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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>> 
> 
> 


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