Hi,
I have a awk script to read a CSV file.
After reading the values i want to call a executable (nameely call_it) with the values what i read from the scv file.
I dont want to use system command inside the awk.
Is there any other way to run the executable from the awk script
Thanks ... (1 Reply)
Hi Jim,
The following script is in working state. But i m having one more problem with awk cmd. Could you tell me how to use any variable inside awk or how to take any variable value outside awk.
My problem is i want to maintain one property file in which i am declaring variable value into that... (12 Replies)
below is the output xml string from some other command and i will be parsing it using awk
cat /tmp/alerts.xml
<Alert id="10102" name="APP-DS-ds_ha-140018-componentFailure-S" alertDefinitionId="13982" resourceId="11427" ctime="1359453507621" fixed="false" reason="If Event/Log Level(ANY) and... (2 Replies)
Hi,
I would like to use grep command inside awk.
Here is my requirement below :
file.txt
col1 col2 col3 col 4 col 5
wrxwrx 124 jun 3 Sensex.EMEA
wrxwrx 120 jun 4 Emex.US
wrxwrx 130 feb 3 passion.AUS
wrxwrx 145 feb 9 lession.AUS
wrxwrx 130 feb 5 pass.US
wrxwrx 130 feb 8... (5 Replies)
Hello
can you please help me with below script which is meant to delete clients from multiple netbackup policies
I want to run a command insdie awk statement
apparelnlty this script is not working for me
for i in $( cat clients_list)
do
bppllist -byclient $i | awk... (6 Replies)
Hi all,
Need an urgent help on the below scenario.
script:
awk -F","
'BEGIN { #some variable assignment}
{ #some calculation and put values in array}
END {
year=#getting it from array and assume this will be 2014
month=#getting it from array and this will be 05
date=#... (7 Replies)
current date command runs well
awk -v t="$(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)
I tried running this.
dsh -w server1 'lsof /audit | awk '{ print $2 }''
It did not like above so I tried to escape the single parenthesis at the end.
dsh -w server1 'lsof /audit | awk '{ print $2 }\''
It then hung so I changed up the parenthesis to this. This worked.
dsh -w server1... (6 Replies)
I am trying to run an awk command inside of ssh and it is not working. These are AIX servers.
for i in `cat servers`; do ssh $i "/bin/hostname; df -g | awk '/dev/ && $4+0 > 70'"; done
server1
server2
server3
server4
I also tried these two methods and they did not work. It just seemed... (5 Replies)
Discussion started by: cokedude
5 Replies
LEARN ABOUT PHP
datetime.sub
DATETIME.SUB(3) 1 DATETIME.SUB(3)DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_sub (DateTime $object, DateInterval $interval)
Subtracts the specified DateInterval object from the specified DateTime object.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $interval
- A DateInterval object
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
EXAMPLES
Example #1
DateTime.sub(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-20');
date_sub($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-10
Example #2
Further DateTime.sub(3) examples
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-19 13:59:30
1992-08-15 19:56:58
Example #3
Beware when subtracting months
<?php
$date = new DateTime('2001-04-30');
$interval = new DateInterval('P1M');
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-03-30
2001-03-02
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.add(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.SUB(3)