Michael Steinmann

16 hours ago, Sérgio Jardim said:

@Soma wrote about his solution related to your problem, take a look:

Thank you @Sérgio Jardim

I will have a look at it and try to get it to work. At least it sounds exactly like what i was trying to do.

I will respond if i am successful.

Hello everyone,

i guess my php knowledge is not good enough to find a solution for the following problem:

I'm building a website where i have projects. Each project has tags like "webdesign, responsive, cms, print, flyer" etc. At the end of a project i want to recommend other projects, which have similar tags. I want to limit the recommendations to three other projects sorted by the hightest amount of same tags.

This is how far i got:

I'm using

<?php $interessantes = $pages->find("tags=$page->tags, id!=$page->id"); ?>

to find other projects with the same tags, excluding the page i'm already on.

Then i'm using three foreach-functions to go through the projects i found, the tags they have and compare them with the tags of the project i'm looking at. If i have two identical tags, i count up a variable.

<?php

// Projects i have found
foreach($interessantes as $interessant):

	$i = 0;

	// Tags of the projects i have found
	foreach($interessant->tags as $tag):

		// Tags of the project i'm looking at
		foreach($page->tags as $tagreferenz):

			if($tag == $tagreferenz):

				$i++;

			endif;

		endforeach;

	endforeach;

endforeach; 

?>

You can see what it looks like in the screenshot with a bit of HTML. I marked the tags of the current project green and the identical tags of other projects red with the amount of hits below. So in this example i would want to have three projects with three similar tags and get rid of those with only two and one.

I guess i have to put my pages into an array (with the number of hits?), sort them and echo them with "limit=3"? Unfortunately i have no idea how to do this. You probably have an even better/shorter solution. Links to other topics are welcome to and i'll try to get my head around it.

If you need further explanation, i'll try my best.

Best regards