Noob Question about basename

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July 9th, 2012, 02:51 PM

poopertropper

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PHP5 - Noob Question about basename

Hey guys,

So I've been really trying my hand at PhP lately (not another noob!) and I ran into a strange issue with basename that I'm wondering if you guys can help me with.

To start out, I've written up the code below to go into a directory and pic out the images there to load into an array. This works great and I've been able to update a clients gallery rather quickly using it.

PHP Code:


		
			
// grab images for stuff



$dir2 = '../fest/'.$year.'/'.$id.'/examples/';

if(!is_dir($dir2)){

    $dir2 = '../fest/'.$year.'/path/ex/'; 



}

$allImgs2 = array();

$dh2 = opendir($dir2);



while (($file = readdir($dh2)) !== false)

{

    if ($file!='Thumbs.db' && $file!='.' && $file!='..' && $file!='.DS_Store' )

    {

        $allImgs2[] = $file; 

    }

}

closedir($dh2);

?>

However I wanted to use the photos file name as a way to title each image in an H1 and I ran into problems. I thought basename would remove the .jpg part and then using CSS I would just set everything to CAPS for styling. So I wrote it like this:

PHP Code:


		
			
<ul id="slider">                                      

<?php

     foreach($allImgs2 as $img2){

     echo '<li><a class="fancybox" href="'.$dir2.$img2.'" title="'.$img2["basename"].'"><img src="'.$dir2.$img2.'"><h1>'.$img2["basename"].'</h1></li></a>';

}

?>

</ul>

I figured using .$img2["basename"]. would simply take the image name and then rip off the .jpg portion. Instead what it's doing is only showing the first letter of the file and nothing else. Am I missing something or using basename incorrectly? Thanks!

July 9th, 2012, 03:09 PM

requinix

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basename() is a function and $img2 is a string.

basename is the complement to dirname: the former will keep the file name (and extension) and discard the path while the latter will keep the path and discard the name.

PHP Code:


		
			
 $file = "/path/to/file.ext";

echo basename($file); // file.ext

echo dirname($file); // /path/to



var_dump($file == dirname($file) . "/" . basename($file)); // true

PHP's basename() can also strip the file extension if you know what it is. Read the manual page to find out how.

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July 9th, 2012, 04:57 PM

poopertropper

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Thanks!!!

I went with

PHP Code:


		
			
<?php

     foreach($allImgs2 as $img2){

     $info = pathinfo($img2);    

     $file_name =  basename($img2,'.'.$info['extension']);

     echo '<li><a class="fancybox" href="'.$dir2.$img2.'" title="'.$file_name.'"><img src="'.$dir2.$img2.'"><h3>'.$file_name.'</h3></li></a>';

                        }

= ?>

This seemed to work ok. Not sure if it's bad coding practice but it did the job. Thanks so much!

July 9th, 2012, 06:02 PM

requinix

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If you would care to take a closer look at pathinfo, you'll see that not only does it tell you the basename of the file, but also the basename without the extension. You don't need basename() at all if you're going that route.

PHP5 - Noob Question about basename

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