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European Lee

11-15-2004 05:56 AM

PHP include Question...

Does anyone know how to change this code..

PHP Code:


		
			
<?include("directory/4/0".date("w").".php");?>

So that an image will appear in the cell where the code is placed?

I have tried this..

PHP Code:


		
			
<?include("directory/4/0".date("z").".jpg");?>

But it spews out a bunch of crap as if it reading the encoding for the image itself.

Anyone know enough about php to answer this?

Regards,

Lee

Pete-KT

11-15-2004 05:57 AM

Sorry i dont carre my magnifying glass with me and with the color and size of that font I CANT read it.

Pete-KT

European Lee

11-15-2004 06:00 AM

Quote:

Originally posted by Pete-KT
Sorry i dont carre my magnifying glass with me and with the color and size of that font I CANT read it.

Pete-KT

Thanks for that highly intelligent, well spelt and helpful reply :thumbsup

Regards,

Lee

etech

11-15-2004 06:02 AM

Hmm..
Why dont you just declare a variable with the file string.

PHP Code:


		
			
$file "directory/4/0".date("w").".php";

<img src=$file border=0>

European Lee

11-15-2004 06:03 AM

Quote:

Originally posted by etech
Hmm..
Why dont you just declare a variable with the file string.

PHP Code:

$file "directory/4/0".date("w").".php"; <img src=$file border=0>

Wont that mean i have to create 365 individual .php files too?

Regards,

Lee

etech

11-15-2004 06:04 AM

Quote:

Originally posted by etech
Hmm..
Why dont you just declare a variable with the file string.

PHP Code:

$file "directory/4/0".date("w").".php"; <img src=$file border=0>

sorry, you need to print or echo the img code

European Lee

11-15-2004 06:07 AM

Quote:

Originally posted by etech
sorry, you need to print or echo the img code

And i would do that how? LOL

Sorry i know less about .php than i do about sex with women :1orglaugh

Regards,

Lee

etech

11-15-2004 06:08 AM

Not sure what you mean,
you could just make 1 php file, and then
call it like *.php?date=xxxxx

and in the file you just feth that variable

PHP Code:


		
			
$date = $_GET["date"];

etech

11-15-2004 06:10 AM

hit me up at ICQ, i got a few minutes :)

Nathan

11-15-2004 06:11 AM

lee,

what you want to do is this:

PHP Code:


		
			
<img src="directory/4/0<?=date("z")?>.jpg">

etech

11-15-2004 06:33 AM

Quote:

Originally posted by Nathan
lee,

what you want to do is this:

PHP Code:

<img src="directory/4/0<?=date("z")?>.jpg">

true,
i just like to keep it in variables :thumbsup

European Lee

11-15-2004 06:39 AM

Thanks guys got it sorted out :thumbsup

Regards,

Lee

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