06-28-2006
comparing files to contents of a file
Hi I have a problem trying to run a while statement.
I have files under one directory that i need to compare to a value in filex and update that file with the result
files in the directory are
DFC1.
DFC5.
DFC345.
DFC344.
DFC9.
The program i am trying to run will take the number form the filename
and compare it to the value in filex
it should run through each value in the directory compare it to the value in filex and if greater then replace the value in filex and remove the filename from the directory
i.e. compare DFC1. the value 1 to the value in filex = 0
it is greater so replace value in filex with 1 delete DFC1. and do next filename DFC5. same for DFC345. and DFC344 (but this one should fail)
any ideas
trying this #! /usr/bin/ksh
cd /swm/test/apps/creditors/temp
pwd
var1=$(ls -t DFC*.)
echo $var1
#while read var1
#do
echo cat $var1
var2=$(echo $var1|cut -d. -f1|cut -c5-)
echo $var2 > /swm/test/apps/creditors/temp/valuex
if
[$var2 -ge var3=$(cat valuex)]
then
rm $var1
else
echo "the current value is greater than the file $var1
thanks for any help
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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 output
SYNOPSIS
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)