I have a bash script and I need it to fulfill some conditions if it is 1st day of month.
I have written this code
ifStart=`date '+%d'`
if [$ifStart == 01]
then
test=`/bin/date --date='1 day ago' +'%Y-%m'`
echo $test
fi
I expect it to show 2013-03 today, but I get an errormessage:
Line 2 command not found.
test=`/bin/date --date='1 day ago' +'%Y-%m'`
this part works well without if.
Any suggestions?
The command that's not being found is actually due to your if statement. You need spaces:
if [ $ifStart == 01 ]
Otherwise [$ifStart will be interpreted as a command.
Related
We have a system that would have a cron job that deletes files up to two months ago. I'm trying to write a script to automate this, but I'm fairly new to bash scripting and was wondering if anyone would be able to help. Our files are in %m%Y format and I would be moving them to another directory and then deleting that directory. So for instance since we are in August (082020), I want to move all files up to June (062020) starting this year in Jan (012020).
Here is my script so far, I am basically just trying to print 012020-062020, can anyone let me know if I am on the right track?
#!/bin/bash
MONTHYEAR=$(date +%m%Y)
DELUPTO=$(expr $(date +%m%Y) - 20000)
CURRENTYEAR=$(date +%Y)
for (( i=$DELUPTO; i>=01 + $CURRENTYEAR; $(expr $i - 10000) ))
do
echo "$i"
done
You should loop from the format yyyydd, so start with
for (( i=202006; i>=202001; i-- )); do
echo "${i:4:2}${i:0:4}"
done
It is up to you how you want to achieve this:
yearmonth=$(date +%Y%m)
or
MONTHYEAR=$(date +%m%Y)
yearmonth=${MONTHYEAR:2:4}${MONTHYEAR:0:2}
You know the month and year, extract those values and then turn it into a stamp, but you will need to insert a day value so I would make it say the 1st:
Example of converting timestamps:
# date -d "8/1/2020" +"%s"
1596254400
# date -d #1596254400 +"%b %d %Y %H:%M:%S"
Aug 01 2020 00:00:00
Then create a time stamp of now minutes X days:
date +%s -d "60 days ago"
Once you have common values to compare, Then compare them and if less than 60 days delete Pseudo code:
del_date=$(date +%s -d "60 days ago")
for each file in directory:
#get month and day from file name here, then
file_date=$(date -d "${fmonth}/1/${fyear}" +"%s")
if [[ $file_date -lt $del_date ]] ;then
echo "Older than 60 days by name"
fi
done
Note: It would probably be better to delete files by checking their ages in the system using stat command opposed to reading the details of file name.
A few months ago I wrote a some bash to get dates. I needed these dates as facts in an ansible script to later use them to get data from the database. This worked fine until today here is the code:
- name: Set date variables
set_fact:
first_day_last_month: "{{lookup('pipe','date -d \"-1 month -$(($(date +%d)-1)) days\" +%Y-%m-%d')}}"
last_day_last_month: "{{lookup('pipe','date -d \"$(date +%Y-%m-01) -1 day\" +%Y-%m-%d')}}"
first_day_current_month: "{{lookup('pipe','date -d \"-$(($(date +%d)-1)) days\" +%Y-%m-%d')}}"
last_day_current_month: "{{lookup('pipe','date -d \"-$(date +%d) days +1 month\" +%Y-%m-%d')}}"
However when I run this now I get an error:
/bin/sh: 1: arithmetic expression: expecting EOF: "08-1"
I tried debugging it on the bash command line:
seven#monstermachine:~$ echo $(date -d "-$($(date +%d)-1)")
08-1: command not found
Mit Nov 8 00:00:00 CET 2017
but until now i'm not getting it fixed.
Anyone have any idea?
%d is zero padded which gives base eight or octal number and 08 is not a valid octal number which thows an error.
What you need is something like :
echo $(date -d "$(($(date +%e)-1))")
Wed Nov 8 07:00:00 IST 2017
Note %e will cause date to be space padded which is equivalent to %_d.
To perform a mathematical expression use $(( expression )) format. I didn't understand the actual logic behind your code, so make sure your code fits the logic.
To get first_day_current_month in your script, you could use:
date +%Y-%m-01
similar for the others, for example to get last_day_last_month:
date -d `date +%Y-%m-01`"-1 day" +%Y-%m-%d
Thanks for the help I will make use of both of your answers to come up with a better working command.
I want to write a bash script that will run on a given but process data with next days date, My current approach is to get the unix time stamp and add a days worth of seconds to it, but I cant get it working, and haven't yet found what I'm looking for online.
Here's what I've tried, I feel like the problem is that its a string an not a number, but I dont know enough about bash to be sure, is this correct? and how do I resolve this?
today="$(date +'%s')"
tomorrow="$($today + 86400)"
echo "$today"
echo "$tomorrow"
If you have gnu-date then to get next day you can just do:
date -d '+1 day'
Some of the answers for this question depend on having GNU date installed. If you don't have GNU date, you can use the built-in date command with the -v option.
The command
$ date -v+1d
returns tomorrow's date.
You can use it with all the standard date formatting options, so
$ date -v+1d +%Y-%m-%d
returns tomorrow's date in the format YYYY-MM-DD.
$(...) is command substitution. You're trying to run $today + 86400 as a command.
$((...)) is arithmetic expansion. This is what you want to use.
tomorrow=$(( today + 86400 ))
Also see http://mywiki.wooledge.org/ArithmeticExpression for more on doing arithmetics in the shell.
I hope that this will solve your problem here.
date --date 'next day'
Set your timezone, then run date.
E.g.
TZ=UTC-24 date
Alternatively, I'd use perl:
perl -e 'print localtime(time+84600)."\n"'
echo $(date --date="next day" +%Y%m%d)
This will output
20170623
I like
input_date=$(date '+%Y-%m-%d')
tomorrow_date=$(date '+%Y-%m-%d' -d "$input_date + 1 day")
echo "$tomorrow_date"
It returns the date of the day after $input_date, in format YYYY-MM-DD
You can try below
#!/bin/bash
today=`date`
tomorrow=`date --date="next day"`
echo "$today"
echo "$tomorrow"
SO, I have a whole script that runs every hour with a date as input. Normally, it takes the current time, but now I need it to run for an interval of time in the past, every hour as well.
What I've done so far is:
DEFINING THE OLD DATE
8 start_date=20131218
9 num_hours=5
10 for i in `seq 1 $num_hours`
11 do
12 date=`date -d "${start_date}+${i} hours"`
13 echo $date # Use this however you want!
14
.
.
.
25 done
The starting date is Dec 18, 2013 and then in each iteration it should give me one more hours from the starting time. This part I found it in another article here and it works. The problem comes when I do
echo $(date)
it prints the current time instead of the time that I previously defined. Of course any other variable that I define from the date has the values from the current time. For instance,
18 datestamp=$(date +%F)
19 hourstamp=$(date +%H)
I'm new in shell programming and I have no idea what to do. Any help?
Thanks in advance.
What you want is this:
18 datestamp=$(date -d "${start_date}+${i} hours" +%F)
19 hourstamp=$(date -d "${start_date}+${i} hours" +%H)
As #BMW said, try to avoid use the date as a variable name to avoid ambiguity.
$(date) will run the command date and export the result.
so when echo it, it will return current date/time.
Second, date is the unix command, avoid to use it as a variable. so this will fix your issue"
DATE=`date -d "${start_date}+${i} hours"`
echo $DATE
I'm trying to calculate the date for a dynamic number of days ago in a bash script.
This is what I've done -
#!/bin/bash
STAMP=`date --date='$1 day ago' +%y%m%d`
but when running myscript 2, it says -
date: invalid date `$1 day ago'
How can I use my argument value in this formula?
It works if ' is replaced with " into this command on the script -
STAMP=`date --date="$1 day ago" +%y%m%d`
The clue was the two different character ` and ' used in the error response -
date: invalid date `$1 day ago'
An expert in bash scripting (not me) can probably explain why this has happen.
It's because variable substitution wouldn't happen in single quotes, i.e. '$1' wouldn't expand but "$1" would.
As such, saying
STAMP=`date --date="$1 day ago" +%y%m%d`
or
STAMP=$(date --date="$1 day ago" +%y%m%d)
would work.