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Modulus in a PHP loop

Tags:

php

modulus

I'm currently checking whether an entry in a loop is the third iteration or not, with the following code:

<?php for ($i = 0; $i < count($category_news); $i++) : ?>

    <div class="grid_8">
        <div class="candidate snippet <?php if ($i % 3 == 2) echo "end"; ?>">
            <div class="image shadow_50">
                <img src="<?php echo base_url();?>media/uploads/news/<?php echo  $category_news[$i]['url']; ?>" alt="Image Preview" width="70px" height="70px"/>
            </div>
               <h5><?php echo $category_news[$i]['title']?></h5>
            <p><?php echo strip_tags(word_limiter($category_news[$i]['article'], 15)); ?></p>
            <?php echo anchor('/news/article/id/'.$category_news[$i]['news_id'], '&gt;&gt;', array('class' => 'forward')); ?>
        </div>
    </div>

    <?php if ($i % 3 == 2) : ?>
         </li><li class="row">
    <?php endif; ?>

<?php endfor; ?>

How can I check if the loop is in its second and not its third iteration?

I have tried $i % 2 == 1 to no avail.

like image 684
Udders Avatar asked Nov 15 '11 11:11

Udders


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How can I get modulus of a number in PHP?

The fmod() function returns the remainder (modulo) of x/y.

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The modulus operator returns the remainder of a division of one number by another. In most programming languages, modulo is indicated with a percent sign. For example, "4 mod 2" or "4%2" returns 0, because 2 divides into 4 perfectly, without a remainder.

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The typical usage of mod is for generating values inside a fixed range. In this case, you want values that are between 0 and strlen("abc")-1 so that you don't access a position outside "abc" . The general concept you need to keep in mind is that x % N will always return a value between 0 and N-1 .

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2 Answers

Modulus checks what's the leftover of a division.

If $i is 10, 10/2 = 5 with no leftover, so $i modulus 2 would be 0.
If $i is 10, 10/3 = 3 with a leftover of 1, so $i modulus 3 would be 1.

To make it easier for you to track the number of item i would start $i from 1 instead of 0. e.g.

for($i=1; $i <= $count; $i++)
    if($i % 2 == 0) echo 'This number is even as it is divisible by 2 with no leftovers! Horray!';
like image 115
Shai Mishali Avatar answered Oct 15 '22 11:10

Shai Mishali


When in doubt, write a snippet of code:

for ($j = 1; $j < 4; $j++)
{
   for ($k = 0; $k < $j; $k++)
   {
      echo "\n\$i % $j == $k: \n";

      for ($i = 0; $i < 10; $i++)
      {
         echo "$i : ";
         if ($i % $j == $k)
         {
            echo "TRUE";
         }
         echo " \n";
      }
   }
}

Here is the output. Use it to figure out what you need to use:

$i % 1 == 0: 
0 : TRUE 
1 : TRUE 
2 : TRUE 
3 : TRUE 
4 : TRUE 
5 : TRUE 
6 : TRUE 
7 : TRUE 
8 : TRUE 
9 : TRUE 

$i % 2 == 0: 
0 : TRUE 
1 :  
2 : TRUE 
3 :  
4 : TRUE 
5 :  
6 : TRUE 
7 :  
8 : TRUE 
9 :  

$i % 2 == 1: 
0 :  
1 : TRUE 
2 :  
3 : TRUE 
4 :  
5 : TRUE 
6 :  
7 : TRUE 
8 :  
9 : TRUE 

$i % 3 == 0: 
0 : TRUE 
1 :  
2 :  
3 : TRUE 
4 :  
5 :  
6 : TRUE 
7 :  
8 :  
9 : TRUE 

$i % 3 == 1: 
0 :  
1 : TRUE 
2 :  
3 :  
4 : TRUE 
5 :  
6 :  
7 : TRUE 
8 :  
9 :  

$i % 3 == 2: 
0 :  
1 :  
2 : TRUE 
3 :  
4 :  
5 : TRUE 
6 :  
7 :  
8 : TRUE 
9 :  
like image 35
Gustav Bertram Avatar answered Oct 15 '22 11:10

Gustav Bertram