I am trying to make a simple script in which i take input from shell and then forward the value to
nawk (BEGIN).
but when i run below mention script so it give no output.
echo "Enter TRUNK GROUP:"
read TGR
cat /omp-data/logs/5etr/080422.APX | nawk -F"|" -v P=$TGR '
BEGIN {
TG=P;... (1 Reply)
I need to set up a strange system through which an arbitrary command is sent to a number of different servers (well, actually, VPS accounts). We have a command "vpass" that "passes" a command from the root level to resident VPS accounts. Suppose I wanted each VPS to do some trivial thing, like... (3 Replies)
Hi, I had to create a new thread as the old thread had to much of confusion
I have two files shashi.sh and py.py
I want to pass a variable from shashi.sh to py.py. How do i achieve that ?.
shashi.sh
export X=12
echo "$("pwd")"
echo "$X"
exec python py.py "$(X)"
py.py... (0 Replies)
Hello,
I have a file with 4 columns.
An arbitrary example is shown below:
a Tp 10 xyz
b Tq 8 abc
c Tp 99 pqr
d Tp 44 rst
e Tr 98 efg
Based on the values in col 2 and col 3, I will execute another program.
I have been running this:... (5 Replies)
I am trying to execute a copy command via shell script. However, on occassion, 2 or more files need to copied. How do I code for the multiple arguments? Does it matter how the files are delimited?
Example: I have a script to copy files from 1 dir to another called duplicate.csh
In most... (1 Reply)
I have a script that kicks off several processes in the background and stored their pids in a variable as follows:
PID_DUMP_TRAN=$PID_DUMP_TRAN" "$!
so I then have a list of pids
If I echo $PID_DUMP_TRAN I get back a list of pids e.g. 8210 8211 8212
However I then want to kill all these... (5 Replies)
I have a shell program that calls another shell program
the following code works
. chkTimeFormat.sh "10/9/12 17:51:19:783."|read c
but when I am passing the the time in a variable like in the code below, the shell chkTimeFormat.sh is not returning proper value
time="10/9/12... (9 Replies)
Hello All,
May i please why my shell variable is not getting passed into awk script.
#!/bin/bash -vx
i="1EB07C50"
/bin/awk -v ID="$i" '/ID/ {match($0,/ID/);print substr($0,RSTART,RLENGTH)}' /var/log/ScriptLogs/keys.13556.txt
Thank you. (1 Reply)
Discussion started by: Ariean
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)