Newbie php addind data to mysql

ok i first tried this.....

Code:

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<form action="insert.php" method="post">

UserName: <input type="text" name="username"><br>
Password: <input type="text" name="password"><br>

<input type="Submit">
</form>

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and in insert.php....

PHP Code:

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<?


$database="users";

mysql_connect() or die ("Problem connecting to DataBase");

@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO info VALUES ('$username','$password')"; 
mysql_query($query);

mysql_close();

?>

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i've been looking at these global varible things but they still not making sence to me can anyone tell em how to solve this? all iw ant to do is add data to mysql

Try to change the insert line to:

Code:

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$query = mysql_query("INSERT INTO info VALUES('$username','$password')");

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BTW what is the error message?


Is there only two fields in that table?

You should also make the password input of type password (makes the stars show).

Oh, and I scan the forums almost daily, there's no need to PM me.

Notice: Undefined variable: username in (document path)insert.php on line 11

Notice: Undefined variable: password in (document path)insert.php on line 11

thanks

sry botut th epm its just i've been trying for nerly 5 days to learn :)

You're trying to use the deprecated (for security reasons) and by default disabled feature of PHP to offer all passed data as global variables. As of new PHP versions, you can access the data with
$_GET['fieldname'] (if the form has method="get")
and
$_POST['fieldname'] (if the form has method="post")

Looks werid...if there is a password or something on the database try to connect to it like this:

PHP Code:

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@ $db = mysql_pconnect ("localhost", "name_of_the_user", "password_to_the_database");

if (!$db){
    echo 'I cannot connect to the database because: ' . mysql_error();
    exit;
}

mysql_select_db ("name_of_the_database"); 


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[Edit]Forget about this: Listen to the man that know this...:D..[/Edit]

how would i incorperate that into my problem? sorry for i'm still learning :)

Quote:

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Posted by NoteMe

PHP Code:

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mysql_error(); 


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Ar so thats how you find out what went wrong in a DB call :D. I've been wondering how to find that out, suppose mysql_error() does kinda make sense :rolleyes:. :D

where do i put the post thing? im confused please help me!

Rewrite the $query = line like this:
$query = "INSERT INTO info VALUES ('".$_POST['username']."','".$_POST['password']."')";

It's possible to do it without concatenating strings, but in case of arrays I don't like doing this for several reasons, one of which is that it breaks the syntax highlighting of jEdit :)