短PHP代码不返回数据（return，true或false）

Question

When I run this code, everything works:

<?php
$con = mysqli_connect ("", "", "", "") or die ($connect_error);

$username = 'tijmen';

$query = mysqli_query($con, "SELECT COUNT(*) FROM users WHERE username='$username'") or     die (mysqli_error());
$result = mysqli_fetch_row($query); 
echo $result[0];
?>

The username 'tijmen' exists in my database so the echo returns '1'.

Next thing I tried was:

$con = mysqli_connect ("", "", "", "") or die ($connect_error);

function user_exists($username) {
$query = mysqli_query($con, "SELECT COUNT(*) FROM users WHERE username='$username'") or die (mysqli_error());
$result = mysqli_fetch_row($query);  
return ($result[0] == 1) ? true : false;
}

if (user_exists('tijmen') === true) {
echo 'user exists';
}
else {
echo 'not found';
}

When I run this code nothing happens and I don't know why. Can anyone help me out on this one?

Answer 1

You do not have $con in your function.

replace this line:

if (user_exists('tijmen') === true) {

by :

if (user_exists('tijmen',$con) === true) {

and in your function:

function user_exists($username,$con) {
$query = mysqli_query($con, "SELECT COUNT(*) FROM users WHERE username='$username'") or die (mysqli_error());
$result = mysqli_fetch_row($query);  
return ($result[0] == 1) ? true : false;
}