Uploading a file

Anon

Hi
I am developing a form which containts one file upload option.While uploading i am checking for proper image files like gif,jpg,bmp..etc.It is working fine by checking in javascript.Suppose if the user uploads a file called temp.gif from the path "c:\images\" ,while submitting the form
is that any way to check the file temp.gif exist in the path.Because i am getting a this warning message.
"Unable to open 'none' for reading: No such file or directory".
"Unlink failed (No such file or directory)"

To copy the above image file in the server directory i am using the following way.
copy($image,"/images/$image_name");
unlink($image_name);
This is working fine if the file exist in the specified path.

Thanks in advance for your help.

srini

Anon

Simply Error Checking:
if the value of $image_name is "none" there is no file to upload ... I think your error msg said something like that :-)) so before you do any other copy operations simply do a
if ($image_name != "none")
{
--upload operation
}
else
{
--error operation
}
cya Rob

Anon

Hi
there is a way to check for the file validity and just to make sure someone doesnt type a random name which is not a file name. try "file_exits()" function. this should help you. You can read up more on this function, follow the below link
http://www.php.net/manual/function.file-exists.php

Anon

sure, but as far as I know you can use this function only on your directories (the ones you have acces for)... I don't think you can check a file on a remote client with the function, and thats what srini asked for...

please correct me if I'm wrong, I haven't played much with directories and files in PHP yet ...

cya Rob

Anon

Nope works like a charm, this checks for a file stream existance. All of our ftp backed upload sites running php uses this to check the validity of the uploaded files. When you upload a file using the enctype=multipart/form-data you will receive a stream of the file. And the variable $filename ($image in this case) will hold the stream. And file_exists($filename or $image) will give you true or false for the stream. Its got nothing to do with the remote userpath. Try it for yourself.

Anon

Hi
As you said i tried like this.
if file_exists($image)
{
....
}else
{
echo("file not found...");
}

By doing so,it is working fine without any warning message.

Thanks a lot for your help.

srini

Anon

...and again I've learned something, thanks a lot Kaushik! I'm going to try this out, as soon as I have the next upload-page to be build ...

cya Rob

Anon

The file info that is derived through $userfile_[properties] will only be set after the file had been actually copied to the destination file. Is this correct? Further, what must I do to use the original filename of the file that the user wants to upload as the filename of the uploaded file in the upload destination since the value that the user specified in the input file is renamed to some *.tmp file when the form is submitted?

Anon

Funny the answer to your question is in a thread you started ...
As Habibi said:

Some of nice things on PHP is you can retrieve various info from uploaded file like followings :
$thefile_name --> returns uploaded file name
$thefile_size --> returns uploaded file size
$thefile_type --> returns uploaded file type, this is much more like MIME-type

What if we want to name the file on the server like the file that was uploaded ?? The filename is in $thefile_name ...

rename($thefile, $thefile_name);

if you have more questions concerning this subject, I really suggest you read this column by Darren Beale: http://www.phpbuilder.com/columns/bealers20000904.php3

cya Rob

Anon

Yes, I know that essentially the filename should be in $thefile_name variable. But the problem is there's no value set for this variable even though the $thefile variable is set. That's why I wondered if this is really the way or what.