Display a condition from an array - php

Question

Need help guys, below is a var_dump array, I am very noob in php and I am very confused on how to do this, basically all I need is a condition
$fittin_colour_image = get_field('featured_images', get_the_ID());
Above is a code using ACF
var_dump($fittin_colour_image);
below is the result
array (size=2)
0 =>
array (size=3)
'fitting_colour' =>
array (size=1)
0 => string 'Black' (length=5)
'image' => string 'http://localhost/mysite.com/wp-content/uploads/2018/04/image-black.jpg' (length=101)
'image_description' => string '' (length=0)
1 =>
array (size=3)
'fitting_colour' =>
array (size=1)
0 => string 'White' (length=5)
'image' => string 'http://localhost/mysite.com/wp-content/uploads/2018/05/image-white.png' (length=100)
'image_description' => string '' (length=0)
I want to have a condition like this
if this array has a color "Black" then display its image
I am currently using this code
if ($fittin_colour_image[0]['fitting_colour'][0] == 'Black') {
echo $fittin_colour_image[0]['image'];
}
its a bit of a hassle since every time I change the condition I change the [0], sorry but I am confused now, don't know what to say, please help

Answer 1

You're very close. Instead of hard-coding the index (0 in your case) you can, instead, loop over the array and display the image:
foreach ($fittin_colour_image as $image) {
if ($image['fitting_colour'][0] == 'Black') {
echo $image['image'];
}
}

Answer 2

test this
$fittin_colour_image = get_field('featured_images', get_the_ID());
if(!empty($fittin_colour_image)){ // Test if array not empty
foreach ($fittin_colour_image as $image) {
if(!empty($image['fitting_colour'][0]) && $image['fitting_colour'][0] === 'Black'){
echo '<img src="'.$image['image'].'" alt="'.$image['image_description'].'" />';
}
}
}