Mysqli Prepared Statement in bindParam() not working

Question

Just as usual i was looking around best practices with PHP, and prepared statements seems the kind of stuff i should now how do with my eyes closed. So i started playing around with some examples i've found.

I've got this error when running the script:

Fatal error: Call to a member function bindParam() on a non-object in /opt/lampp/htdocs/phpSecurity/PreparedStatments/Insert-Multi-Binded-Params/Insert Simple Method.php on line 10

Here it goes the code.

Insert Simple Method.php

<?php
require_once '../config.php';

$stmt = $db->prepare("INSERT INTO coisas (nome, telefone, bi) VALUES (?, ?, ?)");

$nome = 'Fabio Antunes';
$telefone = 916810641;
$bi = 123093456;

$stmt->bindParam(1, $nome);
$stmt->bindParam(2, $telefone);
$stmt->bindParam(3, $bi);

$stmt->execute();

$stmt->close();

$db->close();
?>

config.php

<?php
$server_host = 'localhost';
$server_user = 'root';
$server_password = '';
$server_db = 'PreparedStatements';
$db = new mysqli($server_host, $server_user, $server_password, $server_db);
?>

Not sure what i'm doing wrong here, this is similar example found at php.net, why isn't working? PS: I think the mysqli connection isn't the problem because I've used it to do some prepared statements with SELECT SQL commands. And worked pretty well.

---

EDIT

The Resolution and why.

Well in the example i should use bind_param() for each value in the query. But thanks to Bart, he managed to solve the problem with my code.

Where it is:

$stmt->bindParam(1, $nome);
$stmt->bindParam(2, $telefone);
$stmt->bindParam(3, $bi);

It should be:

$stmt->bind_param("sii", $nome, $telefone, $bi);

Now for those who might wondering what is "sii".

Well bind_param for what i see it binds the "$var" to each question mark "?" in order.

So with one bind_param() i can bind them all at the same time, and the normal use of bind_param() requires to specify the type of data being binded.

My first value to be binded is $nome a String, specified by the "s";

And the others $telefone and $bi are Integers for that he have "i";

For others that have a similar problem here it goes other data types (from php.net).

i = Integer;

s = String;

d = Double;

b = Blob;

If someone as a better explanation please post it or comment. So i can improve my own.

Thanks.

Answer 1

You may think there's nothing wrong with the connection, but you should check to make sure:

$db = new mysqli($server_host, $server_user, $server_password, $server_db);
if (mysqli_connect_errno()) {
    printf("DB error: %s", mysqli_connect_error());
    exit();
}

EDIT:

What happens when you do:

$stmt = $db->prepare("INSERT INTO coisas (nome, telefone, bi) VALUES (?, ?, ?)");
$stmt->bind_param("sii", $nome, $telefone, $bi);
$stmt->execute();

?

Is the table coisas spelled properly?