通过php表单更新SQL表

Question

Query is running however not being sent to SQL server. My Current Register Script.

$link = mysqli_connect("$server", "$user", "$pass", "$webdb");
$username = mysqli_real_escape_string($link, (string) $_POST['username']);
$displayname = mysqli_real_escape_string($link, (string) $_POST['display_name']);
$email = mysqli_real_escape_string($link, (string) $_POST['email']);
$password = sha1((string) $_POST['password']);
$query="INSERT INTO user (`username`, `nicename`, `email`, `password`)
VALUES ('$username', '$displayname', '$email', '$password', '1')";
mysqli_query($link, $query);
mysqli_close($link);
echo $query;
?>

The output I recieve from the Query:

INSERT INTO user (username, nicename, email, password) VALUES ('orion5814', 'Orion5814', '[e-mail]', '72f2ac484bee398758e769530dd56228d905884d', '1')

I've checked all my link variables and they're all set correctly as far as having the right information in place, so I don't know where else to go from here. Sorry for all the questions; you can view it at doxramos.org if you think it would help at all.

Answer 1

The query is flawed. You name 4 columns (username, nicename, email, password), but you list 5 values ('orion5814','Orion5814','[e-mail]','72f2ac484bee398758e769530dd56228d905884d','1') If you remove the last value, the query should work.

Also, you could simplify your code by using the object oriented interface to mysqli like this:

$username = $link->real_escape_string($_POST['username']);

and

$link->query($query);
$link->close();

You also don't need to explicitly cast the variables as strings since that is done automatically if needed for your code.