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

Posted on 2014-04-08
5
Medium Priority
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1,141 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);
							  }
		}


?>

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0
Comment
Question by:sabecs
  • 3
  • 2
5 Comments
 
LVL 60

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/>";
  }
}
?>

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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

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Also, would it be possible to use a wild card and copy say "front_line_plus_for_cats_762_2*" ?
0
 
LVL 60

Accepted Solution

by:
Julian Hansen earned 2000 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/> ";
  }
}
?>

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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 60

Expert Comment

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

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