Help Me .... Up Load Picture ???

[Anon]

i building a catalog system now,
when i create a form to add a new item,
i have to upload a item picture in form,
i use 'browse' button get the picture,

how to get the 'picture file name' when i submit the form ?

[flipis]

you have to:

first: put yor form as ENCTYPE="multipart/form-data", or else it might not upload anything.

second: When uploading a file, you have the file name (say you named the FILE field in the form as $the_file), and some other info about it calling the variable with suffixes:

$the_file_name = The name of the TMP file. (use basename($the_file) to get the REAL name)

When you have all of the info, use copy like this:

copy($the_file,$the_new_file);

So, if the $the_file name is "me.txt" and $the_new_file is "you.txt", the file will be saved as two.txt

Hope it helps

fLIPIS

[snafufans]

$Hello,

I don't know how u get the filename but i had the same problem and did it this way
On the page where the upload button comes place this form

<FORM ENCTYPE="multipart/form-data" ACTION="result.php?id=<?php echo $id ?>" METHOD=POST>

<table style='border: 1 solid #D9D9D9'>
	<tr>
		<td style='background-color: #005883;'>

			<INPUT TYPE="hidden" name="MAX_FILE_SIZE" value="10485760">
			<font style='font-family: Tahoma, Verdana, Arial; font-size: 16px; color: #ffffff; font-weight: bold'>Upload file: </font>
		</td>
		<td style='background-color: #DADADA;'>
			<INPUT NAME="userfile" TYPE="file">
			<INPUT TYPE="submit" VALUE="Upload">
		</td>
</table>

</FORM>

//then in a new file: result.php place the following code:
<?php
if (is_uploaded_file($userfile)) {
copy($userfile, "files/mailings/mailing_" . $id . "/" . $userfile_name );
header("Location: Upload.php?id=" . $id);
} else {
echo "Possible file upload attack: filename '$userfile'.";
}
?>

Ofcourse u maybe don't need to pass the $id but i did..

Greetz,
$nafu

[flipis]

You call the name of the file with

$userfile_name

Its size by

$userfile_size

Its type by

$userfile_type

Look here:

http://www.php.net/manual/en/features.file-upload.php

Hope it helps

fLIPIS

[Anon]

This is my form :

<form enctype="multipart/form-data" method="post" action="<? echo $PHP_SLF ?>">
<TABLE BORDER=0><TR><TD>
<font size="2" face="Verdana"><Bupload</B></FONT>
</TD></TR><TR><TD>
<input type="file" name="img1" size="30">
</TD></TR><TR><TD>
<input type="submit" name="submit" value="submit">
</TD></TR></TABLE>
</form>

after i submit how can i get the picture file name ?

bcos i have try last time in my result i get 'c:/file/picture.jpg'

how to get the result like 'http://mydomain/com/file/picture.jpg'

Thank

[flipis]

You won't. $file_name refers to the name ON YOUR LOCAL DRIVE. The http:// name will only exists AFTER the file has been uploaded.

You use copy like this:
copy($file1,$file2);

say you name them the same ($file1="me.txt", $file2="me.txt"). So then, when uploading, you can call http://youdomain.com/me.txt.

But only AFTER the upload is complete. What PHP really does is to create a temporary file storing all the data in it and then, moves it to the new directory and renames it.

Hope it helps

fLIPIS

[Anon]

...

[flipis]

You're very welcome, mate. :-)

I really hope i was useful and not confusing.

fLIPIS

[Anon]

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