Sponsored Content
Full Discussion: date in awk
Top Forums UNIX for Dummies Questions & Answers date in awk Post 302130461 by matrixmadhan on Monday 6th of August 2007 09:56:49 AM
Old 08-06-2007
Quote:
Originally Posted by Sukhendu Naskar
Actually date commands outputs a multiword string. So u have to quote the variable proprly.

v=`date`
echo 1 |awk -v var="$v" '{print var}'
I don't think its necessary.

It works fine for me.

Code:
zsh
Fedora Core 5

 

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

need date in awk

Hello all: Running SunOS 5.6 on a SPARCstation-5...and using bourne shell. This is a line from a script I am writing: LOG=/opt/msplib/sys/logs/alarm.log$* grep "Call Answered" $LOG |wc -l | awk '{ print "Call Answered\t" $1 "\tIVR\t" "AR-a"} It works great, and produces the... (5 Replies)
Discussion started by: cdunavent
5 Replies

2. UNIX for Dummies Questions & Answers

Use of system date in awk

i'm facing problems in using system date in a awk script... it should display the date in a report in mm/dd/yy format.. please help me in this case.. thanks (3 Replies)
Discussion started by: Manish4
3 Replies

3. UNIX for Dummies Questions & Answers

awk for date

Hello All, in my script i want to get the month part of date. I need the month part because i will grep logs every month using it. When i run the "date" command i got the below result: serverprod{root}>date Mon Aug 24 14:01:20 EEST 2009 but when i type the below command i got Aug24... (5 Replies)
Discussion started by: EAGL€
5 Replies

4. Shell Programming and Scripting

using awk to print date

i want to use awk to print the first,second and add date as third column in a file awk -F"|" ' { print $1,$2,current date }' tom.unl >> top.txt how can i achieve this,i need the comma's to seperate them and finally print current date and time as the third. i want output like... (5 Replies)
Discussion started by: tomjones
5 Replies

5. UNIX for Dummies Questions & Answers

Adding date when using awk

Hi, I want to print the number of lines of a file along with filename and today's date. Ex: XXX|07-22-2010|8 I am using as wc -c -l file.txt | awk '{print "XXX|",date +"%m-%d-%Y","|",$1}' But this one prints as AAA| 0 | 8 Can anyone please help me on this for printing the date? ... (3 Replies)
Discussion started by: aeroticman
3 Replies

6. UNIX for Dummies Questions & Answers

awk, splitting date

can any1 explain me hw is below wrking: wat is substr and dd,mmyear used for wat values will go in dese? sdt='31122010235959' sdate=`validate_date $sdt` validate_date() { dt="$1" set `echo $dt | nawk '{ print... (2 Replies)
Discussion started by: musu
2 Replies

7. Shell Programming and Scripting

Passing date into awk...

Hi All.... I need to pass date into awk and parse logfile based on that.... i used both awk and /usr/xpg4/bin/awk... both are throwing up error..... So here is the stuff... when i use /usr/xpg4/bin/awk : DATE=`date '+%Y %b %d'` START=00 END=23 /usr/xpg4/bin/awk -v DATE={"$DATE"} -v... (3 Replies)
Discussion started by: Nithz
3 Replies

8. Shell Programming and Scripting

Find week of the year for given date using date command inside awk

Hi all, Need an urgent help on the below scenario. script: awk -F"," 'BEGIN { #some variable assignment} { #some calculation and put values in array} END { year=#getting it from array and assume this will be 2014 month=#getting it from array and this will be 05 date=#... (7 Replies)
Discussion started by: vijaidhas
7 Replies

9. HP-UX

awk command in hp UNIX subtract 30 days automatically from current date without date illegal option

current date command runs well awk -v t="$(date +%Y-%m-%d)" -F "'" '$1 < t' myname.dat subtract 30 days fails awk -v t="$(date --date="-30days" +%Y-%m-%d)" -F "'" '$1 < t' myname.dat awk command in hp unix subtract 30 days automatically from current date without date illegal option error... (20 Replies)
Discussion started by: kmarcus
20 Replies

10. Shell Programming and Scripting

awk IF date comparison help

Hey everyone, I'm trying to create a script using awk and if that will list all of our aws tapes that have archived date that is past 90 days from todays current date, so that I can pass that to my aws command to remove. The fifth column is the creation date in epoch/seconds, so I'm... (13 Replies)
Discussion started by: beyondmondays
13 Replies
OSINFO-QUERY.C(1)					      Virtualization Support						 OSINFO-QUERY.C(1)

NAME
osinfo-query - Query information in the database SYNOPSIS
osinfo-query [OPTIONS...] TYPE [CONDITION-1 [CONDITION-2 ...]] DESCRIPTION
The "osinfo-query" command allows extraction of information from the database. TYPE can be one of "os", "platform", "device", or "deployment". With no conditions specified, all entities of the given type will be listed. # List all operating systems $ osinfo-query os Short ID | Name ... ----------------------+----------- centos-6.0 | CentOS 6.0 ... centos-6.1 | CentOS 6.1 ... ... Conditions allow filtering based on specific properties of an entity. For example, to filter only distros from the Fedora Project, use # List all operating systems $ osinfo-query os vendor="Fedora Project" Short ID | Name ... ----------------------+-------------- fedora1 | Fedora Core 1 ... fedora2 | Fedora Core 2 ... ... The set of fields which are printed can be controlled using the "--fields" command line argument: # List all operating systems $ osinfo-query --fields=short-id,version os vendor="Fedora Project" Short ID | Version ----------------------+---------- fedora1 | 1 fedora2 | 2 ... OPTIONS
-s PROPERTY, --sort-key PROPERTY Set the data sorting key. Defaults sorting the first column -f PROPERTY1,PROPERTY2,..., --fields PROPERTY1,PROPERTY2,... Set the visibility of properties in output PROPERTY NAMES
OS Valid property names for the "os" type are: short-id The short OS identifier name The long OS name version The OS version string family The OS kernel family vendor The OS vendor release-date The OS release date eol-date The OS end-of-life date codename The OS code name id The OS identifier PLATFORM Valid property names for the "platform" type are: short-id The short platform identifier name The long platform name version The platform version string vendor The platform vendor release-date The platform release date eol-date The platform end-of-life date codename The platform code name id The platform identifier DEVICE Valid property names for the "device" type are: name The device name product The device product name product-id The device product ID string vendor The device vendor name vendor-id The device vendor ID string class The device type class bus The device bus type id The device identifier DEPLOYMENT Valid property names for the "deployment" type are: id The deployment identifier EXIT STATUS
The exit status will be 0 if matching entries were found, or 1 if not matches were found SEE ALSO
"osinfo-db-validate(1)", "osinfo-detect(1)" AUTHORS
Daniel P. Berrange <berrange@redhat.com> COPYRIGHT
Copyright (C) 2012 Red Hat, Inc. LICENSE
"osinfo-query" is distributed under the termsof the GNU LGPL v2+ license. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE libosinfo-0.2.7 2013-02-06 OSINFO-QUERY.C(1)
All times are GMT -4. The time now is 05:45 PM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy