mysqli_error() expects exactly 1 parameter, 0 given [duplicate]

This question already has answers here: Warning: mysqli_error() expects exactly 1 parameter, 0 given error (4 answers) Closed 2 years ago.

I am trying to get my head around mysql. Can someone tell my why this mysql query is not working? I am getting the following error:

Warning: mysqli_error() expects exactly 1 parameter, 0 given in /home/freebet2/public_ht开发者_如何学Cml/test.php on line 11

test.php

<?php
        require_once($_SERVER['DOCUMENT_ROOT'].'/includes/db.php');
        $conn = db_connect();
        $result = $conn->query("ALTER TABLE users ADD COLUMN refer_old INT(10) AFTER refer_id");

              if(!$result){
                 echo "Error with MySQL Query: ".mysqli_error();
             }
        ?>

db.php

 <?php

    function db_connect() {
       $result = new mysqli('localhost', 'user', 'password', 'db');
       if (!$result) {
         throw new Exception('Could not connect to database server');
       } else {
         return $result;
       }
    }

    ?>

If I change the alter string to something like : $result = $conn->query("SELECT * FROM users refer_id"); I get no error for some reason.

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You are mixing the object-oriented and the procedural styles of the mysqli API :

You are using object-oriented :

$result = new mysqli('localhost', 'user', 'password', 'db');

And, then, procedural :

echo "Error with MySQL Query: ".mysqli_error();

You should use either OO, or procedural -- but not both ; and if you choose procedural, the functions expect the link identifier passed as a parameter.

For instance, mysqli_error should be called either using the object-oriented API :

$link = new mysqli(...);
echo $link->error;

Or the procedural API :

$link = mysqli_connect(...);
echo mysqli_error($link);

(Of course, it will not change the fact that you are having an error in your SQL query, but it'll allow you to get the error message, which should help finding the cause of that error)

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As far as the sql error is concerned, does 'user' have permissions to alter the table?

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Use mysqli_error($result) as mysqli_error expects the connection to be passed as a parameter.