asked on

How can I display part of the filename?

Hi,

I have a little photo gallery.

All photos are numbered starting with 01 (leading 0 included for photos 1-9)
01.jpg, 02.jpg, 03.jpg, 04.jpg, 05.jpg, 06.jpg, 07.jpg, 08.jpg, 09.jpg, 10.jpg.......

I want to display that number in the "DISPLAY PHOTO NUMBER HERE (02 FORMAT)" area of the code attached.

Just the number "02", with out the file extension.

How can I do that?

Thanks!

<?
	for ($i = $start; $i < $limit; $i++) {
		if (!isset($photos[$i])) break; // No more photos
		echo '<a class="vlightbox1" href="' . $eventinfo['photo_path_url'] . '/web/' . $photos[$i] . '"><img src="' . $eventinfo['photo_path_url'] . '/thumbs/' . $photos[$i] . '" alt="Photo" border="0"/>'. $product_title .'DISPLAY PHOTO NUMBER HERE</a>';
	}
	?>

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Asim Nazir

// variable "theString" has the contents "filename.mp3"
filenameOnly = theString.substring(0,theString.lastIndexOf(".");

mayank_joshi

Example

<?php
$path = "/testweb/home.php";

//Show filename with file extension
echo basename($path) ."<br/>";

//Show filename without file extension
echo basename($path,".php");
?>

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The output of the code above will be:
home.php
home

Reference:-
http://www.w3schools.com/PHP/func_filesystem_basename.asp

Computer Guy

ASKER

Could I please have an example with my code?

Bardobrave

Try this in your code:

$photoName = substr($photos[$i],strrchr($photos[$i],"."),strlen($photos[$i]));

I'm not familiar with php, so maybe you'll need to substract 1 from strlen depending on the starting counting value (0 or 1)

Mathiyazhagan

Hi,

Try this,

<?php
	for ($i = $start; $i < $limit; $i++) {
		if (!isset($photos[$i])) break; // No more photos
		$lastPosition = strrpos($photos[$i], '.');
		$photoNumber = substr($photos[$i], 0, $lastPosition);
		echo '<a class="vlightbox1" href="' . $eventinfo['photo_path_url'] . '/web/' . $photos[$i] . '"><img src="' . $eventinfo['photo_path_url'] . '/thumbs/' . $photos[$i] . '" alt="Photo" border="0"/>'. $product_title.$photoNumber.'</a>';
	}
?>

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mayank_joshi

using basename function:-

<?
	for ($i = $start; $i < $limit; $i++) {
		if (!isset($photos[$i])) break; // No more photos
                $photoNumber=basename($photos[$i],".php");
		echo '<a class="vlightbox1" href="' . $eventinfo['photo_path_url'] . '/web/' . $photos[$i] . '"><img src="' . $eventinfo['photo_path_url'] . '/thumbs/' . $photos[$i] . '" alt="Photo" border="0"/>'. $product_title .$photoNumber.'</a>';
	}
	?>

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ASKER CERTIFIED SOLUTION

mayank_joshi

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Mathiyazhagan

Hi,

The accepted solution will given error file names when the file extensions other than .jpg.

If you only use ".jpg" files you can have mayank joshi code. Otherwise my code is the bestway to get the filename with any extension.