how to get day of the year in shell?
From the coreutils date manual:
%j day of year (001..366)
How do I get the first day of the given year's day of the week in Bash Shell?
Please see the below code which will give you 1 first day of the year passed:
# !/usr/bin/bash
year=2020
firstday=`date -d "01 Jan $year" +'%A,%d'`
echo "First day of the year '$year' is : '$firstday'"
Output:
First day of the year '2020' is : 'Wednesday,01'
If you want to just get the day , then remove the %d option.
firstday=`date -d "01 Jan $year" +'%A`
Output :First day of the year '2020' is : 'Wednesday'
Shell script: How to find what day of the year is today?
Try the %j
format specifier:
$ date +%j
016
The standard says this about %j
:
%j
Day of the year as a decimal number [001,366].
Get the date (a day before current time) in Bash
Sorry not mentioning I on Solaris system.
As such, the -date switch is not available on Solaris bash.
I find out I can get the previous date with little trick on timezone.
DATE=`TZ=MYT+16 date +%Y-%m-%d_%r`
echo $DATE
Get Month & Day from Date
In sh
or bash
:
D="2013/01/17"
DAY=$(date -d "$D" '+%d')
MONTH=$(date -d "$D" '+%m')
YEAR=$(date -d "$D" '+%Y')
echo "Day: $DAY"
echo "Month: $MONTH"
echo "Year: $YEAR"
shell script to get year, date and month from YYYY-MM-DD format
except for processing the string as text(with grep/sed/awk/cut/...), you could do with with date command:
kent$ date -d '2013-09-06' +%Y
2013
kent$ date -d '2013-09-06' +%m
09
kent$ date -d '2013-09-06' +%d
06
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