On Thu, 2003-06-19 at 20:19, Mark Tehara wrote:
?php
$myimage=..//..//site//images//2.jpg;
header(Content-type: image/jpeg);
fopen ($myimage, r);
readfile($myimage) ;
fclose($myimage);
?
This seemed to do the trick for me. Thank you.
/ Mark
Even better, you don't even need the
Ahh, the Docs don't tell you that.
They are a little weard wehen it comes to that
- Original Message -
From: Lars Torben Wilson [EMAIL PROTECTED]
To: Mark Tehara [EMAIL PROTECTED]
Cc: PHP [EMAIL PROTECTED]
Sent: Friday, June 20, 2003 10:53 PM
Subject: Re: [PHP] [Newman] Passing
On Fri, 2003-06-20 at 12:07, Mark Tehara wrote:
Ahh, the Docs don't tell you that.
They are a little weard wehen it comes to that
How so?
int readfile ( string filename [, bool use_include_path [, resource
context]])
It does say 'filename', not 'resource'. Seems pretty unequivocal.
try this:
?php
header(Content-type: image/jpeg);
fopen (d:/crushme/images/1.jpg, r);
?
- Original Message -
From: Mark Tehara [EMAIL PROTECTED]
To: PHP [EMAIL PROTECTED]
Sent: Thursday, June 19, 2003 10:58 AM
Subject: [PHP] [Newman] Passing an image through a php file.
I would like to
try this:
?php
header(Content-type: image/jpeg);
fopen (d:/crushme/images/1.jpg, r);
Just my opinion, but maybe you should try reading the file and echoing the
data instead of just opening it. :)
---John Holmes...
I would like to pass an image throught
PROTECTED]; Mark Tehara
[EMAIL PROTECTED]; PHP [EMAIL PROTECTED]
Sent: Friday, June 20, 2003 1:14 AM
Subject: Re: [PHP] [Newman] Passing an image through a php file.
try this:
?php
header(Content-type: image/jpeg);
fopen (d:/crushme/images/1.jpg, r);
Just my opinion, but maybe you
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