This is what I tried:
if (is_uploaded_file($HTTP_POST_FILES['image1']['image1']) {
move_uploaded_file($HTTP_POST_FILES['image1']['image1'],
"$DOCUMENT_ROOT/images/$file");
}
and also this:
if (is_uploaded_file($_FILES['image1']['image1']) {
move_uploaded_file($_FILES['image1']['image1'], "$DOCUMENT_ROOT/images/$file");
}
and this:
if (is_uploaded_file($_FILES['image1']['image1'])) {
copy($_FILES['image1']['image1'], "$DOCUMENT_ROOT/images");
}
,the image (which was within the size range) was never uploaded. I have
a feeling that I am makeing 1 or 2 of the same mistakes in all three but
through my experimenting and reading I am having no luck. I also bought
Programming PHP but I trust it less and less as I even find simple
mistakes like missing quotes in example code. So what's going wrong?
Thank you again,
Kyle
On Sun, 12 Jan 2003 15:17:42 +1100, "Justin French"
<[EMAIL PROTECTED]> said:
> Hi,
>
> the files themselves are available in the $_FILES array, but it's not as
> simple as that.
>
> may i recommend you start by copying the Examples 18-1 and 18-2 from this
> page:
> http://www.php.net/manual/en/features.file-upload.php
>
> Once you've got THAT code working smoothly and understand what's
> happening,
> THEN start modifying it to a 3-file form.
>
> I'd personally push forms through using POST method rather than get
> whenever
> possible -- especially when dealing with files... the manual does it this
> way too :)
>
>
> Justin
>
>
>
> on 12/01/03 12:18 PM, Kyle Babich ([EMAIL PROTECTED]) wrote:
>
> > I just broke skin with php and I'm learning forms, which I'm not good
> > with at all. These are snippets from post.html:
> >
> > <form method="get" action="process.php" enctype="multipart/form-data">
> > and
> > <input type="file" name="image1" maxlength="750" allow="images/*"><br>
> > <input type="file" name="image2" maxlength="750" allow="images/*"><br>
> > <input type="file" name="image3" maxlength="750" allow="images/*"><br>
> >
> > but how would I pass the actual image on because when I do something like
> > this:
> > <?php echo "{$_GET[image1]}";
> > ?>
> > as you could probably guess only the filename prints.
> >
> > So how do I take the image instead of just the filename?
> >
> > Thank you,
> > --
> > Kyle
>
>
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