Simplified anarchy "if statement" - php

Question

Is there a way to code this function/executable..
if($img1 > "")
{
$show = $img1;
}
elseif($img2 > "")
{
$show = $img2;
}
elseif($img3 > "")
{
$show = $img3;
}
else
{
$show = 'default.jpg';
}
echo $show;
..in a more simple way? Thx.

Answer 1

Any time you end up using variables named 1,2,3,etc it means you should be using an array.
If you had it like this, for example:
$img = [
1=> '',
2=> '',
3=> 'image.jpg'
];
You could then process this like so:
// Remove empty values
$img = array_filter($img);
// Echo the first value found (image.jpg)
echo current($img);
Or to add your default:
echo (count($img)) ? current($img) : 'default.jpg';
This uses a ternary operator to echo 'default.jpg' if count($img) is 0.

Answer 2

If you know the id number of the image, you could do something like this:
<?php
$images = array(
'myimg.jpg',
'my_awesome_img.jpg',
'awesome_awesome.jpg'
);
$show = $images[$x];
?>

Answer 3

You can just do
if(!empty($img1)) {
echo $img1;
} elseif(!empty($img2)) {
echo $img2;
} elseif(!empty($img3)) {
echo $img3;
} else {
echo 'default.jpg';
}
given that you only ever want to display one of these.

Answer 4

Did not test, but something like this:
<?php
$img1 = 'hello';
$img2 = 'world';
$array = [1,2,3];
for($i=1; $i<count($array); $i++){
$v = ${"img{$i}"};
if( !empty($v) ){
$show = $v;
break;
}
}
echo $show;
No idea what you're trying to accomplish or why the code looks like that, seems poorly constructed from the get-go.