input type = "image" - where's the variable?

[backpacklondon.com]

Hello,

I'm trying to find out which form button a user clicked.

Because all of my from buttons are images I'm using:

<form action ="?">
<input type ="image" src="button1.gif" name = "Next" >
<input type ="image" src="button2.gif" name = "Prev" >
</form>

Then when my php script receives this form, how do I test for which button was pressed please?

I've tried:

if ($Next):
echo "next was pressed<br>";
endif;

if ($Prev):
echo "Prev was pressed<br>";
endif;

and also.....


if ($Next.x):
echo "next was pressed<br>";
endif;

if ($Prev.x):
echo "Prev was pressed<br>";
endif;

but none of these identify the button uniquely.

Does anyone know if it's possible to determine which button was pressed using input type ="image" please?

Thank you,

Jason

[freddydoesphp]

Jason I do belive it is
if ($Next_x > 0) {

}

[Son Nguyen]

Yeah, why is the variable's name weird, Freddy?
I've never tried that before, but if s.o. does some tests, please let me know if it works

[freddydoesphp]

Just tested it and it works.

[Son Nguyen]

But why it has an _x after the name? Is that a rule? And how would you find such documentation? Thanks, Freddy

[freddydoesphp]

Actually I found out in another forum a while back, when I needed to do it, I couldn't find it in the docs, and yes I do belive its a rule, but I think its more of an html-server-side image map thing, basically your clicking on an image and the server knows how to get the coordiantes of where you clicked somehow.

[freddydoesphp]

But what is weird is that if you use a get method in your form you will see imagename.x and imagename.y with the coordinates to where you clicked on the image, but you have to use imagename_x and imagename_y when you are checking to see whther the image was clicked, weird????

[backpacklondon.com]

Thank you Freddy!

Regards,

Jason