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9.0

PHP Upload Script Giving New Error

Asked by kyleldi in PHP Scripting Language, PHP Frameworks, Databases Miscellaneous

Tags: PHP Mysql database

I created a script months ago that would upload five different upload fields to our server and posts the filename to our sql database, and all has been working great.  Now all the sudden it's throwing errors saying that the field cannot be empty, which I don't have anything in the script itself telling it that it cannot be.  Here's what the error is specifically.


Warning: copy() [function.copy]: Filename cannot be empty in process.php on line 79

Warning: copy() [function.copy]: Filename cannot be empty in process.php on line 83

Warning: copy() [function.copy]: Filename cannot be empty in process.php on line 87

Warning: copy() [function.copy]: Filename cannot be empty in process.php on line 91

I guess what i'm assuming happened as our server provider probably updated its version of PHP and now part of my code isn't correct.  The script still works, but it throws these errors every time unless all five fields are full.  The end of this script redirects to a thank you page.

Note:  The SQL Database itself does allow null entries into these fields. 
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// Uploading File 01 //  // LINE 72 //
$image_part = date("")."".$HTTP_POST_FILES['file01']['name'];
$image_list[14] = $image_part;
copy($HTTP_POST_FILES['file01']['tmp_name'], "$dirname/".$image_part);
// Uploading File 02 //
$image_part = date("")."".$HTTP_POST_FILES['file02']['name'];
$image_list[15] = $image_part;
copy($HTTP_POST_FILES['file02']['tmp_name'], "$dirname/".$image_part);
// Uploading File 03 //
$image_part = date("")."".$HTTP_POST_FILES['file03']['name'];
$image_list[16] = $image_part;
copy($HTTP_POST_FILES['file03']['tmp_name'], "$dirname/".$image_part);
// Uploading File 04 //
$image_part = date("")."".$HTTP_POST_FILES['file04']['name'];
$image_list[17] = $image_part;
copy($HTTP_POST_FILES['file04']['tmp_name'], "$dirname/".$image_part);
// Uploading File 05 //
$image_part = date("")."".$HTTP_POST_FILES['file05']['name'];
$image_list[18] = $image_part;
copy($HTTP_POST_FILES['file05']['tmp_name'], "$dirname/".$image_part);  // LINE 91 //
[+][-]11/04/09 05:39 AM, ID: 25739128Accepted Solution

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About this solution

Zones: PHP Scripting Language, PHP Frameworks, Databases Miscellaneous
Tags: PHP Mysql database
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Solution Provided By: mikeada
Participating Experts: 2
Solution Grade: A
 
[+][-]11/04/09 06:16 AM, ID: 25739456Expert Comment

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[+][-]11/04/09 06:48 AM, ID: 25739807Expert Comment

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[+][-]11/04/09 09:31 PM, ID: 25747002Expert Comment

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[+][-]11/14/09 05:35 AM, ID: 25820638Administrative Comment

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