PHP - Displaying random output more than once, prevent duplicate

Question

I have the following code that I am using to randomly display PHP widgets from a folder:

<?php 
function random_widget($dir = 'wp-content/themes/zonza/elements')
{
    $files = glob($dir . '/*.*');
    $file = array_rand($files);
    return $files[$file];
}
?>

<?php include random_widget();?>
<?php include random_widget();?>
<?php include random_widget();?>

random_widget(); outputs a URL, which I then use in the include function to display the widget.

The code randomly chooses between 6 php files and displays one randomly. I include it 3 times to get 3 widgets. However, I get the same widget displayed more than once sometimes.

What can I do to modify the code to prevent this from happening?

Sean Johnson · Accepted Answer · 2012-06-27 02:46:56Z

2

Try this:

<?php 
function random_widget($dir = 'wp-content/themes/zonza/elements')
{
    static $files = false;
    if(!$files) $files=glob($dir . '/*.*');
    $key = array_rand($files);
    $file=$files[$key];
    unset($files[$key]);
    return $file;
}
?>

It works by removing the file returned from $files, and maintaining $files over multiple function calls (it only globs() on the first time you call the function)

edited Jun 27, 2012 at 2:46

answered Jun 27, 2012 at 2:32

Sean Johnson

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2 Comments

Zach Nicodemous Over a year ago

This code reports an error. Parse error: syntax error, unexpected '(', expecting ',' or ';' on line 4

Sean Johnson Over a year ago

updated the code, I forgot that static variables in PHP cannot be initialized with an expression. Should work fine now.

Lusitanian · Accepted Answer · 2012-06-27 02:32:50Z

0

Declare files at the beginning of the page $files = glob($dir . '/*.*'); In random_widget, do this unset($files[$file]) after you pull the value.

answered Jun 27, 2012 at 2:32

Lusitanian

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2 Comments

Sean Johnson Over a year ago

don't forget to pull $files into the namespace of the function with global $files; if you try this method

Lusitanian Over a year ago

Yes, although a static variable actually would be better.

Jong Bor Lee · Accepted Answer · 2012-06-27 02:47:16Z

0

array_rand takes a number $num_req as an optional second parameter which specifies the number of entries you want to pick. So add this parameter to random_widget, pass it to array_rand to get an array of keys instead of a single key, return the array of files, and then iterate over this array to include the widgets (instead of calling random_widget three times).

<?php 
function random_widget($num_req, $dir = 'wp-content/themes/zonza/elements')
{
    $files = glob($dir . '/*.*');
    $keys = array_rand($files, $num_req);

    $chosen = array();

    foreach($keys as $key) {
        $chosen[] = $files[$key];
    }

    return $chosen;
}

$widgets = random_widget(3);
foreach($widgets as $widget) {
    include $widget;
}

?>

An advantage of this solution over the ones proposed in the other answers is that it is stateless: you can reuse the function in different contexts as much as you want.

Source: http://php.net/manual/en/function.array-rand.php

edited Jun 27, 2012 at 2:47

answered Jun 27, 2012 at 2:34

Jong Bor Lee

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1 Comment

Sean Johnson Over a year ago

I like the idea of one function call to get as many widgets as you want - however this really only works well if all of the widgets are together in one spot. My solution allows for random_widget() to be called from anywhere without much fuss.

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