hi,
This may sound very newbie. But since html comment uses !-- -- which doesn't comment
out the
php code. And the php comment /**/ // apparently doesn't work on html. So how do I
comment out
a chunk of html/php mixed code?
Many thanks,
Wei Wang
and be stuck at this one.
Any advice would be greatly appreciated.
Wei Wang
!DOCTYPE HTML PUBLIC -//W3C//DTD HTML 4.01 Transitional//EN
html
head
titleUntitled Document/title
meta http-equiv=Content-Type content=text/html; charset=iso-8859-1
/head
body
h2Simple Form Example/h2
? function
, 2003 1:22 AM
Subject: Re: [PHP] simple online form stuck at !isset($first) condition
* Thus wrote Wei Wang ([EMAIL PROTECTED]):
? function show_form($first=,$last=) { ?
form action=formprac.php method=POST
note the method --
First Name:
input type=text name
of this code.
I'd highly appreciate it if anyone could drop me a line with any advice.
Many thanks,
Wei Wang
?
/*
PHP Guestbook 1.1
Written by Tony Awtrey
Anthony Awtrey Consulting
See http://www.awtrey.com/support/dbeweb/ for more information
1.1 - Oct. 20, 1999 - changed the SQL statement that reads
a very fundamental point
that led to the failure of this code.
I'd highly appreciate it if anyone could drop me a line with any advice.
Many thanks,
Wei Wang
?
/*
PHP Guestbook 1.1
Written by Tony Awtrey
Anthony Awtrey Consulting
See http://www.awtrey.com/support/dbeweb/ for more information
hi, all,
I have a piece of code stuck at the point of if (($REQUEST_METHOD=='post')).
var_dump($REQUEST_METHOD); shows that the variable doesn't exist.
Since I grabbed the code from a 2001 update, is there any chance that this is already
obsolete? I checked the documentation but didn't find
I am not sure if this is the right place to ask this naive question.
I got a simple form addform.html and add.php look like the following.
But everytime I got empty value from firstname and lastname. It seems like
the input value in the html page was not passed on to the php variable $firstname
Do you mind telling me where this php.ini is?
And I tried the second attempt like this:
$query = INSERT INTO friends (id, firstname, surname) values
(nextval('friends_id_seq'), $_POST['firstname'], $_POST['surname']);
which returns error:
Parse error: parse error, unexpected
Create a php script containing just ?php phpinfo(); ?. That page will tell
you where your php.ini is (or should be).
it told me it's in php/lib, but it's not there.
Try this...
$query = INSERT INTO friends (id, firstname, surname) values
(nextval('friends_id_seq'),
Sorry. When I input Andras as firstname in the form, ...
On Sat, 1 Jun 2002 17:22:02 +0100
[EMAIL PROTECTED] (Wei Wang) wrote:
Create a php script containing just ?php phpinfo(); ?. That page will tell
you where your php.ini is (or should be).
it told me it's in php/lib, but it's
Is there anything I should do to make the php.ini come into effect? I did
put the register_global = On but still got empty values.
And my $_POST version is like this:
?php
$db = pg_connect(dbname=friends);
$query = INSERT INTO friends (id, firstname, surname) values
Sorry.
I restared apache and php.ini came into effect.
On Sat, 1 Jun 2002 17:29:13 +0100
[EMAIL PROTECTED] (Stuart Dallas) wrote:
On Saturday, June 1, 2002 at 5:22:02 PM, you wrote:
Create a php script containing just ?php phpinfo(); ?. That page will tell
you where your php.ini is (or
Changed the query and get this error when input dave as firstname:
ERRORERROR: Attribute 'dave' not found
add2.php is:
html
body
?php
$db = pg_connect(dbname=friends);
$query = INSERT INTO friends (id, firstname, surname) values
(nextval('friends_id_seq'), .$_POST['firstname'].,
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