loginHelp please!

[Action_Man]

Hi i am having difficulty creating a login system for a project. The error i get is

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\wamp\www\project\studentregister.php on line 5
Database error: Access denied for user 'ODBC'@'localhost' (using password: NO)

my php code is

studentregister.php

<?php

require "student_db.php";

$link = mysql_connect( $db_host, $db_username, $db_password ); // connection to database

if ( !$link )

exit ( "Database error: " . mysql_error() );

mysql_select_db ( $db_name, $link ); // selects database name


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

$result = mysql_query( "INSERT INTO studentreg (username, password) VALUES('$username', '$password')", $link ); // insert post into db

if( !$result )
    exit ( "Query failed: " . mysql_error() );
?>

student_db.php

<?php

$db_name = "project";
$db_host = "localhost";
$db_user = "root";
$db_pass = "";

?>

in my html file i have

<form name="login" method="post" action="studentregister.php">																					
																				<br>


									  <table width="560" border="1" cellpadding="2" cellspacing="2" bordercolor="#000000" class="text">
												<tr>
													<td class="text" align="center">Student Registration <BR>

														<BR>
												  </td>
												</tr>
												<TR>
													<TD>
														<TABLE class="tablesalt" id="Table1" cellSpacing="0" cellPadding="1" width="560" border="0">
															<TR>
																<TD class="boldtext" height="10" width="194"></TD>
																<TD height="10"></TD>
																<TD height="10"></TD>
															</TR>
															<TR>
																<TD class="boldtext" height="10" width="194"></TD>
															</TR>
															<TR height="10">
																<TD class="boldtext" width="194">Username</TD>
																<TD><input name="username" type="text" maxlength="20" id="Username" class="textboxes" style="width:150px;" /></TD>
																<TD></TD>
															</TR>

															<TR class="Rows">
																<TD class="boldtext" width="194">Password</TD>
																<TD><input name="password" type="password" maxlength="20" id="Password1" class="textboxes" style="width:150px;" /></TD>

[bpat1434]

NOTE: Your post was moderated. If you had read the notice after creating the thread, you would have seen that some posts (particularly those with certain key words or phrases and by new members) are automatically queued for moderator's approval. That does not mean you should keep posting the same topic multiple times.

Now, your problem is not in your HTML, but in your PHP code for authentication. The user "root" which in windows is ODBC it seems, is not allowed to connect. What username did you give MySQL when you installed WAMP? Was it "root", or something else?

[Action_Man]

Sorry for that i didnt realise it. I have left my wamp details as default so it should be root?

[Joseph07]

You have defined the following as your variables to connect to the database.

$db_name = "project";
$db_host = "localhost";
$db_user = "root";
$db_pass = ""; 

However, you have used a variable other than the above 4 to connect to the database.

// Note $db_username and $db_password
$link = mysql_connect( $db_host, $db_username, $db_password ); // connection to database

if ( !$link )

exit ( "Database error: " . mysql_error() );

mysql_select_db ( $db_name, $link ); // selects database name 

You should change $link to

$link = mysql_connect( $db_host, $db_user, $db_pass ); // connection to database

[Action_Man]

Great thanks that solved that problem, just one more error is appearing

Query failed: Duplicate entry '0' for key 1

i presume this is to do with my mysql table?

[CODE]Table structure for table studentreg

CREATE TABLE studentreg (
userid int(25) NOT NULL,
username varchar(25) NOT NULL,
password varchar(255) NOT NULL,
PRIMARY KEY (userid)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--

-- Dumping data for table studentreg

INSERT INTO studentreg (userid, username, password) VALUES
(0, '', ''); [/CODE]

[bpat1434]

username and password can not be null. You have to put something in there. Plus, all keys in MySQL tables start with 1, not 0.

[_uk_stuff]

your MySQL table does not contain an auto_increment on the userid column so will default to 0. Therefore if you try to insert more than one record without specifying the userid it will assume 0 and complain about duplicate keys... as per the error message.

I would suggest to change the create statement to be as follows:

CREATE TABLE studentreg (
userid int(25) NOT NULL AUTO_INCREMENT,
username varchar(25) NOT NULL,
password varchar(255) NOT NULL,
PRIMARY KEY (userid)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

This should then control the userid aspect for you.

Hope it helps...

[Action_Man]

That worked great they are all saving now, i have started to add in the rest of my form fields but am stuck on one

i have a date of birth and it uses <select name="day" month and year etc,

how do i get that working a mysql field. single input names are grand but im not sure how to display multiple select names. any ideas?

[_uk_stuff]

You would need to 'glue' all of the select fields into a single variable and insert that into the DoB field.

Or failing that you could create separate the DoB into the three sections within the table itself.

My recommendation would be to glue them together and insert them i.e.

$DoB = $_POST['day']."-".$_POST['month']."-".$_POST['year'];

You need to make sure that the format that you insert the date in is also correct... I know that Microsoft SQL expect the format to be YYYY-MM-DD, yet I think MySQL uses DD-MM-YYYY.

[Action_Man]

cheers for that gonna try it now.

[bpat1434]

MySQL date format expects YYYYMMDD for Date-stamps. You can use a unix-timestamp instead as well. If I were to do this, I'd use the MySQL default format (YYYYMMDD) and store it in a DateTime type field in the table.

Then when I want to get the age I can let MySQL do the work:

$query = "SELECT TIMESTAMPDIFF(YEAR, `birthdate_column`, NOW()) AS age
FROM `table`";

Then Age is the number of years old that person is. Simple calculation in MySQL that won't slow you down and you don't have to worry about in code 😉