How can I learn date and time from the internet using bash without installing anything extra.
I am basically looking for an equivalent of bash $ date, but using an NTP (or any other way) to get the correct date and time from the internet. All the methods I find (such as ntpd) are meant to correct the system time, which is not my purpose.
date has a lot of options for formatting, but I'm assuming that you just want the date and time:
ntpdate -q time.google.com | sed -n 's/ ntpdate.*//p'
(or any other time server)
If you have ntpd installed & configured then you can use the NTP Query command ntpq -crv which will return;
associd=0 status=04ff leap_none, sync_uhf_radio, 15 events, stale_leapsecond_values,
version="ntpd 4.2.6p5#1.2349-o Mon Feb 6 07:22:46 UTC 2017 (1)",
processor="x86_64", system="Linux/4.10.13-1.el6.elrepo.x86_64", leap=00,
stratum=1, precision=-23, rootdelay=0.000, rootdisp=1.000, refid=PPS,
reftime=dd2c9f10.f25911ee Wed, Aug 2 2017 19:57:20.946,
clock=dd2c9f11.f4251b0a Wed, Aug 2 2017 19:57:21.953, peer=6516, tc=4,
mintc=3, offset=-0.005, frequency=-17.045, sys_jitter=0.110,
clk_jitter=0.007, clk_wander=0.003, tai=37, leapsec=201701010000,
expire=201706010000
You want the line starting clock which gives the time, date etc - you would be best parsing this out with awk or something if you just want the date stamp rather then everything else.
You do not need to be a root user to run the command. It won't set anything, but will query your local server (presuming your running ntp) and present the details.
Related
I am using NetScaler FreeBSD, which recognizes many of the UNIX like commands, grep, awk, crontab… etc.
I run the following command to get the number of connected users that we have on the system
#> nsconmsg -g aaa_cur_ica_conn -d stats
OUTPUT (numbered lines):
Line1: Displaying current counter value information
Line2: NetScaler V20 Performance Data
Line3: NetScaler NS11.1: Build 63.9.nc, Date: Oct 11 2019, 06:17:35
Line4:
Line5: reltime:mili second between two records Sun Jun 28 23:12:15 2020
Line6: Index reltime counter-value symbol-name&device-no
Line7: 1 2675410 605 aaa_cur_ica_conn
…
…
From above output - I only need the number of connected users (represented in Line 7, 3rd column (605 to be precise), along with the Hostname and Time (of the running script)
Now, to extract this important 3rd column number i.e. 605, along with the hostname, and time of data collected - I wrote the following script:
printf '%s - %s - %s\n' "$(hostname)" "$(date '+%H:%M')" "$(nsconmsg -g aaa_cur_ica_conn -d stats | grep aaa_cur_ica_conn | awk '{print $3}')"
The result is perfect, showing hostname, time, and the number of connected users as follows:
Hostname - 09:00 – 605
Now can anyone please shed light on how I can:
Run this script every day - 5am to 5pm (12hours)?
Each time scripts runs - append a file on a remote Unix share with the output?
I appreciate this might be a bit if a challenge... however would be grateful for any bash scripts wizards out there that can create magic!
Thanks in advance!
I would suggest a quick look into the FreeBSD Handbook or For People New to Both FreeBSD and UNIX® so that you could get familiar with the operating system and tools that could help you achieve better what you want.
For example, there is a utility/command named cron
The software utility cron is a time-based job scheduler in Unix-like computer operating systems.
For example, to run something all days between 5am to 5pm every minute, you could use something like:
* 05-17 * * * command
Try more options here: https://crontab.guru/#*_05-17_*_*_*.
There are more tools for scheduling commands, for example at (https://en.wikipedia.org/wiki/At_(command)) but this something you need to evaluate and read more about it.
Now regarding the command, you are using to get the "number of connected users", you could avoid the grep and just used awk for example:
awk '/aaa_cur_ica_conn/ {print $3}'
This will print only column 3 if line contains aaa_cur_ica_conn, but as before I invite you to read more about the topic so that you could bet a better overview and better understand the commands.
Last but not least, check this link How do I ask a good question? the better you could format, and elaborate your question the easy for others to give an answer.
I have two timestamps in formats :
Timestamp 1 (Variable - RunStartDate): Thu May 3 14:12:54 CDT 2018
Timestamp 2 (Variable - RunEndDate): Thu May 3 18:11:46 CDT 2018
I want the difference of number of hours between these two timestamps in UNIX shell. (I.e. RunEndDate - RunStartDate in hours)
Please help, I am new to UNIX and it is throwing me errors when I just try to subtract the two.
You have a few options here, such as calling out to Perl or Python and using a date/time library to do the math for you. Another option is to use the date program to convert the dates to seconds, subtract the values, and then convert back to hours. Unfortunately, you can't do floating-point math in Bash, so we'll have to call out to a helper program to do that, too.
START=$(date -d "$RunStartDate" +"%s")
END=$(date -d "$RunEndDate" +"%s")
HOURS=$(bc -l <<< "($END - $START) / 3600")
Note that this will only work on GNU systems (e.g. Linux).
This question already has an answer here:
bash "date" returns "invalid date" error for a specific date string
(1 answer)
Closed 7 years ago.
I have found a strange behaviour for the date() function when using bash.
When I use date -d "2008-10-12 +1 days", the date function returned
date: invalid date ‘2008-10-12 +1days’
Even it returned error when trying:
date -d "2008-10-12"
--> date: invalid date ‘2008-10-12’
However, it works for:
date -d "2008-10-13 +1days"
--> Tue Oct 14 00:00:00 CLST 2008
date -d "2008-10-11 +1days"
--> Sun Oct 12 01:00:00 CLST 2008
date -d "2007-10-12 +1days"
--> Sat Oct 13 00:00:00 CLT 2007
... and so on
It seems the date() does not recognize "just" 2008-10-12. Is it a bug? Am I doing something wrong?
I am using CENTOS 7
All the best
In RHEL 7 there were many changes happened to somany commands. To change the time and date earlier we used to use date command and now we need to use timedatectl command (preferred) instead.
http://www.freedesktop.org/software/systemd/man/timedatectl.html
So as there is a new latest command with more features included they might have limited it (Not sure on that). But from this version I have been using timedatectl command which is really great one.
I would like to automate a download of an image from a third party server with a CRON job and then upload the image to my website.
I have 2 issues:
First, the third party site changes the image name every day using the following logic:
http://thirdpartysite.com/ImageFinder.aspx?ReportID=FILENAME
where FILENAME is 26601 +14 for each day after 6 Oct 2014 (so 7 Oct would be 26615, 8 Oct would be 26629 etc).
How do I build this into a simple Linux bash script for use with wget?
Second, how do I upload this to my site via FTP (or similar) with the same script.
NOTE: I have permission to host the file on my site and have linked the original site / placed credit for the image.
Following the suggestions of #Abhay, first get the timestamp of Oct 6, let's store it in the variable $d0:
d0=$(date +%s -d 20141006)
Then store the timestamp of a target date, say Oct 8 and store it in $d1:
d1=$(date +%s -d 20141008)
Then you can calculate the difference and apply the required arithmetic operations in $((...)), like this:
echo $((26601 + 14 * (d1 - d0) / 60 / 60 / 24))
# outputs: 26629
The date command has one very good format: %s, which prints number of seconds since "epoch", which is the fixed date 1 January 1970, 00:00 UTC. I'll call it "timestamp". In combination with this, you can use the -d date-string, so that it prints the given date as number of seconds. Now you can take today's timestamp, subtract the timestamp of "6 Oct 2014" from it, and you get number of seconds between the two times. Now you can divide it by (60 * 60 * 24) to get it in number of days, and do further arithmetic to get the desired number, and make a file name out of it.
The date string formats that -d option takes are flexible, but as of now I am not sure whether it takes "6 Oct 2014" as is. Try a few permutations, or better, check the "info" page.
I am trying to shift the dates of a series of files by 9 hours. I've reached as far as this:
for i in *.MOV; do touch -r "$i" -d "-9 hours" "$i"; done
This should work in recent systems, but the touch command in OSX seems to be a bit outdated and not to support the -d switch.
I'm using Snow Leopard. Any idea on the best option for doing this with a single line command? I don't want to create a script for this.
Ok, sorted it out. OSX comes with a gtouch command, that knows the -d switch. It's part of GNU coreutils. See the comments below for information regarding availability on specific MacOS versions.
For more information on using relative dates with the -d switch see the manual.
Looking at the Wikipedia Page for Touch, it appears you're accustomed to the GNU version of Touch. Which MacOS isn't using.
For what you want to do, look into the "SetFile" command, which gets installed with XCode tools. You have -d and -m options, which reset the Created and Modified dates & times respectively.
http://developer.apple.com/library/mac/#documentation/Darwin/Reference/ManPages/man1/SetFile.1.html
Donno OS X, but it should be easy enough to
get curr time stamp on the file
convert it to seconds
subtract 9 hours (9*60*60 secs) from it
convert it back to the format accepted by touch's -t option
run touch command
All this of course can be done in a single for loop on command line.
Here are simple examples from WikiPedia showing back and forth conversion.
# To convert a specific time stamp to Unix epoch time (seconds since 1970-01-01):
date +"%s" -d "Fri Apr 24 13:14:39 CDT 2009"
# 1240596879
# To convert Unix epoch time (seconds since 1970-01-01) to a human readable format:
date -d "UTC 1970-01-01 1240596879 secs"
# Fri Apr 24 13:14:39 CDT 2009
# Or:
date -ud #1000000000
# Sun Sep 9 01:46:40 UTC 2001
# or: Haven't tested this but should work..
date -d #1000000000 +%y%m%d%%H%M%S
# 010909014640