PHP dealing with Invalid argument and expects parameter's - php

Question

Good Evening, i am trying to teaching myself php, as i go and decided to try build something for our LAN server at work.
i have the following code that works and displays the images from a directory, i am using this as i am building a booking in system for work and naming the image by the job number.
for this i'm naming the images test but having a issue if there inst any images.
<?php
$directory = "saved_images/";
$images = glob("" . $directory . "test*.jpg");
$imgs = '';
foreach($images as $image){ $imgs[] = "$image"; }
$imgs = array_slice($imgs, 0, 20);
$result = count($imgs);
if ($result == 0)
{
$img="No Photos";
echo $img;
}
} else {
foreach ($imgs as $img) {
echo "<img src='$img' /> ";
}}
?>
the issue is, if there is not any photos i would like it to echo no photos instead of the following errors
array_slice() expects parameter 1 to be array,
refering to this line
$imgs = array_slice($imgs, 0, 20);
and
Invalid argument supplied for foreach()
refering to this line
foreach ($imgs as $img)
I have seen someone with a similar problem, but sadly they was advised to ignore the issue and turn off the error reporting, which didn't seem right, I am only asking as this inst causing any problem for the rest of the project and like to know how to fix this problem so if i encounter it again i know what to do.

Answer 1

All you need to do is set $imgs as an array (not a string), and check whether $imgs is empty or not. While you're at it, you will want to check if $images is empty or not, too.
$directory = "saved_images/";
$images = glob("" . $directory . "test*.jpg");
// since the entire script relies on $images not being empty,
// we should check for that to be sure before moving on
// you can also test for glob() returning FALSE on error, if you anticipate that it might
if ( ! empty($images) ) {
$imgs = []; // this should be set as an array, not a string
foreach ($images as $image)
{
$imgs[] = $image;
}
if ( empty($imgs) ) {
echo 'No Photo';
}
else {
$imgs = array_slice($imgs, 0, 20);
$result = count($imgs);
foreach ($imgs as $img)
{
echo "<img src='$img'> ";
}
}
}
else {
echo 'No images in ' . $directory . ';
}

Answer 2

Why are you initializing $imgs as a string?
$imgs = '';
and then treating it as an array?
foreach($images as $image){ $imgs[] = "$image"; }
If you'd initialize it as an array, e.g.
$imgs = array();
Then even if the foreach doesn't add anythign to the array, it'll still be an (empty) array when you pass it into array_slice.
Basically, you create a pizza, then wonder why PHP is complaining that it's not a chocolate cake.

Answer 3

wrap
$imgs = array_slice($imgs, 0, 20);
inside an if(isset()) statement, like so
if(isset($imgs)){
$imgs = array_slice($imgs, 0, 20);
}
If it still doesn't work add
&& count($imgs) > 0
to it
Edit:
And what #MarcB said, replace $imgs = ''; with $imgs = array();

Answer 4

You can use the php function is_array($myArrayVar) (PHP Net) to test if $img is an array before the line where you are using array_slice($imgs, 0, 20). Something like:
if (is_array($imgs)){
// $imgs is an array
} else {
//$imgs is not an array
}