I am receiving an elif error on line 13 and I can not figure out the reasoning behind it. I have added the then statement that I was initially missing. Any help would be great.
#The purpose of this script is for the end user to be able to enter a positive number
#User enters a number
NUM=$1... (4 Replies)
Script Gurus,
Need your help in getting this script to come out of logical error : I have pasted the script below: This script finds disk utilzation ... the script is written for both AIX and SUN OS... and option will be given in the initial to select DB or Non DB... its required for my prj...... (1 Reply)
I am getting below error from this code (which is at line 24):
if ] #this is line24 in code
then
mv $File_source_path/$File_name $File_name'_'`date '+%d%m%y'`
Error:
line 24:
Any help with the syntax. I am putting 2 condition with 'AND' clause.
This is bash shell. (2 Replies)
I am in need of some help. I am not an advance scripter or programmer, I have novice knowledge. I cannot get a script to work as expected. The script basically does an FTP (which I have work perfectly) and writes to a log. However, I want the script to echo "Not Connected" if the FTP fails to... (2 Replies)
Use and complete the template provided. The entire template must be completed. If you don't, your post may be deleted!
1. The problem statement, all variables and given/known data:
This is my problem for the class.
Write a script that asks for the user's age. If it is equal to or higher... (6 Replies)
Hi All,
I write an script will have some functions... I am getting elif not expected error..
Even tried by using set -x for debug but no use..
Could you please help me out in this
Variables
....
....
....
TEST_SRC()
{
cat ${filesrc} >> ${filetgt}
if ]; then
echo... (12 Replies)
Hi there, new to this forum and I recently encounter this problem:
I tried to use if-elif loop in a while-read loop, something like this:
#!/bin/bash
while read myline1 myline2
do
if ; then
echo "Successful!"
elif ; then
echo "Failed"
fi
done < $1
Input file looks like this:
200... (13 Replies)
Hi all,i have configured the following script to check if the file exists or not,
#!/bin/sh
sleep 30
{
FILEFULL=$1`date +$2`*
if ;
then
echo $FILEFULL exist
else
echo "$FILEFULL File not Found" | mail -s 'server' myaccount@mydomain.com
fi
}
but i have a problem, i need to... (2 Replies)
Hello all,
I have a scenario where I take user input values and accordingly take actions
say I run a script with
sh scriptname -x GB -e txt (txt can also be text)
I have used if clause for the first input (-x GB)and it is working fine
Now for second the scenario is
if
then
echo... (3 Replies)
Hello experts,
I am having problems with ELIF.
Please see the below code :
read a;
read b;
read c;
if || || ; then
echo "EQUILATERAL"
elif || || ; then
echo "SCALENE"
else
echo "ISOSCELES"
fi
This code works fine for the EQUILATERAL case but fails completely for the... (5 Replies)
Discussion started by: H squared
5 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)