Hi Gurus,
I have a requirement of writting the shell script where it should ask me two values
FND_TOP=/d02/app/oracle/xxx/fnd/11.5.0
CDCRM_TOP=/d02/app/oracle/xxx/cdcrm/11.5.0
and then keep these values stored as variables for the execution of rest of the script.
Because, I have to... (2 Replies)
Hi All,
Shell is ksh
I've given portion of the script here to explain the problem.
It will accept 2 input parameters .
in_file1=$1
in_file2=$2
outbound_dir=/home/outbound
for i in 1 2
do
eval file$i=$outbound_dir/\$in_file$i
eval echo "filename is \$file$i"
... (4 Replies)
Hi,
I need some direction with the following. The below code is semi-psuedo code which will hopefully make it easier to understand what I am trying to achieve:
for i in `echo ${testarray
}`
do
let c=c+1
eval "first$c=$i"
while... (4 Replies)
Hello,
so i'm making a script, using dynamic variables and trying to expand them. So far it hasn't worked out too well so it seems that I need some help from you, the elite.
Example:
#!/bin/sh
counter=0
until (($counter>5))
counter2=1
until (($counter2>6)); do
if ;... (5 Replies)
Greetings all,
Been trying to do my Googling and forum searches but can't seem to lock in on a solution.
I have a script that parses a log and collects all the uniq events to a flat file. Some days might have 50 unique events, other days might have 75. (Hence my reference to dynamic.)
... (2 Replies)
Hi All,
I am really struggling to solve this problem, this might be small but I am not able to, can somebody help me?
I have few directories and these directories receives text files in large amount with in fraction of seconds. So I just want to send all the files in current directory to... (2 Replies)
Hi all,
I want to dynamically set variables in a bash script. I made a naive attempt in a while loop that hopefully can clarify the idea.
n=0; echo "$lst" | while read p; do n=$(($n+1)); p"$n"="$p"; done
The error message is:
bash: p1=line1: command not found
bash: p2=line2: command... (8 Replies)
Hi ,
i am unable to generate dynamic variables can any one please help me on the below issue
j=1
{record_count_"$j"}=`db2 -xselect substr\(job_name,24\) rec_count from $libname.audit_table_nrt where job_name like \'DATAMART_DEL_RUN%\' and STS_FLAG=\'E\' and seq_no=$i`
echo " record... (3 Replies)
Discussion started by: bhaskar v
3 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)