Undefined Variable on mysqli_query - php

Question

I have similar code that works with no errors but when I try and query my database I'm getting the several errors. Can anyone help ?
errors:
Notice: Undefined variable: link in C:\wamp64\www\twitter clone\functions.php on line 22
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\wamp64\www\twitter clone\functions.php on line 22
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\wamp64\www\twitter clone\functions.php on line 24
Code;
<?php
session_start();
$link = mysqli_connect("localhost", "root", "pass", "twitter");
if (mysqli_connect_errno()) {
print_r(mysqli_connect_error());
exit();
}
if (isset($_GET['function']) == "logout"){
session_unset();
}
function displayTweets($type) {
if ($type == 'public'){
$whereClause = "";
}
$query = "SELECT * FROM tweets ".$whereClause." ORDER BY 'datetime' DESC LIMIT 10";
$result = mysqli_query($link, $query);
if (mysqli_num_rows($result) == 0) {
echo "No Tweets";
}
}
?>

Answer 1

The problem is that your $link variable is not in the function's local scope.
A way to fix this is by defining the variable in the function's local scope by adding the following line to the function:
global $link
Please have a look at this page to read more about the variable scope in PHP.
Edit:
An even better way would be to inject your connection by adding it as an argument to your function, which would have your code look something like this:
function displayTweets($type, $connection) {
if ($type == 'public'){
$whereClause = "";
}
$query = "SELECT * FROM tweets ".$whereClause." ORDER BY 'datetime' DESC LIMIT 10";
$result = mysqli_query($connection, $query);
if (mysqli_num_rows($result) == 0) {
echo "No Tweets";
}
}
After this, you would call your function using displayTweets('public',$link), with 'public' being the type, and $link being your defined connection
Also, in your current function, $whereClause could be undefined. I'm guessing your aware of this, just wanted to state it in case you get errors on that.

Answer 2

You have to use $link variable as global variable ($GLOBALS['link']) in displayTweets function or you have to pass $link as parameter.
$result = mysqli_query($GLOBALS['link'], $query);
or
displayTweets('public',$link);