In Linux,
"echo %date% %time% %COMPUTERNAME%"
returns
%date% %time% %COMPUTERNAME%
not
Fri 09/24/2010 10:46:25.42 WXP2010043001
as Windows does.
I need to be able to do this for the logs I'm setting up.
Use the date command with a format like this:
date +"%m/%d/%Y %H:%M:%S $HOSTNAME"
To get hundredths of seconds, you may need to do some text processing like this:
DATE=date +'%m/%d/%Y %H:%M:%S.%N'
DATE=${DATE%???????}
DATE="$DATE $HOSTNAME"
This is because date offers seconds, nanoseconds, and nothing in between!
You can do:
dt=$(date)
echo $dt $HOSTNAME
echo $(date '+%Y %b %d %H:%M') Your output $HOSTNAME
Outputs:
2013 Nov 01 09:11 Your output PEGASUS-SYDNEY-CL2
it is also possible to use backtiks caracters for this:
echo `date` `hostname`
or with (localised) date formating:
echo `date +"%a %x %X"` `hostname`
As a complement: percentage character is not used to reference variables on any Linux shell. You should use the dollar sign for this.
You should probably read an introduction to Bash (here)
Several people have provided answers based on date, but your question requires the short day name (although my UK Win 7 installation doesn't provide this with the ECHO command you specified), which no one has (so far) addressed.
To get this, you will probably want to include %a in the format string:
date "+%a %m/%d/%Y %H:%M:%S $HOSTNAME"
In Linux, there is the date command. If you don't like the default format, it can be modified. See the manpage of date
For hostname, you can use hostname command, or $HOSTNAME environment variable, if it is set.
With system name, it is more complicated. You can use uname -a, sometimes it contains the OS name. Some distributions also have lsb-release, but not all of them.
Related
Currently I am working on a bash script which has hardcoded the following date:
2022-12-08T15:25:30.165965Z
I would like to generate that date in the script and I would like it to be up to date. What I have tried so far is:
date '+%d-%m-%Y %H:%M:%S.%M'
which outputs:
19-12-2022 12:40:13.40
I was wondering if someone knows how I could add the T and Z letters in the time?
You may use:
date -u '+%d-%m-%YT%H:%M:%S.%NZ'
Here:
-u will print UTC date-time
T before %H will print literal T
%N prints nanosecond value
Z after %N will print literal Z
$ date -u '+%FT%T.%6NZ'
2022-12-19T15:04:34.238733Z
You need GNU date for nanoseconds though.
With just bash using the EPOCHREALTIME variable and printf's %(fmt)T directive.
( IFS=.; TZ=UTC printf '%(%FT%T)T.%sZ\n' $EPOCHREALTIME )
# => 2022-12-19T14:28:33.343999Z
I'm using a subshell so that changing IFS does not affect the rest of the script.
Linux v2.4/Bash v3.2/GNU utils/date command version 5.0
I'm struggling with the date command. In a BASH application, the user can set date and time separately, resulting in separate variables for date and time. Further on, these variables are re-combined but this appears not be palatable for the date command: I get a different date back. Time is the same, however. Testing code:
#!/bin/bash
dnow1="$(date)"
echo "1 $dnow1" # --> Sat Sep 14 16:31:48 EDT 2013
#split date and time
dldate="$(date -d "$dnow1" +"%d-%m-%Y")"
echo "2 $dldate" # --> 14-09-2013
dltime="$(date -d "$dnow1" +"%H:%M:%S")"
echo "3 $dltime" # --> 16:31:48
#try to re-combine date and time
string="${dldate} ${dltime}"
echo "4 $string" # --> 14-09-2013 16:31:48
dnow2="$(date -d "$string")"
echo "5 $dnow2" # --> Thu Mar 5 16:31:48 EST 2020
I must be missing something here. Can anyone enlighten me? Thanks!
Note:
I'm working an original XBOX that has few/low resources so there's no room for other solutions like Python. I'm a 'bashist' anyway so it must be BASH!
Edit: corrected time format. Thanks Mat.
As to "$(....)" I have made it a habit to double quote wherever possible.
When getting your date use this format instead:
#split date and time
dldate="$(date -d "$dnow1" +"%Y-%m-%d")"
From GNU date manual
The output of the date command is not always acceptable as a date string, not only because of the language problem, but also because there is no standard meaning for time zone items like ‘IST’. When using date to generate a date string intended to be parsed later, specify a date format that is independent of language and that does not use time zone items other than ‘UTC’ and ‘Z’.
First of all using -d won't work in 14-09-2013 fashion, you can easily set date and time with one command and put it into variable. eg, just try this below on shell and then you can put into shell script.
date --date="Feb 2 2014 13:12:10"
Sun Feb 2 13:12:10 PST 2014
I'm using bash to build a script where I will get a filename in a variable an then with this variable get the file unix last modification date.
I need to get this modification date value and I can't use stat command.
Do you know any way to get it with the common available *nix commands?
Why you shouldn't use ls:
Parsing ls is a bad idea. Not only is the behaviour of certain characters in filenames undefined and platform dependant, for your purposes, it'll mess with dates when they're six months in the past. In short, yes, it'll probably work for you in your limited testing. It will not be platform-independent (so no portability) and the behaviour of your parsing is not guaranteed given the range of 'legal' filenames on various systems. (Ext4, for example, allows spaces and newlines in filenames).
Having said all that, personally, I'd use ls because it's fast and easy ;)
Edit
As pointed out by Hugo in the comments, the OP doesn't want to use stat. In addition, I should point out that the below section is BSD-stat specific (the %Sm flag doesn't work when I test on Ubuntu; Linux has a stat command, if you're interested in it read the man page).
So, a non-stat solution: use date
date, at least on Linux, has a flag: -r, which according to the man page:
display the last modification time of FILE
So, the scripted solution would be similar to this:
date -r ${MY_FILE_VARIABLE}
which would return you something similar to this:
zsh% date -r MyFile.foo
Thu Feb 23 07:41:27 CST 2012
To address the OP's comment:
If possible with a configurable date format
date has a rather extensive set of time-format variables; read the man page for more information.
I'm not 100% sure how portable date is across all 'UNIX-like systems'. For BSD-based (such as OS X), this will not work; the -r flag for the BSD-date does something completely different. The question doesn't' specify exactly how portable a solution is required to be. For a BSD-based solution, see the below section ;)
A better solution, BSD systems (tested on OS X, using BSD-stat; GNU stat is slightly different but could be made to work in the same way).
Use stat. You can format the output of stat with the -f flag, and you can select to display only the file modification data (which, for this question, is nice).
For example, stat -f "%m%t%Sm %N" ./*:
1340738054 Jun 26 21:14:14 2012 ./build
1340738921 Jun 26 21:28:41 2012 ./build.xml
1340738140 Jun 26 21:15:40 2012 ./lib
1340657124 Jun 25 22:45:24 2012 ./tests
Where the first bit is the UNIX epoch time, the date is the file modification time, and the rest is the filename.
Breakdown of the example command
stat -f "%m%t%Sm %N" ./*
stat -f: call stat, and specify the format (-f).
%m: The UNIX epoch time.
%t: A tab seperator in the output.
%Sm: S says to display the output as a string, m says to use the file modification data.
%N: Display the name of the file in question.
A command in your script along the lines of the following:
stat -f "%Sm" ${FILE_VARIABLE}
will give you output such as:
Jun 26 21:28:41 2012
Read the man page for stat for further information; timestamp formatting is done by strftime.
have perl?
perl -MFile::stat -e "print scalar localtime stat('FileName.txt')->mtime"
How about:
find $PATH -maxdepth 1 -name $FILE -printf %Tc
See the find manpage for other values you can use with %T.
You can use the "date" command adding the desired format option the format:
date +%Y-%m-%d -r /root/foo.txt
2013-05-27
date +%H:%M -r /root/foo.txt
23:02
You can use ls -l which lists the last modification time, and then use cut to cut out the modification date:
mod_date=$(ls -l $file_name | cut -c35-46)
This works on my system because the date appears between columns 35 to 46. You might have to play with it on your system.
The date is in two different formats:
Mmm dd hh:mm
Mmm dd yyyy
Files modified more than a year ago will have the later format. Files modified less than a year ago will have to first format. You could search for a ":" and know which format the file is in:
if echo "$mod_date" | grep -q ":"
then
echo "File was modified within the year"
else
echo "File was modified more than a year ago"
fi
I have 2 files with timestamps in the format of MMDDYYYY-HHMMSS.
For eg. 04192012-000623 and 04192012-000854.
I need to be able to find the difference between the 2 in seconds.
Special cases to check for
the dates straddling midnight. For eg: 04172012-115500 & 04182012-000200.
it shouldn't matter which file comes in first, etc.
I am running ksh with no access to the date -d flag. Can anyone point me in the right direction on how to shell script this? (It is going to be a part of a larger shell script so no other languages please)
This is intended to be run on both solaris and linux ksh. Thanks in advance.
As a starter (from my provided link above):
#! /usr/bin/ksh
echo enter first time stamp
read TIME1
echo enter second time stamp
read TIME2
H1=${TIME1%:+([0-9])}
M1=${TIME1#+([0-9]):}
H2=${TIME2%:+([0-9])}
M2=${TIME2#+([0-9]):}
H1=${H1#0}
M1=${M1#0}
H2=${H2#0}
M2=${M2#0}
((MAM1=H1*60+M1))
((MAM2=H2*60+M2))
((MAM1>MAM2)) && ((MAM2=MAM2+1440))
((diff=MAM2-MAM1))
echo diff = $diff
exit 0
$ ./timestamp
enter first time stamp
17:30
enter second time stamp
18:05
diff = 35
$ ./timestamp
enter first time stamp
23:59
enter second time stamp
00:01
diff = 2
$
if this is just for fun, good luck :). if it's for actual practical purposes, see my strptime wrapper at How to get the difference between now and a different date (in minutes) using ksh (or another shell script)?, it should be easily adaptable.
This solution uses gnu-date and bashisms, since I don't know ksh.
d1=04192012-000623
dd1="${d1:4:4}/${d1:0:2}/${d1:2:2} ${d1:9:2}:${d1:11:2}:${d1:13:2}"
d1=04192012-000854
dd2="${d1:4:4}/${d1:0:2}/${d1:2:2} ${d1:9:2}:${d1:11:2}:${d1:13:2}"
echo $(($(date -d "$dd1" +%s) - $(date -d "$dd2" +%s)))
Maybe ${var:from:len} is not available in ksh to cut parts from strings, then you have to replace it with something else, maybe sed.
stat -c followed by one of the following
%X time of last access, seconds since Epoch
%Y time of last data modification, seconds since Epoch
%Z time of last status change, seconds since Epoch
I need something simple like date, but in seconds since 1970 instead of the current date, hours, minutes, and seconds.
date doesn't seem to offer that option. Is there an easy way?
This should work:
date +%s
Just to add.
Get the seconds since epoch(Jan 1 1970) for any given date(e.g Oct 21 1973).
date -d "Oct 21 1973" +%s
Convert the number of seconds back to date
date --date #120024000
The command date is pretty versatile. Another cool thing you can do with date(shamelessly copied from date --help).
Show the local time for 9AM next Friday on the west coast of the US
date --date='TZ="America/Los_Angeles" 09:00 next Fri'
Better yet, take some time to read the man page
http://man7.org/linux/man-pages/man1/date.1.html
Pure bash solution
Since bash 5.0 (released on 7 Jan 2019) you can use the built-in variable EPOCHSECONDS.
$ echo $EPOCHSECONDS
1547624774
There is also EPOCHREALTIME which includes fractions of seconds.
$ echo $EPOCHREALTIME
1547624774.371210
EPOCHREALTIME can be converted to micro-seconds (μs) by removing the decimal point. This might be of interest when using bash's built-in arithmetic (( expression )) which can only handle integers.
$ echo ${EPOCHREALTIME/./}
1547624774371210
In all examples from above the printed time values are equal for better readability. In reality the time values would differ since each command takes a small amount of time to be executed.
So far, all the answers use the external program date.
Since Bash 4.2, printf has a new modifier %(dateformat)T that, when used with argument -1 outputs the current date with format given by dateformat, handled by strftime(3) (man 3 strftime for informations about the formats).
So, for a pure Bash solution:
printf '%(%s)T\n' -1
or if you need to store the result in a variable var:
printf -v var '%(%s)T' -1
No external programs and no subshells!
Since Bash 4.3, it's even possible to not specify the -1:
printf -v var '%(%s)T'
(but it might be wiser to always give the argument -1 nonetheless).
If you use -2 as argument instead of -1, Bash will use the time the shell was started instead of the current date. This can be used to compute elapsed times
$ printf -v beg '%(%s)T\n' -2
$ printf -v now '%(%s)T\n' -1
$ echo beg=$beg now=$now elapsed=$((now-beg))
beg=1583949610 now=1583953032 elapsed=3422
With most Awk implementations:
awk 'BEGIN {srand(); print srand()}'
This is an extension to what #pellucide has done, but for Macs:
To determine the number of seconds since epoch (Jan 1 1970) for any given date (e.g. Oct 21 1973)
$ date -j -f "%b %d %Y %T" "Oct 21 1973 00:00:00" "+%s"
120034800
Please note, that for completeness, I have added the time part to the format. The reason being is that date will take whatever date part you gave it and add the current time to the value provided. For example, if you execute the above command at 4:19PM, without the '00:00:00' part, it will add the time automatically. Such that "Oct 21 1973" will be parsed as "Oct 21 1973 16:19:00". That may not be what you want.
To convert your timestamp back to a date:
$ date -j -r 120034800
Sun Oct 21 00:00:00 PDT 1973
Apple's man page for the date implementation:
https://developer.apple.com/library/mac/documentation/Darwin/Reference/ManPages/man1/date.1.html
use this bash script (my ~/bin/epoch):
#!/bin/bash
# get seconds since epoch
test "x$1" == x && date +%s && exit 0
# or convert epoch seconds to date format (see "man date" for options)
EPOCH="$1"
shift
date -d #"$EPOCH" "$#"