Does anybody have an explanation for the following:
The following scripts runs fine on IRIX64 6.5 but has bugs on Solaris 8.
#! /bin/sh
echo run only on an SGI machine
echo type in linenumber
read j
echo value
read value
awk -f rmspass2 level=$value $j'step1.mlf'
When the script is... (5 Replies)
Hi
I have a unix shell script with an awk statement. I would like to print some of the fields of an input file. However, I would like to print them dynamically, ie by passing the literal $1 $3 into the script to define the output.
I have tried the following:
variable1='$1'
awk... (2 Replies)
I am trying to pass the results from a variable gathered from awk, however when I echo the 'PARSE' and 'SUB', the response is blank. This is my command.
awk -F= '/Unit/''{ PARSE=substr($2,1,5) ; SUB=substr($2,1,1) }' inputfile.lst
Is this a kind of valid attempt or am I obligated to declare... (3 Replies)
Hi.
I need to parse file and assign some values to variables, right now i do like below
MYHOMEDIR=`awk '/Home/ {print $NF}' output.txt`
MYSHELL=`awk '/Shell/ {print $NF}' output.txt`
PRGRP=`awk '/Primary/ {print $NF}' output.txt`
SECGRP=`awk '/Second/ {print $NF}' output.txt`
In this... (10 Replies)
Using ksh to call a function which has awk script embedded.
It parses a long two element list file, filled with text numbers (I want column 2, beginning no sooner than line 45, that's the only known thing) . It's unknown where to start or end the data collection, dynamic variables will be used. ... (1 Reply)
I'm trying to use awk to write new entries to a hosts file if they don't exist. I need to do so depending on the type of system I have. Below is what I have, but it isn't working.
awk -v myip1=$IP1 myip2=$IP2 myhost1=$HOST1 myhost2=$HOST2' BEGIN { mqhost1=0; mqhost2=0; stap1=0; stap2=0; }
... (4 Replies)
Hi all,
I am wanting to pass variables from a file to an awk arithmetic formula.
When I use the formula with the value it works well. As soon as I make these variables I get an inf (infinity) response. I can certainly echo the variables back and they look correct. My googling for answers has... (3 Replies)
Hi guys,
I need to fetch data from logfile between two given dates,i got the below code from our forum.It works perfect,but i need to enter the value dynamically to awk while running.
awk '/2012 Jun/{p=1}!/2012 Jul/ && prev~/2012 Jul/ && p{p=0}{prev=$0}p' file
i tried the below code,but... (4 Replies)
Am looking to pass some Linux environment variables into AWK , can I simply use the -v option ?
awk -F: -v AHOME=$HOME '{ if
{rm AHOME/file.txt
a=2 } }'
config.txt
... (4 Replies)
Discussion started by: alldbest
4 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)