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UPLOAD ERROR: move_uploaded_file FILENAME failed to open stream: Permission denied on line #

Posted on 2004-09-25
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Last Modified: 2013-12-13
I'm getting the error on title when trying to upload a text file to a Windows server. I'm developing on a localhost windows computer, and I can upload internally without problems, however when using the same script to upload a 300K text file in the remote Windows server it sends the warning on the title.

There was a similar problem listed in this forum, but the remote server was Unix, so the answer does not apply here. I tried that solution (ftp a file with the exact same name in the remote location and CHMOD it to 777) but since it is a Windows server CHMOD does not work. I only have FTP access to the remote server.

Help is sincerely appreciated.
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Question by:aixarat
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5 Comments
 
LVL 9

Expert Comment

by:techtonik
ID: 12155086
If you have only FTp access, then you cannot do anything except to contact your hosting support team on this issue so they can change configuration to allow webserver upload/write privileges to your chosen direcory.
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Author Comment

by:aixarat
ID: 12155410
So there's no run-time override?
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LVL 49

Expert Comment

by:Roonaan
ID: 12156197
Well, as the error occurs when executing move_uploaded_file, you could open - in your phpscript - an ftp connection and move the file using this connection from the temporary upload dir to the dir you require it to be. Possible problems could be the max numbers of ftp connections and slow ftp servers. But it sure would be a workaround, which in your case could be handy.

-r-
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Author Comment

by:aixarat
ID: 12160043
The reason that I'm trying to use move_uploaded_file is that I'm trying to build a simple file uploading system for a feature in a particular website. I'm already FTPing to gen my files on place.
I'm thinking about a possible workaround: is there a way to directly open the uploaded file without moving it to a new folder? I suppouse that will also require me to delete it from the temp folder... Any comments?
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Accepted Solution

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Roonaan earned 250 total points
ID: 12160076
One of the items in the $_FILES arrays is the filename of the temporary file.

You could use file() or file_get_contents() to get the contents of this temporary file.

Use print_r($_FILES) to find the correct key, as I don't know it from heart.

-r-
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