Display Image (BUT....)

[Warptweet]

COMPLETE EDIT:

 

Okay, I know the function...

 

<?php
$imagesize = getimagesize(http://www.warptweet.com/warptweetslogo.png)
?>

 

BUT, not how can I do this..?

 

IF $imagesize OVER 75x75

do this action

IF $imagesize UNDER 75x75

do this action

[trq]

Take a look at getimagesize().

[Warptweet]

Yes, I looked at that.

 

But now I have another problem.

 

how can I make it so that IF getimagesize(blahblahblah) is over 75x75 or under 75x75?

[Jessica]

If you actually try to write some code we can help. But all you did was change your question to use the word thorpe posted.

[Warptweet]

Well, i don't really have much of a code since I don't really understand the function..

 

<?php
$imagesize = getimagesize(http://www.warptweet.com/warptweetslogo.png);
if $imagesize < 75{
do this action
}
else
do this action

 

So I basically need help with two things...

the line "if $imagesize <75" probably does not mean 75x75 in size, so I need a way to make the computer understand if $imagesize is over 75x75.

 

And if $imagesize is over 75x75, then do an action.

Else, then do a different action.

[Jessica]

Did you read the manual?

getimagesize returns an array of information. You'll need to compare the width and the height.

[Warptweet]

I read the manual.

 

I'll see what your trying to say, thanks.

[Warptweet]

Yay! It works!

 

Now can you please just help me with one more quick thing?

 

<?php

list($width, $height) = getimagesize("http://www.warptweet.com/warptweetslogo.png");

 

if ($width > 100){

if ($height > 100){

do this action

}

}

else

do this action

?>

 

My question: Is that how I would do it? Or is the > and < thingy wrongly coded?

[Jessica]

if($width > 75 || $height > 75){

is if either or.

if($width > 75 && $height > 75){

is if both.