#$(cut -d_ -f2 <<<"$i")" '+%Y%m%d%H%M%S'; done
The epoch time is provided with the -d #<time> and the desired format is '+%Y%m%d%H%M%S'.
Example:
% for i in foo_*_bar.tar.gz; do date -d "#$(cut -d_ -f2 <<<"$i")" '+%Y%m%d%H%M%S'; done
20160703001158
20160703144102
I have a string from conf file (lets call for example date1):
#!/bin/bash
# it is example
date1="201605250925"
datenow="$(date +%Y%m%d%H%M -d "+1hour")"
date2=$(date +%Y%m%d%H%M -d "$date1 + 1hour")
# NOT WORK?
echo "$date1 --> $date2"
# WORK!
echo "$date1 --> $datenow"
I need to add 1 hour. But getting error like this:
date: invalid date `201605250925 + 1hour'
But its work for datenow.
How can I user addhour for custom date format from string?
You need a format that meets the command date expectations, something like:
2016-05-25 09:25
The space denote the start of time and the time format is HH:MM.
That comes from then international ISO 8601, but using an space instead of a T.
If the format is fixed, we can use bash internal capacities (no external command except date used) to change it like this:
#!/bin/bash
d1="201605250925"
dc="${d1:0:8} ${d1:8:2}:${d1:10:2}+0"
d2=$(date +'%Y%m%d %H:%M' -ud "$dc + 1 hour" )
echo "$d2"
Or POSIXly (dash) with no call needed to sed, awk or cut (faster):
#!/bin/dash
d1="201605250925"
dt=${d1##????????}
dc="${d1%%"$dt"} ${dt%%??}:${dt##??}+0"
d2=$(date -ud "$dc + 1 hour" +'%Y%m%d %H:%M')
echo "$d2"
20160525 10:25
The inclusion of a +0 after the time in dc: 20160525 09:25+0
will ensure that date will interpret the time as with offset 0 (UTC).
The use of the option -u to date will ensure that the value read in UTC also change in UTC, avoiding any Daylight correction or local time change.
If you want to keep the same format as your input string you could use cut or sed to split it:
cut
d1="201605250925"
ds=$(echo $d1 | cut --output-delimiter=" " -c 1-8,9-12); \
d2=$(date '+%Y%m%d%H%M' -ud "$ds + 1hour")
sed
d1="201605250925"
ds=$(echo $d1 | sed 's/\(.*\)\(....\)$/\1 \2/g'); \
d2=$(date '+%Y%m%d%H%M' -ud "$ds + 1hour")
This way the date utility can make sense of the date and time.
Result:
$ echo $d2
201605251025
Is there a simple way to convert Common Log Format (NCSA) to timestamp ?
I found only C++ decision and some perl decisions. And wonder why i can't use standart unix function like date.
> date +"%d/%m/%Y:%H:%M:%S" -d "17/Oct/2013:16:52:28" +"%s"
"17/Oct/2013:16:52:28" -> 1382014348
For example:
Date in iso8601 to timestamp
> date -d "2013-10-17T18:07:39+04:00" +"%S"
1382018859
You can try this ( if you surely want to use date) ,
date -d "$(sed -e 's#/#-#g; s#:# #' <<< '17/Oct/2013:16:52:28')" '+%s'