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Full Discussion: Variable Substitution
Top Forums Shell Programming and Scripting Variable Substitution Post 302383293 by mohanpadamata on Tuesday 29th of December 2009 06:49:36 AM
Old 12-29-2009
MySQL Variable Substitution

Hi ,

I have a variable as follows,

Temp=`cat ABC.txt | cut -c5-`

This will yeild a part of the date. say , 200912.

I would like to substitute this variable's value in a filename.

eg: File200912F.zip

when i say File$TempF.zip , it is not substituting.

Any help ?

Thanks in advance..
 

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ZIPARCHIVE.OPEN(3)							 1							ZIPARCHIVE.OPEN(3)

ZipArchive::open - Open a ZIP file archive

SYNOPSIS
mixed ZipArchive::open (string $filename, [int $flags]) DESCRIPTION
Opens a new zip archive for reading, writing or modifying. PARAMETERS
o $filename - The file name of the ZIP archive to open. o $flags - The mode to use to open the archive. o ZipArchive::OVERWRITE o ZipArchive::CREATE o ZipArchive::EXCL o ZipArchive::CHECKCONS RETURN VALUES
o $Error codes - Returns TRUE on success or the error code. o ZipArchive::ER_EXISTS File already exists. o ZipArchive::ER_INCONS Zip archive inconsistent. o ZipArchive::ER_INVAL Invalid argument. o ZipArchive::ER_MEMORY Malloc failure. o ZipArchive::ER_NOENT No such file. o ZipArchive::ER_NOZIP Not a zip archive. o ZipArchive::ER_OPEN Can't open file. o ZipArchive::ER_READ Read error. o ZipArchive::ER_SEEK Seek error. EXAMPLES
Example #1 Open and extract <?php $zip = new ZipArchive; $res = $zip->open('test.zip'); if ($res === TRUE) { echo 'ok'; $zip->extractTo('test'); $zip->close(); } else { echo 'failed, code:' . $res; } ?> Example #2 Create an archive <?php $zip = new ZipArchive; $res = $zip->open('test.zip', ZipArchive::CREATE); if ($res === TRUE) { $zip->addFromString('test.txt', 'file content goes here'); $zip->addFile('data.txt', 'entryname.txt'); $zip->close(); echo 'ok'; } else { echo 'failed'; } ?> PHP Documentation Group ZIPARCHIVE.OPEN(3)
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