[SOLVED] Somethign Wrong With My Code?

[LostKID]

Hi everyone im having a bit of a problem i cant seem to connect to my database.. well i cant seem to insert any information to be precise. im ok pulling information back from the database which leads me to believe there may be something wrong with my coding.. could you lend me some time to show me where i have gone wrong?

 

PHP Processing code after registering details.

<?php 
// CONNECT TO THE DATABASE
mysql_connect ("localhost", "xxx", "xxx") or die ('Error: ' . mysql_error());
mysql_select_db ("xxx");

// CALL IN VARIABLES
$fname = $_POST['fname'];
$lname = $_POST['lname'];

// THE QUERY
$query = "INSERT INTO user (fname,lname) VALUES ('$fname', '$lname')";

// UPDLOAD
$result = mysql_query($query) or die ('Error updating database');

if($results){
echo "Account createed Successfully, please login with your details.";
}

?>

 

The details are correct it just keeps giving me the error "'Error updating database".

The table exists in the database too.. its doing my head in.

[mikesta707]

echo the mysql_error() and see what happens

[peranha]

Change this

// UPDLOAD
$result = mysql_query($query) or die ('Error updating database');

 

to

 

// UPDLOAD
$result = mysql_query($query) or die ('Error updating database'.mysql_error());

 

See if you get any error messages.

[AviNahum]

i'm not sure, but try this:

<?php 
// CONNECT TO THE DATABASE
mysql_connect ("localhost", "xxx", "xxx") or die ('Error: ' . mysql_error());
mysql_select_db ("xxx");

// CALL IN VARIABLES
$fname = $_POST['fname'];
$lname = $_POST['lname'];

// THE QUERY
$query = "INSERT INTO user ('fname' , 'lname') VALUES ($fname, $lname)";

// UPDLOAD
$result = mysql_query($query) or die ('Error updating database');

if($results){
    echo "Account createed Successfully, please login with your details.";
}

?>

[LostKID]

this is the link btw.. sorry

 

www.brokenbox.co.uk/MY

 

i have added the first 2 fixes will try the 3rd

[cbolson]

Are you sure that it isn't inserting into the database? - ie have you checked your tables directly?

 

You have a small error here:

if($results){
   echo "Account createed Successfully, please login with your details.";
}

You have defined the mysql_query as $result no $results so your message will not be shown.

 

Chris

[LostKID]

okay those last 2 fixes done it, its adding data to my database now i can go back to adding the rest of the code back to how it was lol thanks  you guys your the best.

[LostKID]

Hi again, i decided to reopen this thread up as not solved again instead of making another topic to fill up the forum.

 

i been adding the rest of my code. ie validation and what not and my code to see if the email address is already in use doesnt work. any ideas?

 

<?php 
// CONNECT TO THE DATABASE
mysql_connect ("localhost", "xxx", "xxx") or die ('Error: ' . mysql_error());
mysql_select_db ("xx");

// CALL IN VARIABLES
$username = $_POST['username'];
$password = $_POST['password'];
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$gamert = $_POST['gamert'];

// VALIDATE
if($fname == "" || $lname == ""){
echo("you didnt enter your full name");
exit();
}
if($password == ""){
echo("you didnt enter anything for your password, please try again");
exit();
}
if($email == ""){
echo("you didnt enter anything for your email, please try again");
exit();
}
if(!ereg("^.+@.+\\..+$", $email)){
echo("the email you entered was not valid, please try again");
exit();
}

// CHECK IF EMAIL EXISTS
$sql = "SELECT email FROM user WHERE email='$_POST[email]'";
$result = mysql_query($sql) or die('couldnt execute query'.mysql_error());
$num = mysql_num_rows($result);
if($num == 1){
echo("This email address is already in use, please use another email address");
exit();
}

// THE QUERY
$query = "INSERT INTO user (username,password,fname,lname,email,gamert) VALUES ('$username','$password','$fname','$lname','$email','$gamert')";

// UPDLOAD
$result2 = mysql_query($query) or die ('Error updating database'.mysql_error());

if($result2){
echo "Account createed Successfully, please login with your details. you will now be redirected..";
}

?>
<script type="text/javascript">
<!--
setTimeout('Redirect()',2000);
function Redirect()
{
location.href='index.php';
}
//-->
</script>

[LostKID]

and yes anyone looking over the code and seeing i havent used mp5 for passwords, i will put that in after i know everything is working okay. its just for my debugging. cos im not very good at any of this lol

[cbolson]

You have this:

 

if($num == 1){

 

That is rather limiting as it only returns true if there is exactly one result, what if the email is in the db more than once?

 

You should use this

if($num <>0){

or

if($num >0){

 

Chris

[LostKID]

You have this:

 

if($num == 1){

 

That is rather limiting as it only returns true if there is exactly one result, what if the email is in the db more than once?

 

You should use this

if($num <>0){

or

if($num >0){

 

Chris

 

ding ding we have a winner... damn cant believe that slipped by me lol i shouldnt cleaned out the database.. thats the reason it wasnt working at all. thanks your right ill use the more than expression. thanks! me so tired the little things get passed me. good job there are great people on this forum to lend a helping hand!