Solved

php - display images

Posted on 2012-04-09
12
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Last Modified: 2012-04-10
hello there,
I have about 30 images in one folder and I would like to display 3 random images but sometimes when displaying randomly I get two images that are the same.. how can I prevent from showing the same image?
0
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Question by:XK8ER
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12 Comments
 
LVL 36

Expert Comment

by:Loganathan Natarajan
ID: 37826375
can u post your code?
0
 
LVL 1

Author Comment

by:XK8ER
ID: 37826384
it just reads a random image from a path of images..
0
 
LVL 29

Expert Comment

by:Olaf Doschke
ID: 37826554
You create three random numbers in the range of 1 to filecount and if two or all of them are the same generate one or two new numbers until you have three different numbers.

Bye, Olaf.
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LVL 1

Author Comment

by:XK8ER
ID: 37826596
this is what I have

<?php
ini_set('error_reporting', E_ALL);
ini_set('display_errors', 'On');
ini_set('display_startup_errors', 'On');

function getRandomFromArray($ar) {
    mt_srand( (double)microtime() * 1000000 );
    $num = array_rand($ar);
    return $ar[$num];
}

function getImagesFromDir($path) {
    $images = array();
    if ( $img_dir = @opendir($path) ) {
        while ( false !== ($img_file = readdir($img_dir)) ) {
            if ( preg_match("/(\.gif|\.jpg|\.png)$/", $img_file) ) {
                $images[] = $img_file;
            }
        }
        closedir($img_dir);
    }
    return $images;
}

$root = '';
$path = 'images/andrea/';
$imgList = getImagesFromDir($root . $path);
$img = getRandomFromArray($imgList);

?>

<img src="<?php echo $path . $img ?>" alt="" /> 

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0
 
LVL 29

Expert Comment

by:Olaf Doschke
ID: 37826713
And you call this three times, I suppose.

Stop doing that instead let it create three img tags with three unique images this way and call it once only:

<?php
ini_set('error_reporting', E_ALL);
ini_set('display_errors', 'On');
ini_set('display_startup_errors', 'On');

function getRandomArrayElements($ar,$n) {
    mt_srand( (double)microtime() * 1000000 );

    for ($i=1; $i<=n; $i++)
    {
     $num = array_rand($ar);
     $randomelements[] = $ar[$num];
     unset($ar[$num]);
    }

    return $randomelements;
}

function getImagesFromDir($path) {
    $images = array();
    if ( $img_dir = @opendir($path) ) {
        while ( false !== ($img_file = readdir($img_dir)) ) {
            if ( preg_match("/(\.gif|\.jpg|\.png)$/", $img_file) ) {
                $images[] = $img_file;
            }
        }
        closedir($img_dir);
    }
    return $images;
}

$root = '';
$path = 'images/andrea/';
$imgList = getImagesFromDir($root . $path);
$imgRand = getRandomArrayElements($imgList,3);

for each($imgRand as $img)
{?>

<img src="<?php echo $path . $img ?>" alt="" /> 

<?php
}
?>

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The image list $imgList is reduced with unset(), while random images are taken from it, so no image is taken twice.

Bye, Olaf.

PS: even simple array_rand() can return $n keys at once. And using array_flip() you can get $n elements at once:

function getRandomArrayElements($ar,$n) {
    mt_srand( (double)microtime() * 1000000 );

    $randomelements = array_rand(array_flip($ar), $n);

    return $randomelements;
}

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0
 
LVL 110

Expert Comment

by:Ray Paseur
ID: 37827023
Input: array.  Output: some number of random selections from the array.  Warning: I have never tested this code, so please be prepared to correct my errors or omissions.

If the array elements are unique, the output will be unique.  If the array elements are not unique but you want them to be unique, you can make the array keys and array values into the same data fields. If you do that, the non-unique array keys will overwrite each other.  See also: http://php.net/manual/en/function.array-unique.php
function getSomeRandomFromArray($arr, $n=3) 
{
    $out = array();
    if (!is_array($arr)) return $out
    shuffle($arr);
    if (count($arr) < $n) return $arr;
    while ($n)
    {
        $out[] = array_pop($arr);
        $n--;
    }
    return $out;
}

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0
 
LVL 9

Expert Comment

by:rinfo
ID: 37827077
read all files names in an array  $allFileName
$sel_files = array_rand($allFileName, 3);
$sel_files now should contain 3 distinct filename.
0
 
LVL 9

Expert Comment

by:rinfo
ID: 37827103
You may also use shuffle on $allFileName and
read first three element as random unique filename.
You can be more assured of distinct files names in Shuffle and there could be a chance of
a file name repeated in array_rand.
shuffle($allfilename);
for int $i = 0; $i<3;$i++
   $selFiles = $allfileName[$i];
0
 
LVL 1

Author Comment

by:XK8ER
ID: 37828813
im testing the provide code but its not working and i dont see any errors..
0
 
LVL 9

Expert Comment

by:rinfo
ID: 37829123
can you post the codes you are using
0
 
LVL 9

Expert Comment

by:rinfo
ID: 37829137
can you post the codes you are using
0
 
LVL 9

Accepted Solution

by:
rinfo earned 500 total points
ID: 37829381
These are the codes I have used and worked ok in both the cases
using  array_rand
<?php
  $imageDir = 'FuncIcon';
  $arrAllImages = array();
  $sel_files = array();
  if ($handle = opendir($imageDir)) {    
    while (false !== ($file = readdir($handle))) {
        if(($file == '.') || ($file == '..'))
          {
             continue;
             }
        $arrAllImages[] = $file;
    }
   closedir($handle);
   
   $sel_files = array_rand($arrAllImages, 3);
   echo "$arrAllImages[0].\n.$arrAllImages[1].\n.$arrAllImages[2]";  
   
}

using shuffle

<?php
  $imageDir = 'FuncIcon';
  $arrAllImages = array();
  if ($handle = opendir($imageDir)) {    
    while (false !== ($file = readdir($handle))) {
        if(($file == '.') || ($file == '..'))
          {
             continue;
             }
        $arrAllImages[] = $file;
    }
   closedir($handle);
   
   shuffle($arrAllImages);
   
   $i = 0;
   
   echo $arrAllImages[0].' '.$arrAllImages[1].' '.$arrAllImages[2];  
   
}
 
?>

These codes not very elegant but work in both the cases.
Maybe you have forgotten to take care of the . and ..
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