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Full Discussion: Substitute File name
Top Forums Shell Programming and Scripting Substitute File name Post 302110369 by Deal_NoDeal on Monday 12th of March 2007 01:54:57 PM
Old 03-12-2007
Quote:
Originally Posted by vanand420
Hi all
I am in a small problem pl help me out.
I am having a directory having ZIP files with name starting as :
01.xyz
02.pqr
and so on
I want to run the script-
cat myfile | awk '{print $1, $2}' | while read var1 var2
do
zcat $var2* | grep "^000$var1" >> my_output
done
Where the myfile contains like :
1231231 01
1451515 02
i.e. I want to substitute the var2 for ZIP files in the directory and then search the var1 in that file. But It gives error *.Z : no such file or directory.
Please help me out.
Thanks in advance.
When you use:
zcat $var2* | grep "^000$var1" >> my_output

It is actually translated to something like this:

zcat 01* | grep ....

So, there is the issue.

If you know that your $var2 contains the exact file name, then remove the "*". Use

zcat $var2 | grep "^000$var1" >> my_output

You can also use the file extension to be safe i.e "zcat $var2.zip".

Or if $var2 doesn't contain the exact file name then try using double quotes around $var*. Like

zcat "$var2*" | grep "^000$var1" >> my_output

Hope this helps.
 

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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)
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