Solved

PHP - copy/duplicate multiple image files in same directory?

Posted on 2014-04-08
5
1,080 Views
Last Modified: 2014-04-09
Hi,
I am trying to copy/duplicate some image files in the same directory.

The code below works fine for copying say

'front_line_plus_for_cats_762_1.jpg' to 'front_line_plus_for_cats_1094_1.jpg'

and

'front_line_plus_for_cats_762_2.jpg' to 'front_line_plus_for_cats_1094_2.jpg'

but how can I modify it so it also copies


front_line_plus_for_cats_762_1_med.jpg
front_line_plus_for_cats_762_1_sld.jpg
front_line_plus_for_cats_762_1_sml.jpg
front_line_plus_for_cats_762_1_swp.jpg
front_line_plus_for_cats_762_1_tny.jpg

front_line_plus_for_cats_762_2_med.jpg
front_line_plus_for_cats_762_2_sld.jpg
front_line_plus_for_cats_762_2_sml.jpg
front_line_plus_for_cats_762_2_swp.jpg
front_line_plus_for_cats_762_2_tny.jpg

to

front_line_plus_for_cats_1094_1_med.jpg
front_line_plus_for_cats_1094_1_sld.jpg
front_line_plus_for_cats_1094_1_sml.jpg
front_line_plus_for_cats_1094_1_swp.jpg
front_line_plus_for_cats_1094_1_tny.jpg

front_line_plus_for_cats_1094_2_med.jpg
front_line_plus_for_cats_1094_2_sld.jpg
front_line_plus_for_cats_1094_2_sml.jpg
front_line_plus_for_cats_1094_2_swp.jpg
front_line_plus_for_cats_1094_2_tny.jpg


Thanks

<?php


		for ($i = 1; $i <= 5; $i++) {	
						  $image_{$i} = $row_query_images['image'.$i];	
						  //if a picture exits lets copy it 	
						  if(stristr($image_{$i}, 'nopic') === FALSE) {
							  
							  $newfile = str_replace ( $id, $new_record_id, $image_{$i});
							  
							  $rootd = $_SERVER['DOCUMENT_ROOT'];
							  copy( $rootd.$image_{$i}, $rootd.$newfile);
							  }
		}


?>

Open in new window

0
Comment
Question by:sabecs
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 2
5 Comments
 
LVL 58

Expert Comment

by:Julian Hansen
ID: 39988127
Firstly why will your current code not work?

As I understand it you are simply replacing the numeric value in the string with a new numeric value so the str_replace code should work.
The following code seems to do what you require.
<?php
$input = array (
  'front_line_plus_for_cats_762_1_med.jpg',
  'front_line_plus_for_cats_762_1_sld.jpg',
  'front_line_plus_for_cats_762_1_sml.jpg',
  'front_line_plus_for_cats_762_1_swp.jpg',
  'front_line_plus_for_cats_762_1_tny.jpg',
  'front_line_plus_for_cats_762_2_med.jpg',
  'front_line_plus_for_cats_762_2_sld.jpg',
  'front_line_plus_for_cats_762_2_sml.jpg',
  'front_line_plus_for_cats_762_2_swp.jpg',
  'front_line_plus_for_cats_762_2_tny.jpg'
);
$id = 762;
$new_record_id = 1094;
for ($i = 0; $i < count($input); $i++) {  
  $image_{$i} = $input[$i];  
  //if a picture exits lets copy it   
  if(stristr($image_{$i}, 'nopic') === FALSE) {
    $newfile = str_replace ( $id, $new_record_id, $image_{$i});
//    $rootd = $_SERVER['DOCUMENT_ROOT'];
    $rootd = "";
    echo "copy( " . $rootd . $image_{$i} . "," . $rootd.$newfile . ")<br/>";
  }
}
?>

Open in new window

0
 

Author Comment

by:sabecs
ID: 39989676
Thanks for your help.
 
$image_{$i} gives me the values of
'front_line_plus_for_cats_762_1.jpg',  'front_line_plus_for_cats_762_2.jpg', 'front_line_plus_for_cats_762_3.jpg' ...
which comes from  my MySQL table.

but my uploads folder has different versions/sizes of the front_line_plus_for_cats_762_1.jpg
so I need to also move/copy the _med, _sld, _sml _swp _tny version of files etc

so I was just after a way of simply adding these to my current code?

I am currently out of the office but I might try something similar to below when I get back.

copy( $rootd.$image_{$i}, $rootd.$newfile);  //copy main file
copy( str_replace(".jpg","_sml.jpg",$rootd.$image_{$i}), str_replace(".jpg","_sml.jpg",$rootd.$newfile));  //copy _sml file
copy( str_replace(".jpg","_sld.jpg",$rootd.$image_{$i}), str_replace(".jpg","_sld.jpg",$rootd.$newfile));  //copy _sld file
copy( str_replace(".jpg","_med.jpg",$rootd.$image_{$i}), str_replace(".jpg","_med.jpg",$rootd.$newfile));  //copy _med file
copy( str_replace(".jpg","_swp.jpg",$rootd.$image_{$i}), str_replace(".jpg","_swp.jpg",$rootd.$newfile));  //copy _swp file
copy( str_replace(".jpg","_tny.jpg",$rootd.$image_{$i}), str_replace(".jpg","_tny.jpg",$rootd.$newfile));  //copy _tny file

Open in new window


Also, would it be possible to use a wild card and copy say "front_line_plus_for_cats_762_2*" ?
0
 
LVL 58

Accepted Solution

by:
Julian Hansen earned 500 total points
ID: 39990195
In that case I would use the preg_match function to get the filename prefix
<?php
$input = array (
  'front_line_plus_for_cats_762_1.jpg',
  'front_line_plus_for_cats_762_2.jpg',
);
$id = 762;
$new_record_id = 1094;
for ($i = 0; $i < count($input); $i++) {  
  $image = $input[$i];  
  if (preg_match("/(.*)_(\d+)_(\d+).jpg/", $image, $result)) {
     $target = $result[1] . '_' . $new_record_id . '_' . $result[3]. '.jpg';
     $source = $result[1] . '_' . $result[2] . '_' . $result[3] . 'jpg';
     echo "copy($source, $target);<br/> ";
     $target = $result[1] . '_' . $new_record_id . '_' . $result[3]. '_sml.jpg';
     $source = $result[1] . '_' . $result[2] . '_' . $result[3 ] . '_sml.jpg';
     echo "copy($source, $target);<br/> ";
     $target = $result[1] . '_' . $new_record_id . '_' . $result[3]. '_sld.jpg';
     $source = $result[1] . '_' . $result[2] . '_' . $result[3 ] . '_sld.jpg';
     echo "copy($source, $target);<br/> ";
     $target = $result[1] . '_' . $new_record_id . '_' . $result[3]. '_med.jpg';
     $source = $result[1] . '_' . $result[2] . '_' . $result[3 ] . '_med.jpg';
     echo "copy($source, $target);<br/> ";
     $target = $result[1] . '_' . $new_record_id . '_' . $result[3]. '_swp.jpg';
     $source = $result[1] . '_' . $result[2] . '_' . $result[3 ] . '_swp.jpg';
     echo "copy($source, $target);<br/> ";
     $target = $result[1] . '_' . $new_record_id . '_' . $result[3]. '_tny.jpg';
     $source = $result[1] . '_' . $result[2] . '_' . $result[3 ] . '_tny.jpg';
     echo "copy($source, $target);<br/> ";
  }
}
?>

Open in new window

Still not clear on why you need to create the variable $image_{$i} instead of just $image as shown in the code above?
0
 

Author Closing Comment

by:sabecs
ID: 39990433
Thanks Julian, that is perfect..
 I will also just use $image instead of $image_{$i}
0
 
LVL 58

Expert Comment

by:Julian Hansen
ID: 39990767
You are welcome - thanks for the points.
0

Featured Post

Online Training Solution

Drastically shorten your training time with WalkMe's advanced online training solution that Guides your trainees to action. Forget about retraining and skyrocket knowledge retention rates.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

I imagine that there are some, like me, who require a way of getting currency exchange rates for implementation in web project from time to time, so I thought I would share a solution that I have developed for this purpose. It turns out that Yaho…
Many old projects have bad code, but the budget doesn't exist to rewrite the codebase. You can update this code to be safer by introducing contemporary input validation, sanitation, and safer database queries.
Learn how to match and substitute tagged data using PHP regular expressions. Demonstrated on Windows 7, but also applies to other operating systems. Demonstrated technique applies to PHP (all versions) and Firefox, but very similar techniques will w…
The viewer will learn how to count occurrences of each item in an array.

622 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question