i need output with Year in start field, below command i am using in Linux to get the License details but i am getting date like(start Tue 1/7 9:54) , so want to know the year for this:
lmutil lmstat -a -c user#server_name -f abcd -t
Output : start Tue 1/7 9:54
Looking for start Tue 1/7(with year) 9:54
Please help.
Thanks
I have some problems with ntpd sync on servers. Unless we resolve this issue I have written script to set date on servers manually for those who are not in sync.
For this I have taken one reference machine and I am catching current date of that machine and trying to set it on all other machines.
I'm using following command in script
ssh -i /mnt/keys/g.pem -o StrictHostKeyChecking=no root#$IP 'date --set="$ref_date"'
but when command runs it set wrong date.
e.g ref_date=Sat Sep 24 06:52:17 UTC 2016
when I echo above command it shows following line
ssh -i /mnt/keys/g.pem -o StrictHostKeyChecking=no root#xx.xx.xx.xx 'date --set="Sat Sep 24 06:52:17 UTC 2016"'
but when same command actually runs it gives following output
ssh -i /mnt/keys/g.pem -o StrictHostKeyChecking=no root#xx.xx.xx.xx 'date --set="Sat Sep 24 06:52:17 UTC 2016"'
Sat Sep 24 00:00:00 UTC 2016
Note: I have replaced $IP with xx.xx.xx.xx in above outputs.
Kindly provide solution to this.
ssh -i /mnt/keys/g.pem -o StrictHostKeyChecking=no root#$IP "date --set=\"$ref_date\""
See: Difference between single and double quotes in Bash
The problem is:
After user enter a linux command.
How can I get the output of the first command using another command?
Note: we cannot redirect output of first command to somewhere.
Using history expansion
$ date -d "12:00"
Thu Sep 19 12:00:00 EDT 2013
$ d=$(!!)
$ echo $d
Thu Sep 19 12:00:00 EDT 2013
I am running SunOS.
bash-3.00$ uname -a
SunOS lvsaishdc3in0001 5.10 Generic_142901-02 i86pc i386 i86pc
I need to find Yesterday's date in linux with the proper formatting passed from command prompt. When I tried like this on my shell prompt-
bash-3.00$ date --date='yesterday' '+%Y%m%d'
date: illegal option -- date=yesterday
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
I always get date illegal option, why is it so?
Is there anything wrong I am doing?
Update:-
bash-3.00$ date --version
date: illegal option -- version
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
Try this below thing. It should work
YESTERDAY=`TZ=GMT+24 date +%Y%m%d`; echo $YESTERDAY
Try this one out:
DATE_STAMP=`TZ=GMT+24 date +%Y%m%d`
where GMT is the time zone and you might need to alter the 24 according to the hours difference you have from GMT. Either that or you can change GMT to a time zone more comfortable to you e.g. CST
As larsks suggested, you can use perl:
perl -e 'use POSIX qw(strftime); print strftime "%a %b %e %H:%M:%S %Y",localtime(time()- 3600*24);'
Slightly modified from
http://blog.rootshell.be/2006/05/04/solaris-yesterday-date/
To get YYYYMMDD format use this
perl -e 'use POSIX qw(strftime); print strftime "%Y%m%d",localtime(time()- 3600*24);'
This link explains how to format date and time with strftime
http://perltraining.com.au/tips/2009-02-26.html
A pure bash solution
#!/bin/bash
# get and split date
today=`date +%Y%m%d`
year=${today:0:4}
month=${today:4:2}
day=${today:6:2}
# avoid octal mismatch
if (( ${day:0:1} == 0 )); then day=${day:1:1}; fi
if (( ${month:0:1} == 0 )); then month=${month:1:1}; fi
# calc
day=$((day-1))
if ((day==0)); then
month=$((month-1))
if ((month==0)); then
year=$((year-1))
month=12
fi
last_day_of_month=$((((62648012>>month*2&3)+28)+(month==2 && y%4==0)))
day=$last_day_of_month
fi
# format result
if ((day<10)); then day="0"$day; fi
if ((month<10)); then month="0"$month; fi
yesterday="$year$month$day"
echo $yesterday
TZ=$TZ+24 date +'%Y/%m/%d' in SunOS.
Playing on Solaris10 with non-GMT environment, I'm getting this:
# date
Fri Jul 26 13:09:38 CEST 2013 (OK)
# (TZ=CEST+24 date)
Thu Jul 25 11:09:38 CEST 2013 (ERR)
# (TZ=GMT+24 date)
Thu Jul 25 11:09:38 GMT 2013 (OK)
# (TZ=CEST+$((24-$((`date "+%H"`-`date -u "+%H"`)))) date)
Thu Jul 25 13:09:38 CEST 2013 (OK)
As You colud see, I have and I want to get CEST , but TZ=CEST+24 giving me wrong CEST data; GMT+24 giving me correct data, but unusable.
To get the proper result, I has to use GMT+22 (wrong command, correct result) or CEST+22 (wrong value, but finnaly correct result for correct TZ)
A pure bash solution given by #olivecoder is very reliable compared to any other solution but there is a mistake to be corrected. when the day fall on 1st of the month the script is failing with date saying "last_day_of_month" in day value. #olivecoder has missed $ in
day=last_day_of_month, that it should be
day=$last_day_of_month;
After this correction it works very good.
Using Timezone -24 is having some issue based on time when use it. in some cases it goes to day before yesterday. So I think its not reliable.
What is the *nix command to view a user's default login shell?
I can change the default login shell with chsh, but I don't know how to get what is the user's default shell.
Pseudocode
$ get-shell
/usr/bin/zsh
The canonical way to query the /etc/passwd file for this information is with getent. You can parse getent output with standard tools such as cut to extract the user's login shell. For example:
$ getent passwd $LOGNAME | cut -d: -f7
/bin/bash
The command is finger.
[ken#hero ~]$ finger ken
Login: ken Name: Kenneth Berland
Directory: /home/ken Shell: /bin/tcsh
On since Fri Jun 15 16:11 (PDT) on pts/0 from 70.35.47.130
1 hour 59 minutes idle
On since Fri Jun 15 18:17 (PDT) on pts/2 from 70.35.47.130
New mail received Fri Jun 15 18:16 2012 (PDT)
Unread since Fri Jun 15 17:05 2012 (PDT)
No Plan.
The login shell is defined in /etc/passwd. So you can do:
grep username /etc/passwd
I think what you are looking for is this:
#!/bin/bash
grep "^$1" /etc/passwd | cut -d ':' -f 7
Save that as get-shell somewhere in your path (probably ~/bin) and then call it like:
get-shell userfoo
SHELL variable is used to represent user's current shell
echo $SHELL