[PHP] Loading a File into Variable - How??
I thought this would be fairly easy, but, I can't figure out how to load the
contents of a file into a variable so I can output it later.
The file to be loaded is in my include_path on the server and does contain
some HTML.
file_get_contents() is exactly what I need, but, it only works on a CVS
version of PHP, whatever that is.
I also tried the following function:
function file_get_contents($filename) {
$fd = fopen ($filename, r, 1);
$contents = fread($fd, filesize($filename));
fclose($fd);
return $contents;
}
But it returns nothing. If I use readfile() the file contents is displayed,
but, what I really want to do is store it in a string variable, not output
it directly. How can I do this?
Thanks,
Monty
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Re: [PHP] Loading a File into Variable - How??
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Monty wrote:
But it returns nothing. If I use readfile() the file contents is displayed,
but, what I really want to do is store it in a string variable, not output
it directly. How can I do this?
Look 4 implode() in the function list
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tHe TeSt, YeS iT iS
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Re: [PHP] Loading a File into Variable - How??
Look 4 implode() in the function list Implode isn't really what I need, I just want to load an entire file into a single string variable. However, I figured out the problem shortly after posting that first message (of course). Because the file being opened is in the include_path, it seems filesize() doesn't see those files. So, if I replace the filesize($filename) command with a hard-coded number, it works. Monty -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Loading a File into Variable - How??
Hi Monty, I've been trying to do the same thing with no success. Would you be so kind as to show me how you finally did it? I'm not too clear what you meant by: So, if I replace the filesize($filename) command with a hard-coded number, it works. Tia, Andre On Thursday 18 July 2002 04:28 pm, you wrote: I just want to load an entire file into a single string variable. However, I figured out the problem shortly after posting that first message (of course). Because the file being opened is in the include_path, it seems filesize() doesn't see those files. So, if I replace the filesize($filename) command with a hard-coded number, it works. Monty -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Loading a File into Variable - How??
On Thu, Jul 18, 2002 at 04:28:57PM -0400, Monty wrote:
Look 4 implode() in the function list
Implode isn't really what I need, I just want to load an entire file into a
single string variable.
Yes, it IS what you need. Plus, to work around your other problems
mentioned in later posts...
$string = implode('', file('filename') );
--Dan
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Re: [PHP] Loading a File into Variable - How??
In 4.3 you would use file_get_contents()
In prior versions I would suggest:
$fp = fopen('filename','r');
$string = fread($fp, filesize('filename'));
fclose($fp);
The implode(file()) stuff is very memory-inefficient.
-Rasmus
On Thu, 18 Jul 2002, Analysis Solutions wrote:
On Thu, Jul 18, 2002 at 04:28:57PM -0400, Monty wrote:
Look 4 implode() in the function list
Implode isn't really what I need, I just want to load an entire file into a
single string variable.
Yes, it IS what you need. Plus, to work around your other problems
mentioned in later posts...
$string = implode('', file('filename') );
--Dan
--
PHP classes that make web design easier
SQL Solution | Layout Solution | Form Solution
sqlsolution.info | layoutsolution.info | formsolution.info
T H E A N A L Y S I S A N D S O L U T I O N S C O M P A N Y
4015 7 Av #4AJ, Brooklyn NY v: 718-854-0335 f: 718-854-0409
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Re: [PHP] Loading a File into Variable - How??
Andre, here's the function that worked for me...
function file_get_contents($filename) {
$fd = fopen($filename, r, 1);
$contents = fread($fd, 12000);
fclose($fd);
return $contents;
}
$page_string = file_get_contents(my_file.php);
The third parameter in fopen() [1] can be removed if you don't want to look
for files in your include_path. The second parameter in fread() [12000] is
where I hardcoded the filesize. Increase that number if you'll be opening
larger files.
Originally I had 12000 replaced with filesize($filename) but if the
$filename was opened from the include_path, this seems to always return
zero, which is why I hardcoded the byte size into fread().
Monty
From: [EMAIL PROTECTED] (Andre Dubuc)
Reply-To: [EMAIL PROTECTED]
Newsgroups: php.general
Date: Thu, 18 Jul 2002 17:45:14 -0400
To: Monty [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Loading a File into Variable - How??
Hi Monty,
I've been trying to do the same thing with no success. Would you be so kind
as to show me how you finally did it? I'm not too clear what you meant by:
So, if I replace the filesize($filename) command with a hard-coded number,
it works.
Tia,
Andre
On Thursday 18 July 2002 04:28 pm, you wrote:
I just want to load an entire file into a
single string variable.
However, I figured out the problem shortly after posting that first message
(of course). Because the file being opened is in the include_path, it seems
filesize() doesn't see those files. So, if I replace the
filesize($filename) command with a hard-coded number, it works.
Monty
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Re: [PHP] Loading a File into Variable - How??
Hi Rasmus,
file_gets_contents() doesn't work in my version of PHP (4.2.1). It says not
a valid function or something like that. Also, I discovered that the
filesize() function won't work on files fopened from the include_path. It
returns a value of zero, so, I had to hardcode the bytes into the fread().
From: [EMAIL PROTECTED] (Rasmus Lerdorf)
Newsgroups: php.general
Date: Thu, 18 Jul 2002 15:14:13 -0700 (PDT)
To: Analysis Solutions [EMAIL PROTECTED]
Cc: PHP List [EMAIL PROTECTED]
Subject: Re: [PHP] Loading a File into Variable - How??
In 4.3 you would use file_get_contents()
In prior versions I would suggest:
$fp = fopen('filename','r');
$string = fread($fp, filesize('filename'));
fclose($fp);
The implode(file()) stuff is very memory-inefficient.
-Rasmus
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