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PostPosted: Fri Aug 10, 2007 12:13 pm 
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Posts: 521
did you update

$Query = "SELECT ID FROM #DATABASE TABLE# WHERE username = '$username'";
$Result = mysql_db_query(#DATABASE NAME#, $Query, $Link);

accordingly? everything that is #CAP# is things that need to be changed

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PostPosted: Fri Aug 10, 2007 12:24 pm 
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Joined: Mon Jan 15, 2007 10:20 am
Posts: 18
Location: Charmhaven, Australia
I hadn't but I have now. It now reads

#database table#
#database name#

Yet I still get the error below.

Parse error: parse error, unexpected T_IF, expecting ')' in /home/www/xxxxxx.com/registration2.php on line 9

Regards,
Alan.


Last edited by DumbOne on Fri Aug 10, 2007 12:46 pm, edited 1 time in total.

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 Post subject:
PostPosted: Fri Aug 10, 2007 12:31 pm 
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Posts: 521
on $Result = mysql_db_query(#DATABASE NAME#, $Query, $Link);

change it so it has 'DATABASE NAME'

so its in '

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PostPosted: Fri Aug 10, 2007 12:40 pm 
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Getting near the end! It now reads

Parse error: parse error, unexpected T_ECHO, expecting ')' in /home/www/xxxxxx.com/registration2.php on line 16


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 Post subject:
PostPosted: Fri Aug 10, 2007 12:46 pm 
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Posts: 521
did you make the same changes on the last 2 #CAPS#?

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 Post subject:
PostPosted: Fri Aug 10, 2007 1:00 pm 
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Joined: Mon Jan 15, 2007 10:20 am
Posts: 18
Location: Charmhaven, Australia
This is what I now have

<?php
$Link = mysql_connect ('host,'username','password'); //Fill in the details

$username = $_POST['username']; //Gets username
$password = $_POST['password']; //Gets password

$Query = "SELECT ID FROM 'database table' WHERE username = '$username'";
$Result = mysql_db_query('database name', $Query, $Link);
if ($Row = mysql_fetch_array($Result)){
echo "Username already taken"; //Sends Error Message
die(); //Ends Script
}

$Query2 = "INSERT INTO 'database table' VALUES ('', '$username', '$password')";
if ( mysql_db_query('database name', $Query2, $Link)){
echo "Thank you for registering";
}
mysql_close();
?>

And get the message

Parse error: parse error, unexpected T_STRING in /home/www/xxxxxxxx.com/registration2.php on line 2


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 Post subject:
PostPosted: Fri Aug 10, 2007 1:12 pm 
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Joined: Mon Jan 15, 2007 10:20 am
Posts: 18
Location: Charmhaven, Australia
I just noticed I'm missing the ' after host which I have just added.
Now I get the messages;

Warning: mysql_connect(): Unknown MySQL Server Host 'host' (1) in /home/www/xxxxxx.com/registration2.php on line 2

Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in /home/www/xxxxxx.com/registration2.php on line 8

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/xxxxxx.com/registration2.php on line 9

Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in /home/www/xxxxxx.com/registration2.php on line 15

Warning: mysql_close(): no MySQL-Link resource supplied in /home/www/xxxxxx.com/registration2.php on line 18

I'll have to leave it for now, it's 11:15pm and I've had too many late nights.
Thanks for your all help. All going well I'll be back tomorrow.


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 Post subject:
PostPosted: Fri Aug 10, 2007 1:27 pm 
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Joined: Tue Feb 27, 2007 12:18 am
Posts: 521
well as it doenst know what

host

is, you have to put in your MySQL host!

It will be something like

mysql3.freehostia.com

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 Post subject:
PostPosted: Sat Aug 11, 2007 8:14 am 
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Joined: Mon Jan 15, 2007 10:20 am
Posts: 18
Location: Charmhaven, Australia
Josh, I've had a good long stare at the code and have made some minor changes. The code now looks like this;

<?php
$Link = mysql_connect ('mysql3.freehostia.com','my database name','mysql password'); //Fill in the details

$username = $_POST['username']; //Gets username
$password = $_POST['password']; //Gets password

$Query = "SELECT ID FROM 'logintable' WHERE username = '$username'";
$Result = mysql_db_query('my database name', $Query, $Link);
if ($Row = mysql_fetch_array($Result)){
echo "Username already taken"; //Sends Error Message
die(); //Ends Script
}

$Query2 = "INSERT INTO 'logintable' VALUES ('', '$username', '$password')";
if ( mysql_db_query('my database name', $Query2, $Link)){
echo "Thank you for registering";
}
mysql_close();
?>

I'm now getting the error;

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/xxxxxx.com/registration2.php on line 9

I'm unable to work it out, even though I have tried quite a few different code changes.
Thanks Again,
Alan.


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 Post subject:
PostPosted: Sat Aug 11, 2007 9:13 am 
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Joined: Tue Feb 27, 2007 12:18 am
Posts: 521
Code:
<?php
$Link = mysql_connect ('mysql3.freehostia.com','my database name','mysql password'); //Fill in the details

$username = $_POST['username']; //Gets username
$password = $_POST['password']; //Gets password

$Query = "SELECT ID FROM logintable WHERE username = '$username'";
$Result = mysql_db_query('my database name', $Query, $Link);
echo "Query1: " .  mysql_error(); //Comment out if the script works
if ($Row = mysql_fetch_array($Result)){
echo "Username already taken"; //Sends Error Message
die(); //Ends Script
}

$Query2 = "INSERT INTO logintable VALUES ('', '$username', '$password')";
if ( mysql_db_query('my database name', $Query2, $Link)){
echo "Thank you for registering";
}else{
     echo "Query2: " .  mysql_error(); //Comment out if the script works
}
mysql_close();
?>

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 Post subject:
PostPosted: Sat Aug 11, 2007 10:14 am 
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Joined: Mon Jan 15, 2007 10:20 am
Posts: 18
Location: Charmhaven, Australia
I hate being a nuisense but I'm getting another although different error;

Parse error: parse error, unexpected T_STRING in /home/www/xxxx.com/registration2.php on line 15

Give up if you like!


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 Post subject:
PostPosted: Sat Aug 11, 2007 12:09 pm 
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Joined: Mon Jan 15, 2007 10:20 am
Posts: 18
Location: Charmhaven, Australia
I'm in luck (almost)! It is now working; the usernames & passwords are loading onto my database, however I can put the same username & password in more than once and there is no message "Username already taken". I get the message below after submitting if I use a username once or twice. I have entered six, two being the same. As you can see the last username I entered was "me".

Query1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'username = 'me'' at line 1
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/xxxxx.com/registration2.php on line 10
Thank you for registering

That's it for tonight, Alan.


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