try before the :
if ($submit)
this:
phpself();
the vars are now declared on a different way in the new phph versions
you need to do:
if ($_POST[submit]){
and the insert values are also now :
echo $_POST[phone];
so check phpinfo(); there are all the vars you need...
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Atos Origin
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-Original Message-
From: Josh Jones [mailto:bootsec99;yahoo.com]
Sent: woensdag 13 november 2002 14:25
To: [EMAIL PROTECTED]
Subject: [PHP-DB] PHP 4 inserting data in MySQL 3.23
I'm running the following setup:
Apache 2.0.40-8
MySQL 3.23.52-3
PHP 4.2.2-8.0.5
I have no problems retreiving data from the database
and posting it to a website, but I can't insert data
from the website to the database via forms or
whatever.
I'm new to this, but I've triple checked all the
permissions and the code - it all works fine from the
MySQL command line. I can't find any errors in the
logs either, maybe I'm looking in the wrong place.
Here's a sample of some testing code:
html
body
?php
$db = mysql_connect(localhost, webuser);
mysql_select_db(example,$db);
$result = mysql_query(SELECT * FROM mytable,$db);
echo table border=1\n;
echo trtdName/tdtdPhone/tr\n;
while ($myrow = mysql_fetch_row($result)) {
printf(trtd%s/tdtd%s/td/tr\n, $myrow[0],
$myrow[1]);
}
echo /table\n;
?
/body
/html
html
body
?php
if ($submit) {
// process form
$db = mysql_connect(localhost, webuser);
mysql_select_db(example,$db);
$sql = INSERT INTO mytable (name, phone) VALUES
('$name','$phone');
$result = mysql_query($sql);
echo Thank you! Information entered.\n;
} else{
// display form
?
form method=post action=?php echo $PHP_SELF?
Name:input type=Text name=namebr
Phone:input type=Text name=phonebr
input type=Submit name=submit value=Enter
information
/form
?php
} // end if
?
/body
/html
Any help or direction pointing would be appreciated.
Thanks
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