mysqli_prepare() expects parameter 1 to be mysqli

Question

Having trouble with procedural use of mysqli

Here\'s the function:

db.php

Answer 1

Since $link is being define in db.php probably you should

require db.php;

in function.php file

Answer 2

`enter code here`
need
   <?php
function addItemToCatalog($var1, $var2, $var3, $var4) {
    $sql = 'INSERT INTO catalog (var1, var2, var3, var4, $link) 
            VALUES (?, ?, ?, ?)';

end
<?php
require_once ("db.php");
require_once ("function.php");

$var1 = $_POST['var1']; //showing without filtering methods
$var2 = $_POST['var2'];
$var3 = $_POST['var3'];
$var4 = $_POST['var4'];

if(!addItemToCatalog($var1, $var2, $var3, $var4, $link)){ 
    echo 'some error text';

Answer 3

The way the two files db.php and function.php are glued together I think results in $link being defined as a global variable - you need to use global to access it within function:

function addItemToCatalog($var1, $var2, $var3, $var4) {
    global $link;
    ...
}

or give the $link to the function explicitly through parameter:

function addItemToCatalog($var1, $var2, $var3, $var4, $link) {
    ...
}