Passing it on...

Anon

I got:
1: Form
2: Comfirm
3: Inserting to mySQL db

I got to pass information from 2 to 3.
How should I do that?
You can try take a look at this but I'm not sure if my ip "is in the open"
http://192.168.13.182//horse-2/

Should I use sessions? If I should, then how?
Please give me any suggestions at all, Arni

Anon

When you submit the form, the data is passed through a POST or GET operation.
This sends the data to the server.
If you want to pass the data to a new page, you can send it in the URL;

http://www.something.somewhere/my_script.php?my_var=my_value&my_other_var=my_other_value

Or, you can print all the values you got from the form into a new form,
but make all the values hidden.
The submit button could then be the "CONFIRM" button and you can simply POST the data to your script.

Anon

Thanks, I'll try this...

Anon

Some image I'm trying to upload to the server...

Warning: fopen("\\php2","r") - No such file or directory in c:\apache\htdocs\horse-2\do_form_continue.php3 on line 52

Warning: Supplied argument is not a valid File-Handle resource in c:\apache\htdocs\horse-2\do_form_continue.php3 on line 52

This happens in the third file when I try to upload it...

$binary_junk = addslashes(fread(fopen($img, "r"), filesize($img))); and then I try to insert $binary_junk to blob field, but I get this error and nothing get's inserted ofcourse...

Any suggestions?
Thanks, Arni

Anon

Check the code that generates the "\\php2" because that's a weird filename.

Anon

That's not the filename, I can't explane this... well, here goes:

<form method="post" enctype="multipart/form-data" action="do_form.php3">
To post from file 1

I do in file 2:
if ($img != "none") {
$size = GetImageSize("$img");
checksize($size[0], "450", "width");
checksize($size[1], "450", "height");
echo "$img_size bytes<br>\n";
echo "$img_type<br>\n";
echo "$img_name<br>\n";
} else {
echo "<img src=\"alert.red.gif\">\n";
}

And somewhere on the bottom:
if($img != "none") {
echo "<input type=\"hidden\" name=\"img\" value=\"$img\"\n>";
echo "<input type=\"hidden\" name=\"img_size\" value=\"$img_size\">\n";
echo "<input type=\"hidden\" name=\"img_type\" value=\"$img_type\">\n";
echo "<input type=\"hidden\" name=\"img_name\" value=\"$img_name\">\n";
} else {
echo "<input type=\"hidden\" name=\"img\" value=\"\"\n>";
echo "<input type=\"hidden\" name=\"img_size\" value=\"\">\n";
echo "<input type=\"hidden\" name=\"img_type\" value=\"\">\n";
echo "<input type=\"hidden\" name=\"img_name\" value=\"\">\n";
}

And in file 3:
if($img) {
$binary_junk = addslashes(fread(fopen($img, "r"), filesize($img)));
}

and insert $binary_junk, $img_size, $img_type and $img_name into mySQL db...

The code that generates the error msg is
$binary_junk = addslashes(fread(fopen($img, "r"), filesize($img)));

Got any ideas?
Thanks, Arni

Anon

So $img gets posted from a form?