Solved

How can I display part of the filename?

Posted on 2011-03-16
8
327 Views
Last Modified: 2012-05-11
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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0
Comment
Question by:Computer Guy
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8 Comments
 
LVL 10

Expert Comment

by:Asim Nazir
ID: 35145102
// variable "theString" has the contents "filename.mp3"
filenameOnly = theString.substring(0,theString.lastIndexOf(".");
0
 
LVL 9

Expert Comment

by:mayank_joshi
ID: 35145115
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
0
 
LVL 3

Author Comment

by:Computer Guy
ID: 35145192
Could I please have an example with my code?
0
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LVL 19

Expert Comment

by:Bardobrave
ID: 35146127
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)
0
 
LVL 10

Expert Comment

by:Mathiyazhagan
ID: 35146723
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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0
 
LVL 9

Expert Comment

by:mayank_joshi
ID: 35147160
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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0
 
LVL 9

Accepted Solution

by:
mayank_joshi earned 500 total points
ID: 35147177
sorry i used .php instead of .jpg

Correct Code:-

<?
	for ($i = $start; $i < $limit; $i++) {
		if (!isset($photos[$i])) break; // No more photos
                $photoNumber=basename($photos[$i],".jpg");
		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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0
 
LVL 10

Expert Comment

by:Mathiyazhagan
ID: 35155750
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.

0

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