1633036680022 , This is epoch result i got from elasticsearch.
if i tried to convert this epcho to human-readable date,
So i used epochconverter
And i used bash command to convert this in my terminal,
$ date -d #1633036680022
Tuesday 15 November 53718 05:30:22 PM IST
This output from terminal say the Year 53718, because the epoch '1633036680022' is in milliseconds.
All i want is ,epoch in seconds.
You can divide by 1000 and convert to timestamp.
date -d #"$(echo "1633036680022/1000" | bc)"
Strip milliseconds with bash (output only first 10 digits):
x="1633036680022"
date -d "#${x:0:10}"
Related
I found command like this $(date +%s%3N) to convert current time to milliseconds, but example I have time like this 2018-25-08 12:00 PM. How to convert it to millisecond in bash?
Finally, I found this command to convert static date to milliseconds
date -d 'date -d '08/23/2018 23:59:59' +"%s%3N"
I'm having trouble with checking time since EPOCH. (and late subtract it from another one).
I get the date like this:
var=$(date)
echo $var
wto, 1 mar 2016, 16:00:14 CET
and later I'm trying to turn it into seconds since epoch:
date -d "$var" +"%s"
date: invalid date ‘wto, 1 mar 2016, 16:00:14 CET’
I'm giving this just as an example. Actually I will be reading the date from file, written in default locale format (I'm operating on couple different machines).
if you type date -h there is the reason why you got this error.
the -d option MUST be declared only with TIME and not with complete DATE format
-d,--date TIME Display TIME, not 'now'
so
date -d "23:59:59"
then:
Tue Mar 1 23:59:59 2016
if you need get only the seconds from a date you have to execute this:
date +"%S"
if you use the -d the output will be deplyed in msec
I have date format like 2016-02-26 14:12:36., I have to convert it into unix epoch time. I have the option to convert current time to epoch time using command : date +%s, however how to convert a specific date to epoch unix time
On the command line you convert the date provided to the UNIX epoch time like this:
date -j -f "%Y-%m-%d %H:%M:%S" "2016-02-26 14:12:36" "+%s"
This is on os x. The manual for date have some examples and strftime have all the formatting needed.
You can get seconds since the epoch from date like this ,
$ SECFROMEPOCH=`date --date="2016-02-26 14:12:36" +%s`
And then you can check the value that date gave back to you like this,
$ echo $SECFROMEPOCH | awk '{print strftime("%c",$1)}'
Need a help in unix shell script in calculating date.
I will be getting date value (eg: 20150908) as parameter, now inside the script i need to calculate 7 days ago date (20150908 -7).
something like below:
date=20150908
lastweek_date=20150908 - 7 ---> this should output as 20150901
Could someone help me on this.
Thanks
With GNU date, we can subtract one week:
$ date -d "20150908 - 1 week" '+%Y%m%d'
20150901
Alternatively, we could subtract 7 days:
$ date -d "20150908 - 7 days" '+%Y%m%d'
20150901
And, to show that this works over month boundaries:
$ date -d "20150901 - 1 week" '+%Y%m%d'
20150825
This solution is not OSX/BSD compatible.
A week is 604800 seconds long so to get the number of seconds since the epoch in a portable and POSIX compliant fashion and use it to compute the date 1 week ago do as follows:
PRESENT=$( date +%s )
WEEKAGO=$(( PRESENT - 604800 ))
printf "%s\n" "$WEEKAGO"
At work all the days config files are generated fresh and appended with a
session number. The company went public on Feb 16, and the 86400 is seconds
in one day. The session number is generated by subtracting the company start
day from seconds_since_last_day and adding a few zero's
That is the key to interacting with the days config files. I get this - However I do not
understand the
date -ud "$distance days ago 00:00:00".
Is it the number of seconds since 1970?
if $session; then
# return the session of the last day
seconds_since_day_one=`date -ud "Feb 16 2002" +"%s"`
seconds_since_last_day=`date -ud "$distance days ago 00:00:00" +"%s"`
days_between=`printf "%010d" $(( (seconds_since_last_day - seconds_since_day_one) / 86400 ))`
# Truncate on the left to 9 bytes to leave room
# to append the engine suffix for your environment
echo $days_between | awk '{l=length($1); print substr( $1, (l-8), l )}'
date -ud "$distance days ago 00:00:00" in itself just prints the date a certain amount of days ago in a quite readable format, but when you add the FORMAT string to control the output +"%s" does indeed mean the number in so called Unix Time (number of seconds since 1970-01-01 00:00:00 UTC).
If the variable $distance is set to a number it shows the date that number of days ago, if its set to 0 it means today, 1 it means yesterday, 2 the day before yesterday and so on. To better understand these formats and relative keywords there are rather good documentations in (amongst other places) the GNU coreutils package.
Check these URLs:
http://www.gnu.org/software/coreutils/manual/html_node/Relative-items-in-date-strings.html#Relative-items-in-date-strings
http://www.gnu.org/software/coreutils/manual/html_node/Date-input-formats.html
http://www.gnu.org/software/coreutils/manual/html_node/date-invocation.html#date-invocation
Wikipedia explanation of Unix Time:
http://en.wikipedia.org/wiki/Unix_time
The option -d to date provides a generic string to obtain the date.
So, for example, date -d yesterday will print yesterday's date, and date -d 'yesterday 12:00 AM' will print yesterday's date with the time set to 12:00 AM.
So, date -d 6 days ago 00:00:00 will print the date from 6 days ago, with the time set to 00:00:00. I hope it answers your question.
The format +"%s" tells date to print the number of seconds from 1970, instead the date.
mktime and strftime in awk can be used to get the date of the time.
http://www.gnu.org/software/gawk/manual/html_node/Time-Functions.html
For instance, strftime("%A",mktime("YYYY MM DD 00 00 00"))
should give you the day.