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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-----Original Message-----
From: Josh Jones [mailto:bootsec99@yahoo.com]
Sent: woensdag 13 november 2002 14:25
To: php-db@lists.php.net
Subject: 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 "<tr><td>Name</td><td>Phone</tr>\n";
while ($myrow = mysql_fetch_row($result)) {
printf("<tr><td>%s</td><td>%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="name"><br>
Phone:<input type="Text" name="phone"><br>
<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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