Array_slice returning the value "array"

Question

I'm trying to display an image on screen by reading the contents and array and using that value to fill in part of the file name.

In the code below the array has 3 values, 100, 101 and 102. I'm trying to take the first value "100" to output

If I just set $image to the value of "100" it works fine. When I set $image as an array_slice, and echo $image, it returns "array" instead of "100".

<?php

$dirname = "images/wrf/";
$array=array("100", "101", "102");
$image = array_slice($array, 0, 1);

echo '<img src="' . $dirname . 'wrf' . $image . 'medium.jpg">';

//out put for debugging 
print_r($array);
echo "contents of image===";
echo $image;

?>

php.net/array_slice array array_slice ( array $array , int $offset [, int $length = NULL [, bool $preserve_keys = false ]] ) it clearly says that it returns an array. — Charlotte Dunois
– Charlotte Dunois, Commented Apr 18, 2016 at 16:10

Henders · Accepted Answer · 2016-04-18 16:12:37Z

4

array_slice result is array according to the docs,

if you'd like to get one element of array use [] operator instead

$image = $array[0];

edited Apr 18, 2016 at 16:12

Henders

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answered Apr 18, 2016 at 16:02

Kamil Karkus

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3 Comments

DigitalFirefly Over a year ago

Thanks, that did it.

DigitalFirefly Over a year ago

Is there a way to increment the 0 in $image = $array[0]; ? I'd like to take each value in the array and display the image for each one.

DigitalFirefly Over a year ago

perfect, that did it.

Jakir Hossain · Accepted Answer · 2016-04-18 16:15:03Z

1

$image = array_slice($array, 0, 1);

output of $image

Array
(
    [0] => 100    
)

$image = $array[0];

output of $image

100

see edited code.

$dirname = "images/wrf/";
$array=array("100", "101", "102");
$image = $array[0];

echo '<img src="' . $dirname . 'wrf' . $image . 'medium.jpg">';

answered Apr 18, 2016 at 16:15

Jakir Hossain

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