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DDStudios
says

I want to target these element for my 3 collumn portfolio page and I did something like this :

jQuery('.page-template-portfolio-full-php #portfolioItem li:nth-child(3n)').css({ marginRight: '0px'});
        jQuery('.page-template-portfolio-full-php #portfolioItem li:nth-child(6n)').css({ marginRight: '0px'});
         jQuery('.page-template-portfolio-full-php #portfolioItem li:nth-child(9n)').css({ marginRight: '0px'});
          jQuery('.page-template-portfolio-full-php #portfolioItem li:nth-child(12n)').css({ marginRight: '0px'});

I know it’s not the best way and I know I could do something like i++ or something but im not sure exactly what it is ..

I want to do the same thing for the 2 collumn .. only this time target every 2nd, 4th, 6th element etc etc ..

How is it done?

Thanks guys!

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Ivor
Envato team
says

There you go.

jQuery('#portfolioItem li:nth-child(3n+3)').css({ marginRight: '0px' });
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MCStudios
says

Hi DD

I think it’s easier with i++ you can do it this way:

$counter =0;

query_posts  <--- your normal query

if ( have_posts() )  ,etc. you know the rest

++$counter;
      if($counter == 3) {
        $postclass = 'last';
        $counter = 0;
      } else { $postclass = ''; }
<li class="&lt;?php echo $postclass; ?&gt;">
       your code.....
</li>

This will add the class last to every 3rd, 6th, 9th, etc elements. and then you just need your css.

.last{margin-right: 0 !important;}

Edit: Ivor was faster :)

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theMOLITOR
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Ivor
Envato team
says

Hi DD

I think it’s easier with i++ you can do it this way:

$counter =0;

query_posts  <--- your normal query

if ( have_posts() )  ,etc. you know the rest

++$counter;
      if($counter == 3) {
        $postclass = 'last';
        $counter = 0;
      } else { $postclass = ''; }
<li class="&lt;?php echo $postclass; ?&gt;">
       your code.....
</li>

This will add the class last to every 3rd, 6th, 9th, etc elements. and then you just need your css.

.last{margin-right: 0 !important;}

Edit: Ivor was faster :)

Mine is a single line :D

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DDStudios
says

Damn, I knew it was simple :)

Thanks my friend.

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MCStudios
says

Mine is a single line :D

Man i need to learn more about jQuery :)

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Ivor
Envato team
says


Mine is a single line :D
Man i need to learn more about jQuery :)

But your solution gave me an idea for a problem I’m having here haha! Thanks.

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Siddharth
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Yep, the nth child selector is the way to go. But keep in mind that this take a performance hit if the number of portfolio items is huge. [Actually huge.]

In those cases, you’re better of checking the position of an entry and assigning an appropriate class through server side code, PHP in this case, I’m assuming. Then you can merely use the class selector to manipulate specific elements in the front end, as needed.

This way, computation is offloaded to the server instead of the client having to do the job. Again, this only applies if you have a huge number of elements to parse. Otherwise, the simpler method noted above should do.

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revaxarts
says
ok, here comes mine:
jQuery('#portfolioItem').find('li').each(function(i,e){
   if(!i%3) $(e).css({ marginRight: '0px' });
});

The child selector isn’t good for large list. The id selecter is the fastest and the find method is fast too.

Far better is to assign a class which is slightly faster than the css:

jQuery('#portfolioItem').find('li').each(function(i,e){
   if(!i%3) $(e).addClass('last');
});

Anyway, I recommend the latest jquery version but at least 1.4.4.

Edit: and in one line:
jQuery('#portfolioItem').find('li').each(function(i,e){if(!i%3)$(e).addClass('last');});
;)
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