Return current position of a page in page array

[Henrik]

How can I identify the current position of a page in a page array? I just want to know if the position is odd or even, to alternately place an image left or right. 

Lets say I have this code:

if($page->children) {
	foreach($page->children as $child) {
		// ...
		if($child->... position is odd ...) { 
			echo "<div class='image_left'> ...";
        	}
        	if($child->... position is even ...) { 
			echo "<div class='image_right'> ...";
        	}
        	// ...
	}
}

How can I identify if the current position of the page in the page array is odd or even?

Thank you 

[arjen]

Hey Hendrik,

In this thread you'll find different solutions.

No pun intended and I don't know if you've searched, but next time you could try Google first. The forum is loaded with great examples. Try some variations on keywords.

[diogo]

See Arjen's post for more info, but for your case would be more than enough to use the PHP classical way:

$i = 1;

foreach($page->children as $child) {

    if($i % 2 == 0) { 
		// even
    } else {
        // odd
    }
          
    $i++;
}

 

[Henrik]

Thanks, Arjen for your help! I found a solution in your post: 

I guess this is developer-basics. Now I learned this one too. 
Thank you!

And yes, you are right. I've indeed searched, but not so thoroughly I probably should. I'm a little bit in a hurry right now. Sorry and Thanks again! 

@diogo: Thank you too  Didn't saw your post. This was what I did. 

[arjen]

No problem mate. Glad you got it working. 

[LostKobrakai]

There's not even the need to manually count the number. 

foreach($page->children->getValues() as $key => $child) {
    if($key % 2 == 0) { 
		// even
    } else {
        // odd
    }
}