i am using the below script and trying to move files in that directory in that pattern to archive. But it doesn;t seem to take the metacharacters. Please sugggest.
Code
Debug output: (1 Reply)
hi there,
can somebody give me a definition for daemons, or example what are they !!
and what the use for?
i've done some research and all what i found is /etc/...
or /usr/bin/...
and i haven't quietly got the concept.
any ideas !!
Thanks. (5 Replies)
Could someone please direct me to a link that gives the definitions for each of the letters from the results of the $- environment variable? It would be nice to know what shell options each of the letters represents, but I am specifically looking for the shell option for 'c' (lowercase c). Thank... (12 Replies)
Hi all,
I have a variable say var1 (output from somewhere, which I can't change)which store something like this:
echo $var1
name=fred
age=25
address="123 abc"
password=pass1234
how can I make the variable $name, $age, $address and $password contain the info?
I mean do this in a... (1 Reply)
Hi,
I am trying to create a macro in FTP to rename multiple files.Below given is the codewhich i tried. I wanted to pass the files from the file "$FTPTXRENAMESUCLIST`" to the renfiles macro.
Your help will be really appreciated.
ftp -i -v -n << endftp > $FTPTXLOG 2> $FTPER
open... (0 Replies)
A byte is the smallest unit of storage which can be accessed in a computer's memory- either in RAM or ROM.It also holds exactly 8 bits.But its old view one byte was sufficient to hold one 8 bit character.Modern days especially on .NET or international versions of Win 32, 16 bits is needed.
... (2 Replies)
I have many headers with huge amount of structures in them, typical one looks like this:
$ cat a.h
struct Rec1 {
int f1;
int f2;
};
struct Rec2 {
char r1;
char r2;
};
struct Rec3 {
int f1;
float k1;
float ... (6 Replies)
I'm a bit confused about the term ‘environment variables'.
Within your shell you can set two types of variables:
1. Shell variable - affecting functionality within your shell
2. User defined variable
When using the ‘export' command on a variable you make sure it's being inherited by new sub... (2 Replies)
Hello,
I am running ubuntu16.04.
By means of @Rudic's help, I have below command.
What I need to do is to replace video_id by value of video_id in which WHERE clause is matched:
{print "INSERT INTO video_series_files (id, video_id, file_type, protocol, \
url, languages, quality, accessed... (12 Replies)
Discussion started by: baris35
12 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)