To all,
The following is a file upload question using PHP.
I am able to upload a file to my server with the following script. (see
below)
It's in two parts - the first is a html form and the second is the php
script which takes the variable $img1_name.
However, I'd like to check if, in fact, the file was uploaded to the server.
The only way I can check this is to copy and paste the pathname that I
upoloaded in the browser's window like:
c:/Program Files/Apache Group/Apache/htdocs/reservoirphoto.jpg
When I do that the photo does load - however it loads in a paint program I
use like PaintShopPro - not as a file within the browser.
The question is this...in the textbook I'm reading (Julie Meloni - PHP Fast
and Easy - page 174), it says that:
"(I can) use the File / Open Page / menu item in your web browser to
navigate through your filesystem and find the file you uploaded."
The screenshot in the book has the .jpg file photo *within* the browser
screen - not by itself in a paint program.
Can anybody assist me in this?
Thanking all in advance.
Tony Ritter
............................................................................
..........................
file://Script 1 - upload_form.html
<HTML>
<HEAD>
<TITLE>Upload a File</TITLE>
</HEAD>
<BODY>
<H1>Upload a File</H1>
<FORM METHOD="post" ACTION="do_upload.php" ENCTYPE="multipart/form-data">
<p><strong>File to Upload:</strong><br>
<INPUT TYPE="file" NAME="img1" SIZE="30"></p>
<P><INPUT TYPE="submit" NAME="submit" VALUE="Upload File"></p>
</FORM>
</BODY>
</HTML>
...................................
file://Script 2 - do_upload.php
<?
if ($img1_name != "")
{
copy("$img1", "c:/Program Files/Apache Group/Apache/htdocs/$img1_name") or
die("Couldn't copy the file.");
}
else
{
die("No input file specified");
}
?>
<HTML>
<HEAD>
<TITLE>Successful File Upload</TITLE>
</HEAD>
<BODY>
<H1>Success!</H1>
<P>You sent: <? echo "$img1_name"; ?>, a <? echo "$img1_size"; ?>
byte file with a mime type of <? echo "$img1_type"; ?>.</P>
</BODY>
</HTML>
........................
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