John Davidson

Message:

I'm just trying to loop through the array and echo it 3 times inside the div but here I got confused and why is it throwing me an undefined array key. If I put fixed numeric value inside $arr[i] then its running properly but why isn't it picking up the value of i? What am I missing?
Please help me to solve the problem and Thanks in advance.
Here's the code below.

   <?php



      $arr = array("Gallery-1.jpg","Gal-2.jpg","Gal-3.jpg","gal-4.jpg","gal-5.jpg");



      $i = 0;



      for ($i=0; $i<3; $i++) {

        echo "<div class='col-lg-4 col-md-6 portfolio-item filter-programs'>

        <img src='assets/img/portfolio/.$arr[i].' class='img-fluid' alt=''>

        <div class='portfolio-info'>

          <h4>Programs 1</h4>

          <p>Programs</p>

          <a href='assets/img/portfolio/.$arr[i].' data-gallery='portfolioGallery'

            class='portfolio-lightbox preview-link' title='App 1'><i class='bx bx-plus'></i></a>

          

        </div>

      </div>";

      }

      ?>

---

My Answer:
It looks like you are trying to access the array key using the variable `$i` inside the loop. However, in PHP, you need to use square brackets `[]` to access array elements, not parentheses `()`.

Here is the corrected code:

php
$arr = array("apple", "banana", "cherry");

for ($i = 0; $i < count($arr); $i++) {
echo "

" . $arr[$i] . "

";
}


This code will loop through the array and echo each element inside a `