Echo php array value

Question

I have the following code that reads the content of a folder

PHP Code:

<?
//PHP SCRIPT: getimages.php
header('content-type: application/x-javascript');

//This function gets the file names of all images in the current directory
//and ouputs them as a JavaScript array
function returnimages($dirname="./images") {
     $pattern="(\.jpg$)|(\.png$)|(\.jpeg$)|(\.gif$)"; //valid image extensions
     $files = array();
     $curimage=0;
           if($handle = opendir($dirname)) {
                while(false !== ($file = readdir($handle))){
                     if(eregi($pattern, $file)){ //if this file is a valid image 
//Output it as a JavaScript array element
                   echo 'galleryarray['.$curimage.']="'.$file .'";' . "\n";
             $curimage++;
                    } 
            }

            closedir($handle);
}
        sort($files);
        return($files);
}

echo 'var galleryarray=new Array();' . "\n"; //Define array in JavaScript
returnimages() //Output the array elements containing the image file names

?>

which spits out this the following code:

var galleryarray = new Array();
galleryarray[0] = "image1.jpg";
galleryarray[1] = "image5.jpg";
galleryarray[2] = "image2.jpg";
galleryarray[3] = "image4.jpg";
galleryarray[4] = "image8.jpg";
galleryarray[5] = "image7.jpg";
galleryarray[6] = "image0.jpg";
galleryarray[7] = "image3.jpg";
galleryarray[8] = "image6.jpg";

Now in my html file I want to echo something like

<img src="images/image0.jpg" />
<img src="images/image1.jpg"/>
...
...

How can I do that?

Dion · Accepted Answer · 2012-04-13 17:19:57Z

1

foreach($galleryarray as $img) {
    echo "<img src='images/$img' />";
}

edited Apr 13, 2012 at 17:19

answered Apr 12, 2012 at 8:05

Dion

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Comments

phsaires · Accepted Answer · 2012-04-12 09:52:32Z

0

maybe this help u

$pattern="(\.jpg$)|(\.png$)|(\.jpeg$)|(\.gif$)"; //valid image extensions
           if($handle = opendir("./images")) {
                while(false !== ($file = readdir($handle))){
                     if(eregi($pattern, $file)){ ?>
                   <img src="/images/<?php echo $file;  ?>">
            <?php 
                    } 
            }
           }

answered Apr 12, 2012 at 9:52

phsaires

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Comments

Community · Accepted Answer · 2023-11-17 19:55:19Z

0

assuming that you want to sort image name by value and then show the images. you first need to sort values by using function "asort"

    asort($galleryarray);
    foreach ($galleryarray as $key => $val) {
        echo "<img src='images/{$val}'/>"
    }

if you want to print and array

print_r($array); or if you want nicely formatted array then

echo ''; print_r($array); echo '';
use var_dump($array) to get more information of the content in the array like datatype and length.
you can loop the array using php's foreach(); and get the desired output. more info on foreach in php's documentation website https://www.php.net/manual/en/control-structures.foreach.php

edited Nov 17, 2023 at 19:55

CommunityBot

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answered Jun 8, 2014 at 15:33

Muhammad Tahir

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Comments

BenMorel · Accepted Answer · 2014-06-26 16:14:55Z

0

Try changing the echo line in the function to this:

echo '<img src="images/' . $file . '" />' ;

Hope this will help :)

OR

Outside the function, use it like this:

$images = returnimages(); //will get the array containing the images
foreach($images as $img)
{
  echo '<img src="images/' . $img . '" />';
}

Also, you have to include this line inside the if condition in the while loop:

$files[] = $file; //insert the file into the array. This array is what we are going to return from the function

Update

I have tested this code and it's working:

    //PHP SCRIPT: getimages.php
    header('content-type: application/x-javascript');

    function returnimages($dirname="./images") {
         $pattern="([^\s]+(\.(?i)(jpg|png|gif|bmp))$)";     //  http://www.mkyong.com/regular-expressions/how-to-validate-image-file-extension-with-regular-expression/
         $files = array();
         if($handle = opendir($dirname)) {
            while(false !== ($file = readdir($handle))){
                if(preg_match($pattern, $file)){ //if this file is a valid image 
                    $files[] = $file;
                } 
            }

            closedir($handle);
        }
        //sort($files);         // http://php.net/manual/en/function.sort.php
        natcasesort($files);    // case insensitive "natural order" algorithm :: http://php.net/manual/en/function.natcasesort.php

        return($files);
    }

    $images = returnimages(); //will get the array containing the images
    foreach($images as $img)
    {
      echo '<img src="images/' . $img . '" />';
    }

In my images folder, I have put 5 files for testing. And upon running it, it would give the following:

<img src="images/a.jpg" />
<img src="images/ba.jpg" />
<img src="images/ca.jpg" />
<img src="images/Copy (3) of a.jpg" />
<img src="images/Copy (4) of a.jpg" />

edited Jun 26, 2014 at 16:14

BenMorel

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answered Apr 12, 2012 at 8:07

Akhilesh B Chandran

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4 Comments

user1325414 Over a year ago

is there a way to sort the file name alphabetically?

user1325414 Over a year ago

no i tried sort($file) and sort($files) it didnt work. where should i put them?

user1325414 Over a year ago

the code works great. how can i print or echo the output in my html file?

Akhilesh B Chandran Over a year ago

The last 5 lines of code will do the trick. First we are calling the function returnimages() and then assigning the array returned by the function into a variable $images. This would be an array. The next line, that is the foreach loop will iterate through each element in the array. Inside the loop, I'm echoing the images wrapping HTML img tags.

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