[SOLVED] MYSQL INSERT INTO help!

[fonecave]

Why is this code not adding anything?

 

It connects ok but doesnt add a recordset

 

<?php
if (isset ($_POST['submit']))

{
$name=$_POST["name"];
$address=$_POST["address"];
$password=$_POST["password"];

$con = mysql_connect("localhost","my_db","password"); //these are correct login details table is called Members
if (!$con)
  {
  die('Contact support@fonecave.co.uk - Could not connect: ' . mysql_error());
  }
mysql_select_db("fonecave_00001", $con);
mysql_query("INSERT INTO members (name, email, password) VALUES ('$name', '$address', '$password')");

echo "Registration Complete!";

?>

[DeFActo]

try this

<?php
if (isset ($_POST['submit']))

{
$con = mysql_connect("localhost","my_db","password"); //these are correct login details table is called Members
if (!$con)
  {
  die('Contact support@fonecave.co.uk - Could not connect: ' . mysql_error());
  }
mysql_select_db("fonecave_00001", $con);
mysql_query("INSERT INTO members (name, email, password) VALUES ('$_POST[name]', '$_POST[address]', '$_POST[password]')");

echo "Registration Complete!";

?>

[fonecave]

I Have tried that i get this

 

 

Parse error: syntax error, unexpected T_ELSE in /www/110mb.com/f/o/n/e/c/a/v/e/fonecave/htdocs/login.php on line 438

 

[DeFActo]

<?php
if (isset ($_POST['submit']))

{
$con = mysql_connect("localhost","my_db","password"); //these are correct login details table is called Members
if (!$con)
  {
  die('Contact support@fonecave.co.uk - Could not connect: ' . mysql_error());
  }
mysql_select_db("fonecave_00001", $con);
mysql_query("INSERT INTO members (name, email, password) VALUES ('$_POST[name]', '$_POST[address]', '$_POST[password]')");

echo "Registration Complete!";
}
?>

what about now?

 

[fonecave]

i get the successful text echoed on the page,

 

but nothing appears in the database not even blank records

 

but it must be connecting fine can you think of anything else

[DeFActo]

Could you post your form?

[bravo81]

Sure the DB Table isnt Members? and not members like you wrote.

 

Sounds stupid but could be it.

 

Have you set the fields to the right settings? Like..your not trying to isntert text into an ENUM field etc.

 

screenshot of the DB? Form HTML?

 

Hope that helps.

[fonecave]

** Deleted for security purposes **

[fonecave]

Table structure for: Members

 

  Field Name                              Type                    Null                  Key                      Default                      Extra

  name                                      varchar(20)            No                                            Unknown 

  email                                      varchar(30)            No                  PRI 

  password                                varchar(20)            No

 

Ihave tried members and Members neither work

[fonecave]

I Think there was an error with the db i have receated the table and it now work,

 

What code would i use to search for a duplicate email address before adding?

[bravo81]

This is raw code from one of my projects:

 

$sql_email_check = mysql_query("SELECT email FROM users 

            WHERE email='$email' AND status='Alive'"); 

$sql_username_check = mysql_query("SELECT username FROM users 

            WHERE username='$reg_username'"); 



$email_check = mysql_num_rows($sql_email_check); 

$username_check = mysql_num_rows($sql_username_check); 



if(($email_check > 0) || ($username_check > 0)){ 

    echo ""; 

    if($email_check > 0){ 

        $message= "*This email address has already been used by another player!"; 

        unset($email); 

    } 

    if($username_check > 0){ 

        $message="*That desired username is already in use!"; 

        unset($reg_username); 

    } 

[fonecave]

ok i get this now:

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /www/110mb.com/f/o/n/e/c/a/v/e/fonecave/htdocs/login.php on line 406

 

$sql_email_check = mysql_query("SELECT email FROM members WHERE email='$address'"); 

$email_check = mysql_num_rows($sql_email_check);   //line406

if($email_check > 0) {

echo 'Registration Failed, Already Registered!';
}
else {

mysql_select_db("fonecave_00001", $con);
mysql_query("INSERT INTO members (name, email, password) VALUES ('$name', '$address', '$password')");
echo "Registration Complete!";}

[fonecave]

is this because the result is 0?

[AndyB]

Replace:

$sql_email_check = mysql_query("SELECT email FROM members WHERE email='$address'"); 

 

With this:

$query = "SELECT email FROM members WHERE email='$address'";
$sql_email_check = mysql_query($query) or die("Error: ". mysql_error(). " with query ". $query); 

 

If it's not obvious what needs to be fixed from the resulting error, post more information.

[fonecave]

Thanks for the tip, i must use that more.

 

Heres what i got...

 

Error: No database selected with query SELECT email FROM members WHERE email='user@fonecave.co.uk'

 

it was obvious what i had to do so now fixed!

 

Thank You all so much for your help yet again!