b) output an appropriate header for the file type c) pass through the actual file contents
if I were doing an HTML file - if I were doing a PDF file?
Todd
Justin French wrote:
Same way
<img src='image.php?file=myimage.gif' />
image.php would do simular things to what view.php does in the script below.
Justin
on 10/03/03 2:07 PM, Mark Tehara ([EMAIL PROTECTED]) wrote:
On that note, how would i load an image from outside the document root?
----- Original Message ----- From: "Justin French" <[EMAIL PROTECTED]> To: "Todd Cary" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]> Sent: Monday, March 10, 2003 4:06 PM Subject: Re: [PHP] Displaying a file
Change your link to something like:
<a href="view.php?file=raceschedule.pdf" ...>...</a>
view.php will NOT be a "HTML page" -- it will be responsible for:
a) some conditional stuff, like checking for a logged in user b) output an appropriate header for the file type c) pass through the actual file contents
You would actually want to store the target files outside the doc root, or forbid apache to serve them directly over http.
There's a decent article here: http://www.zend.com/zend/trick/tricks-august-2001.php
Justin French
on 10/03/03 1:17 PM, Todd Cary ([EMAIL PROTECTED]) wrote:
I want to display a file under program control in the same manner as one would with using an >a> tag e.g.
Click <a href="files/raceschedule.pdf" Name="Race Schedule" Target="_blank">here</a> to open the Race Schedule
In other words, if certain conditions are met, then I want to display the file. What is the best way to do that?
Many thanks.....
Todd
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