Update SQL tables through php form

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', '[email protected]', '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.

G-Nugget · Accepted Answer · 2012-10-09 21:41:50Z

2

The query is flawed. You name 4 columns (username, nicename, email, password), but you list 5 values ('orion5814','Orion5814','[email protected]','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.

answered Oct 9, 2012 at 21:41

G-Nugget

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1 Comment

Morgan Green Over a year ago

You were correct on that and I also appreciate the input on simplifying strings. On my list of things to do; thank you very much

alan978 · Accepted Answer · 2012-10-09 21:47:59Z

2

As jordi12100 suggested it is good pratice that you check errors while you connecting to database or executing queries.

You can do it like this:

$link = mysqli_connect("$server", "$user", "$pass", "$webdb") or die( "Error:" . mysqli_connect_error());

mysqli_query($link, $query) or die ("Error:" . mysqli_error($link));

This can give you idea what you did wrong. Hope this helps.

answered Oct 9, 2012 at 21:47

alan978

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Morgan Green Over a year ago

Major Help; I'm getting used to mysqli and having something give me error numbers will help drastically.

Jordi Kroon · Accepted Answer · 2012-10-09 21:54:25Z

1

Probarly an error in your query. Catch the error with mysqli_error();

edited Oct 9, 2012 at 21:54

answered Oct 9, 2012 at 21:40

Jordi Kroon

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