Hello,
I have a problem with trying to run a shell script that reads in user input, validates, and sets to a 'default' value if the input is not valid. I cannot get the portion of resetting to a default value to work. These lines are skipped, and the
value of x is still whatever the user... (1 Reply)
Greetings all,
I'm in the midst of writing a login component for a series of shell scripts. What my login script does is this:
1. Prompt for username and read in username
2. Prompt for destination host and read in destination host
3. run ssh username and destination host
4. After user keys... (0 Replies)
Hi all,
Just like to ask if it is possible to do the following:
1. Have a shell script that calls ssh username@destinationhost
2. Upon successful verification, we ssh into the destination host and automatically use ksh to run a shell script that resides in the destination host. (Hopefully no... (8 Replies)
Hi
newbeeeee alarm
i want to send a little script over ssh
this script mus download a report.tar then rename and move. the report name format is report_<host.with.dot>-10-09-20-11:55:25.tar
function remote_cmd_mv
{
_host=$1
ARCHROOTDIR='/tmp'
... (8 Replies)
HI Unix Gurus,
I an stuck in an interesting issue, where I am trying to execute a script on remote server after ssh.
The script on remote server is interactive,. Whenever it is called it hangs where it expects input from terminal and I have to terminate it.
I have searched through fourm... (12 Replies)
Hi All,
In a Shell scriipt with a SQL block I want to issue a query against a local DB and a remote DB on a remote server. The shell script is running locally.
This is how I connect to the local server. But I want the query to reference remote table in the join. Question can I specify a... (1 Reply)
I have a script like this (Yes, I know the DAY6 number isn't right - I'm just testing at this point):
DAY0=`date -I`
DAY1=`date -I -d "1 day ago"`
DAY6=`date -I -d "2 days ago"`
if
then
ssh root@synology1 nohup rm -rf "/volume1/Fileserver/$DAY6"
fi
I've tested the line to remove the... (5 Replies)
I need to run a local shell script on a remote machine. I am able to achieve that by executing the command
> ssh -qtt user@host < test.sh
However, when I try to pass arguments to test.sh it fails.
Any pointers would be appreciated. (7 Replies)
Hello,
I need to create a shell script which will copy files - which are created on particular date and starting with particular name - to local windows XP machine.
Is this possible.?
Currently it is being done manually using winscp (1 Reply)
local script:
cat > first.sh
cd /tmp
echo $PWD
echo `whoami`
cd /tmp/123
tar -cvf 789.tar 456
sleep 10
except script:
cat > first
#!/usr/bin/expect
set ip 10.5.15.20
set user "xyz123"
set password "123456"
set script first.sh
spawn sh -c "ssh $user@$ip bash < $script" (1 Reply)
Discussion started by: Aditya Avanth
1 Replies
LEARN ABOUT PHP
debug_zval_dump
DEBUG_ZVAL_DUMP(3) 1 DEBUG_ZVAL_DUMP(3)debug_zval_dump - Dumps a string representation of an internal zend value to outputSYNOPSIS
void debug_zval_dump (mixed $variable, [mixed $...])
DESCRIPTION
Dumps a string representation of an internal zend value to output.
PARAMETERS
o $variable
- The variable being evaluated.
RETURN VALUES
No value is returned.
EXAMPLES
Example #1
debug_zval_dump(3) example
<?php
$var1 = 'Hello World';
$var2 = '';
$var2 =& $var1;
debug_zval_dump(&$var1);
?>
The above example will output:
&string(11) "Hello World" refcount(3)
Note
Beware the refcount
The refcount value returned by this function is non-obvious in certain circumstances. For example, a developer might expect the
above example to indicate a refcount of 2. The third reference is created when actually calling debug_zval_dump(3).
This behavior is further compounded when a variable is not passed to debug_zval_dump(3) by reference. To illustrate, consider a
slightly modified version of the above example:
Example #2
<?php
$var1 = 'Hello World';
$var2 = '';
$var2 =& $var1;
debug_zval_dump($var1); // not passed by reference, this time
?>
The above example will output:
string(11) "Hello World" refcount(1)
Why refcount(1)? Because a copy of $var1 is being made, when the function is called.
This function becomes even more confusing when a variable with a refcount of 1 is passed (by copy/value):
Example #3
<?php
$var1 = 'Hello World';
debug_zval_dump($var1);
?>
The above example will output:
string(11) "Hello World" refcount(2)
A refcount of 2, here, is extremely non-obvious. Especially considering the above examples. So what's happening?
When a variable has a single reference (as did $var1 before it was used as an argument to debug_zval_dump(3)), PHP's engine opti-
mizes the manner in which it is passed to a function. Internally, PHP treats $var1 like a reference (in that the refcount is
increased for the scope of this function), with the caveat that if the passed reference happens to be written to, a copy is made,
but only at the moment of writing. This is known as "copy on write."
So, if debug_zval_dump(3) happened to write to its sole parameter (and it doesn't), then a copy would be made. Until then, the
parameter remains a reference, causing the refcount to be incremented to 2 for the scope of the function call.
SEE ALSO var_dump(3), debug_backtrace(3), References Explained, References Explained (by Derick Rethans).
PHP Documentation Group DEBUG_ZVAL_DUMP(3)