Re: [PHP] Simple Form Processing
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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Simple Form Processing
Did you read that link at all?
adapting the manual's code to suit your form names and directory
preferences:
if (is_uploaded_file($_FILES['image1']['tmp_name'])) {
copy($_FILES['image1']['tmp_name'],
$DOCUMENT_ROOT/images/$_FILES['image1']['name']);
} else {
echo Couldn't do it!;
}
So, to summarise what i *think* is wrong with your code:
1. $_FILES['image1']['image1'] should have been
$_FILES['image1']['tmp_name']
2. $DOCUMENT_ROOT/images/$file :: what is $file where is that defined???
I *think* you wish to save the file with same name as the user had it
named??? if so $DOCUMENT_ROOT/images/$_FILES['image1']['name'] is what you
want, or better still, for clarity,
{$DOCUMENT_ROOT}/images/{$_FILES['image1']['name']}
If all else fails do a print_r($_FILES) and make sure that the file is at
least being uploaded.
Like I said before, there is a perfect working example in the manual... get
it working THEN try to adapt it to suit your needs :)
Cheers,
Justin
on 12/01/03 11:39 PM, Kyle Babich ([EMAIL PROTECTED]) wrote:
This is what I tried:
if (is_uploaded_file($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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Simple Form Processing
On Sunday 12 January 2003 20:39, Kyle Babich wrote:
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?
So what *does* $_FILE contain?
Hint: if your code doesn't behave as expected always print_r() your variables
so you can whether they contain what you expected them to contain.
And if $_FILE doesn't contain anything then check php.ini to ensure that you
have enables file uploads.
--
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
/*
Wilcox's Law:
A pat on the back is only a few centimeters from a kick in the pants.
*/
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Simple Form Processing
I just broke skin with php and I'm learning forms, which I'm not good
with at all. These are snippets from post.html:
form method=get action=process.php enctype=multipart/form-data
and
input type=file name=image1 maxlength=750 allow=images/*br
input type=file name=image2 maxlength=750 allow=images/*br
input type=file name=image3 maxlength=750 allow=images/*br
but how would I pass the actual image on because when I do something like
this:
?php echo {$_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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Simple Form Processing
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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
