Accessing element four places forward in array

[deshg]

Hey everyone,

I am recursing through an array (by using while (list($key, $value) = each($gidarray))). I am trying to reorder the array so at certain points as well as the current element I need to access the element four places forward from the current one. I then need to delete that element so it is not accessed when the loop continues.

If I use numerical keys (eg$i=0, while($gidarrary[$i]), $i++ etc) then I can just use $gidarrary[$i+4] to access the current element +4, however if I then use unset($gidarrary[$i+4]) it (understandably) removes the element (and key) completely and hence $i+4 then refers to a key that no longer exists. As a result I am using list/each so as to traverse the array so the key values are irrelevant. The problem is I cannot work out how to access an element four places on, the only thing I can see would work is to use the next() command 4 times, access the element and key value then use previous() to return to the current position, but this seems completely illogical and I’m sure there’s a more sensible way to do it.

I would really appreciate any help as this is really confusing me, obviously if you need more info then please say, I just didn’t want to put too much unnecessary stuff in here to start with.

Thanks so much as ever everyone,

Dave

[halojoy]

Can you use this isset() test inside the loop?

Then I would try use the continue; variant
http://php.net/manual/en/control-structures.continue.php

// $i loop start
if(!isset( $gidarrary[$i] ))
    continue;
//do something
//here
//loop end


//     or more normal way:

// $i loop start
if(isset( $gidarrary[$i] )){
    //do something
    //here
}
//loop end

[laserlight]

deshg wrote:

I am trying to reorder the array so at certain points as well as the current element I need to access the element four places forward from the current one.

What kind of reordering are you trying to do?

[deshg]

Thanks for your reply, i'm not sure if i can really use isset and gidarray[$i] because when i then unset($gidarray[$i+4]) and the iteration continues, when $i=$i+4 it will be unset and the loop will stop rather than continuing to the end if you see what i mean? I'm also not convinced even if this is possible (and i'm wrong which is quite possible!) that using continue will help, i still want the loop to complete that iteration and keep the pointer the same, i just need to delete some items from further up the array. Initially i wanted to do this, but i don't think it's possible:

$array=array('one', 'two', 'three', 'four');
hence:
array[0]='one';
array[1]='two';
array[2]='three';
array[3]='four';
CODE TO DELETE array[1] which removes element and updates numerical keys
array ends up as (notice key changes):
array[0]='one';
array[1]='three';
array[2]='four';

I don't think this is possible and i can see why, but i thought i'd mention it as maybe it is or maybe it makes what i'm trying to do clearer?

Thanks so much for your help, it really is greatly appreciated,

Dave

[deshg]

laserlight;10942995 wrote:

What kind of reordering are you trying to do?

I'm making a gallery of thumbnails whereby if one user has more than 3 it will show them all in the same row (each row has 3 columns) and reorder the other thumbnails around it so the grid looks the same (and there aren't spaces). It;s a bit of a mathematical nightmare as it's ordered by date, but groups galleries by the same user together and then if there are 3 galleries by the same user it reorders the other gallery thumbs around it. I'm not sure if that makes sense or makes it any clearer!!!

Thanks!!

[laserlight]

Ah, this sounds clearer now 🙂

deshg wrote:

I'm making a gallery of thumbnails whereby if one user has more than 3 it will show them all in the same row (each row has 3 columns) and reorder the other thumbnails around it so the grid looks the same (and there aren't spaces).

How exactly is this re-arrangement of "the other thumbnails around it so the grid looks the same" supposed to look like?

Basically, I suspect that if you described in detail the bigger picture of what you are trying to do, someone may come up with a workable solution to your problem, which is better than giving you an idea of a hack to make your not-so-elegant solution work.

[deshg]

Fair point! 🙂 The layout is 3 columns by 4 rows with 1 thumbnail in each. If it encounters 3 thumbnails in a row from the same user then it should reorder the surrounding content so this starts on it's own row by pulling thumbnails from after it to fill up the space. Eg:

A | B | C
D | E | E
E | F | G
H | I | J

In the example above each letter represents a thumbnail from a specific user, E has 3 thumbnails and everyone else has 1. In the example above i am tryin to reorder it so that it is:

A | B | C
D | F | G
E | E | E
H | I | J

Does that make sense or is that even more confusing?

Thanks so much for your help,

Dave

[johanafm]

Get your initial thumbnail data into an array of users, where each user has an array of thumbnails

$user_thumbs[A][] = 'thumbnail';
$user_thumbs[D][] = 'thumbnail';
$user_thumbs[E][] = 'thumbnail';
$user_thumbs[E][] = 'thumbnail';
$user_thumbs[E][] = 'thumbnail';
$thumbs[F][] = 'thumbnail';

Then iterate over user_thumbs and for each user check how many thumbnails they have

$three_thumb_users = array();
$i = 0;
foreach ($user_thumbs as $user => $thumbs) {
	// could of course skip the $new_row variable and do this in the if check below.
	if ($i % 3 == 0)
		$new_row = true;
	else
		$new_row = false;
	# it's a new row. if we've previously encountered users with 3 thumbs they are found
	# in $three_thumb_users now.
	if ($new_row == true && !empty($three_thumb_users)) {
		$thumbs = array_shift($three_thumb_users);
		# use $thumbs for output. contains three thumbs.
	}
	# not a new row, so skip users having 3 thumbs
	else if (count($thumbs) == 3) {
		$three_thumb_users[$user] = $thumbs;
		unset($thumbs[$user]);
	}
	# neither a new row or user having three thumbs. so just output away.
	else {
		output the one or two thumbs.
	}
	++$i;
}
if (!empty($three_thumb_users)) {
	# left over user with three thumbs. needs to be output as well.
}