Didn't have the same permission's. Thats what it was. I have one more problem. I'm trying to execute a function an put a "if/then" statement within the single line of ssh command.
Code:
echo "`date +%D:%H:%M:%S`" ; read -p " User ID"= UserID
echo "`date +%D:%H:%M:%S`"
read -p "Enter Server(s)"= servertext
for i in $servertext
do
echo "$i"
echo "`date +%D:%H:%M:%S`"
ssh -t -t $i 'if [ $? -eq 0 ] ; then printf "User Exist" -a pcffunc ; else echo " user does not exist" ; fi '
pcffunc() {
functions.....
}
done
The script will do the first part but won't execute the function. Any thoughts?
Last edited by fpmurphy; 02-12-2012 at 11:41 AM..
Reason: code tags please!
I am using Net::SSH::Expect to connect to the device(iLO) with SSH. After the $ssh->login() I'm able to view the prompt, but not able to send any coommands.
With the putty I can connect to the device and execute the commands without any issues.
Here is the sample script
my $ssh =... (0 Replies)
Hi,
I want to use ssh to add a register key on remote ssh server. Since there are space characters in my register key string, it always failed. If there is no space characters in the string, it worked fine. The following is what I have tried. It seems that "ssh" command doesn't care about double... (9 Replies)
I have write a script which contains
ssh -p 12345 dcplatform@10.125.42.50
ssh 127.0.0.1 -p 5555 "$CMD"
ssh root@$GUEST_IP "$CMD"
before I use public key, it works well, now I want to change to "expect", BUT I don't want to change above code and "parameter position"
I can post a... (1 Reply)
I'm trying to setup a link between my home pc (work-machine) and a server at work (tar-machine) that is behind a gateway (hop-machine) and not directly accessible.
my actions:
work-machine$ ssh -L 1234:tar-machine:22 hop-machine
work-machine$ ssh -p 1234 user@127.0.0.1
- shh access on... (1 Reply)
Hi,
I want to validate ssh connection one after one for multiple servers..... password less keys already setup but now i want to validate if ssh is working fine or not...
I have .sh script like below and i have servers.txt contains all the list of servers
#/bin/bash
for host in $(cat... (3 Replies)
Hi,
I am trying to complete my bash script in order to find which SSH servers on LAN are still active with the ssh keys, but i am frozen at this step:
#!/bin/bash
# LAN SSH KEYS DISCOVERY SCRIPT
</etc/passwd \
grep /bin/bash |
cut -d: -f6 |
sudo xargs -i -- sh -c '
&& cat... (11 Replies)
Discussion started by: syrius
11 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)