parse error her is my code

deolsabh

<head>
<title>web site</title>
</head>
<body>
<h1>web stie </h1>
<?
#if (!$name || !$last || !$street || !$city || !$zip || !$dob )
#{
#echo " You have not entered all the required details.<br>"

."Please go back and try again.";

#exit;
#}
$name = addslashes($name);
$last = addslashes($last);
$email = addslashes($email);
$address = addslashes($address);
$city = addslashes($city);
$zip = addslashes($zip);
$state = addslashes($state);
$phone = addslashes($phone);
$dob = addslashes($dob);

@ $db = mysql_pconnect("localhost", "user", "password");

if (!$db)
{
echo "Error : Could not connect to database. Please try again later.";
exit;
}

mysql_select_db("bday");
$query = "insert into customers values
('".$name."', '".$last."', '".$email."', '".$address."', '".$city."', '".$zip."', '".$state."', '".$phone."', '".$dob."')";
$result = mysql_query($query);

die('MySQL error: ' . mysql_error() . '<br>Query: ' . $query);

if (!$name || !$last || !$street || !$city || !$zip || !$dob )
{
echo " You have not entered all the required details.<br>"
."Please go back and try again.";
exit;
}

if ($result)
echo mysql_affected_rows(). " persons added into database.";
?>

</body>
</html>
~

leatherback

Hi,

It would make life much easier for all of us if you tell us WHERE the parse error should be (It is in the error you get) and pointing us to that line.

I suppose this is the problem:
$query = "insert into customers values
('".$name."', '".$last."', '".$email."', '".$address."', '".$city."', '".$zip."', '".$state."', '".$phone."', '".$dob."')";

There is just way too many quotes in it, as far as I can tell.'

try:

$query = "insert into customers values 
('$name', '$last', '$email', '$address', '$city', '$zip', '$state', '$phone', '$dob')";

I'd also suggest to use a full insert command, which will prevent you from inserting in the wrong colums:

$query = "insert into customers (name, last, email, address, city, zip, state, phone, dob) values 
('$name', '$last', '$email', '$address', '$city', '$zip', '$state', '$phone', '$dob')";

deolsabh

thanks a lot that worked but now i get following error from my code
echo " You have not entered all the required details.<br>"
."Please go back and try again.";
exit;

i commented this set of code and then i was success full but no inofrmation was submitted to database

please help

leatherback

Where do the values for the insert come from? If they are the result of a submitted form, and you are using a newer version of PhP, you most likely have to extyract the values from the $_POST array. This is the result of a change in security settings (REGISTER_GLOBALS is by default switched of, instead of on, which was the default before 4. ??? ).

So, instead of using $name, you have to extract and define $name:

$name = $_POST[name];

Do this for all variables.

To check what has been passed to your script, add:

echo "<pre>";
print_r($_POST);
echo "</pre>";

This will show you the structure of the $POST array, and show the values of the var's in it.
(You can use the print_r for all arrays).

One other thing: Pace, for debugging purposes:

echo "<p>" . $query . "</p>";

to your script, after defineing the $query var. That way you can see what is passed to the database server.

deolsabh

This works great
thank you very much for your help

leatherback

Hi,

If the problem is solved, please mark this thread resolved (See link at bottm of page).,
Then others won't have to check it anymore.

Cheers.,

Jelle

Happy coding.

deolsabh

one last question
now all field are ok in my database but address field i dont see any information
it is as

address varchar not null
i changed it to
address test not null
but still nothing goes to database.

leatherback

IF only address doesn't get included..

Is there a value for it in the query?

The value of address is not passed from the form (Different name on the form perhaps)
The columname might be different (Capital letters?)
You did not fille the variable (Again: Spelling/caps?)
Have a carfull look at the query which is echod to the screen. If you have PhPmydamin: Paste that code in the sql command section and run i. It should tell you what is going wrong!

deolsabh

I changed the name of the it works now
thanks
prithvi