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Add month to a variable date in shell script
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I am getting the current year and month via bash. However when I try to find the next year and month I am getting errors.
prints 2019 and 05
mydate=$(date +'%Y');
mymonth=$(date +'%m');
Both fail
nexty=$(date +'%Y') -d next year;
nextm=$(date +'%m') -d 'next month';
The closing parenthesis are misplaced. Try the following:
nexty=$(date +'%Y' -d 'next year');
nextm=$(date +'%m' -d 'next month');
I'm not sure, if you really want "next year and next month" or "year+month of next month". This is a solution for the second one, but "Examine the current date only once" is good for the first one too:
Examine the current date only once:
date=($(date +%Y\ %m -d 'next month'))
echo "year=${date[0]}, month=${date[1]}"
or
date=$(date +%s -d 'next month')
year=$(date -d#$date +%Y)
month=$(date -d#$date +%m)
echo "$year $month"
Otherwise you can run into trouble at day change. Example: 1 command executed in last second of December, the other one in first second of January.
I have this script I am building and need a bit of help. I want to use as minimal lines of code as possible. However what I want the script to do is the following.
Backup a specified file such as authlog into a new file with the keywords I specify.
However. I also want it to back up the file if current backed up file is older than 2+ days ago and if it is newer. Leave it alone. if (file) is older than +2 days old delete it and replace it with an updated one.
I'm getting integer expression expected and I'm not sure how to fix it
#!/bin/bash
authlog=/home/(myhomedir)/logs/backups/authlog-${current_date}
backup="$(cat /var/log/auth.log | grep -e failed -e invalid > /home/(myhomedir)/logs/backup)"
dayold=$(date -d '1 day ago' "+%Y-%m-%d")
current_date=$(date +"%m_%d_%Y")
if [ $authlog -lt $dayold ] ;
then ${backup}
else
echo $authlog is newer than $current_date $authlog has not been updated
if [ $authlog -eq ${current_date} ]
then "$(rm {authlog})"
fi
fi
You can't use -lt to compare non-numeric strings. This is why you're getting "integer expression expected". You're trying to compare a filename to a date string.
The easiest way to compare dates is by converting them to seconds since epoch (Jan 1, 1970 00:00:00 UTC). As an example, suppose I have a file called foo.bar and I want to check if it was last modified 3 or more days ago:
filetime=$(stat -c %Y foo.bar)
three_days_ago=$(date -d "3 days ago" +%s)
if [ $filetime -lt $three_days_ago ]; then
echo "foo.bar was last modified more than 3 days ago"
fi
Also, your authlog variable does not contain a date. I'm assuming there's a missing step that you forgot.
I want to get sysdate -1 and sysdate -2 in variable and echo it. I am using below query which gives todays date as output.
#! /bin/bash
tm=$(date +%Y%d%m)
echo $tm
How to get yesterday and day before yesterdays date?
Here is another one way,
For yesterday,
date -d '-1 day' '+%Y%d%m'
For day before yesterday,
date -d '-2 day' '+%Y%d%m'
Yesterday date
YES_DAT=$(date --date=' 1 days ago' '+%Y%d%m')
Day before yesterdays date
DAY_YES_DAT=$(date --date=' 2 days ago' '+%Y%d%m')
For any date you can use below one default it take 1 days. If its passing value that day before it take
ANY_YES_DAT=$(date --date=' $1 days ago' '+%Y%d%m')
You can get the yesterday date by this:
date -d "yesterday 13:00 " '+%Y-%m-%d'
and day before yesterday by this:-
date -d "yesterday-1 13:00 " '+%Y-%m-%d'
For older versions of BSD date (on old versions of macOS for example) which don't provide a -v option, you can get yesterdays date by subtracting 86400 seconds (seconds in a day) from the current epoch.
date -r $(( $(date '+%s') - 86400 ))
Obviously, you can subtract 2 * 86400 away for the day for yesterday etc.
Edit: Add reference to old macOS versions.
I have a shell script that runs on Linux and uses this call to get yesterday's date in YYYY-MM-DD format:
date -d "1 day ago" '+%Y-%m-%d'
It works most of the time, but when the script ran yesterday morning at 2013-03-11 0:35 CDT it returned "2013-03-09" instead of "2013-03-10".
Presumably daylight saving time (which started yesterday) is to blame. I'm guessing the way "1 day ago" is implemented it subtracted 24 hours, and 24 hours before 2013-03-11 0:35 CDT was 2013-03-09 23:35 CST, which led to the result of "2013-03-09".
So what's a good DST-safe way to get yesterday's date in bash on Linux?
I think this should work, irrespective of how often and when you run it ...
date -d "yesterday 13:00" '+%Y-%m-%d'
Under Mac OSX date works slightly different:
For yesterday
date -v-1d +%F
For Last week
date -v-1w +%F
This should also work, but perhaps it is too much:
date -d #$(( $(date +"%s") - 86400)) +"%Y-%m-%d"
If you are certain that the script runs in the first hours of the day, you can simply do
date -d "12 hours ago" '+%Y-%m-%d'
BTW, if the script runs daily at 00:35 (via crontab?) you should ask yourself what will happen if a DST change falls in that hour; the script could not run, or run twice in some cases. Modern implementations of cron are quite clever in this regard, though.
Here a solution that will work with Solaris and AIX as well.
Manipulating the Timezone is possible for changing the clock some hours.
Due to the daylight saving time, 24 hours ago can be today or the day before yesterday.
You are sure that yesterday is 20 or 30 hours ago. Which one? Well, the most recent one that is not today.
echo -e "$(TZ=GMT+30 date +%Y-%m-%d)\n$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
The -e parameter used in the echo command is needed with bash, but will not work with ksh.
In ksh you can use the same command without the -e flag.
When your script will be used in different environments, you can start the script with #!/bin/ksh or #!/bin/bash. You could also replace the \n by a newline:
echo "$(TZ=GMT+30 date +%Y-%m-%d)
$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
date -d "yesterday" '+%Y-%m-%d'
To use this later:
date=$(date -d "yesterday" '+%Y-%m-%d')
you can use
date -d "30 days ago" +"%d/%m/%Y"
to get the date from 30 days ago, similarly you can replace 30 with x amount of days
Just use date and trusty seconds:
As you rightly point out, a lot of the details about the underlying computation are hidden if you rely on English time arithmetic. E.g. -d yesterday, and -d 1 day ago will have different behaviour.
Instead, you can reliably depend on the (precisely documented) seconds since the unix epoch UTC, and bash arithmetic to obtain the moment you want:
date -d #$(( $(date +"%s") - 24*3600)) +"%Y-%m-%d"
This was pointed out in another answer. This form is more portable across platforms with different date command line flags, is language-independent (e.g. "yesterday" vs "hier" in French locale), and frankly (in the long-term) will be easier to remember, because well, you know it already. You might otherwise keep asking yourself: "Was it -d 2 hours ago or -d 2 hour ago again?" or "Is it -d yesterday or -d 1 day ago that I want?"). The only tricky bit here is the #.
Armed with bash and nothing else:
Bash solely on bash, you can also get yesterday's time, via the printf builtin:
%(datefmt)T
causes printf to output the date-time string resulting from using
datefmt as a format string for strftime(3). The corresponding argu‐
ment is an integer representing the number of seconds since the
epoch. Two special argument values may be used: -1 represents the
current time, and -2 represents the time the shell was invoked.
If no argument is specified, conversion behaves as if -1 had
been given.
This is an exception to the usual printf behavior.
So,
# inner printf gets you the current unix time in seconds
# outer printf spits it out according to the format
printf "%(%Y-%m-%d)T\n" $(( $(printf "%(%s)T" -1) - 24*3600 ))
or, equivalently with a temp variable (outer subshell optional, but keeps environment vars clean).
(
now=$(printf "%(%s)T" -1);
printf "%(%Y-%m-%d)T\n" $((now - 24*3600));
)
Note: despite the manpage stating that no argument to the %()T formatter will assume a default -1, i seem to get a 0 instead (thank you, bash manual version 4.3.48)
You can use:
date -d "yesterday 13:55" '+%Y-%m-%d'
Or whatever time you want to retrieve will retrieved by bash.
For month:
date -d "30 days ago" '+%Y-%m-%d'
As this question is tagged bash "DST safe":
And using fork to date command implie delay, there is a simple and more efficient way using pure bash built-in:
printf -v tznow '%(%z %s)T' -1
TZ=${tznow% *} printf -v yesterday '%(%Y-%m-%d)T' $(( ${tznow#* } - 86400 ))
echo $yesterday
This is a lot quicker on more system friendly than having to fork to date.
From bash version 5.0, there is a new variable $EPOCHSECONDS
printf -v tz '%(%z)T' -1
TZ=$tz printf -v yesterday '%(%Y-%m-%d)T' $(( EPOCHSECONDS - 86400 ))
echo $yesterday
What is the syntax to extract the day of the week from a stored date variable?
The dateinfile format is always [alphanum]_YYYYMMDD.
In this pseudocode example, trying to get dayofweek to store Saturday:
#! /bin/bash
dateinfile="P_20090530"
dayofweek="$dateinfile -u +%A"
[me#home]$ date --date=${dateinfile#?_} "+%A"
Saturday
Or, to put it as you've requested:
[me#home]$ dayofweek=$(date --date=${dateinfile#?_} "+%A")
[me#home]$ echo $dayofweek
Saturday
date -d $(echo $dateinfile | cut -f2 -d_) -u +%A
The inner expression separates the 20090530 from P_20090530, and the outer one extracts the day of week from that date
I needed the same output recently and googled upon this, but I had the same problem stating Bad Substition error.
I then read the date manual and made my version up as follows:
#! /bin/sh
dateinfile="P_20090530"
dayofweek=`date --reference $dateinfile +%A`
echo $dayofweek