PHP if else confusion - php

Question

i'm trying to make an image appearance and disappearance based on 3 condition,
condition A = when user is logged in and it's username fits the displayname(by using the GET function) then it should echo "yes"
condition B = When user is logged in and it's username does not fits the displayname then it should echo "no"
condition C = when user is not logged in then it should echo "no" too
(i swapped the image with yes and no for easier referencing)
By logging in, the user has a cookie which is set like below
setcookie("user", $user, $expire);
setcookie("loggedin", 1, $expire);
First i get the cookie which i set when user logins.
$user1 = $_COOKIE["user"];
$loggedin = $_COOKIE['loggedin'];
$user = strtoupper($user1);
then i get my player's name
$playername = $_GET['player'];
Now i do the conditions
$uplayername = strtoupper($playername);
function showplusicon(){
global $uplayername;
if(($loggedin = "1") and ($user == $uplayername)){
echo "yes";
}
else if (($loggedin = "1") and ($user != $uplayername)){
echo "no";
}
else{
echo "no";
}
}
I don't see what's the problem but it keeps being registered as condition B.

Answer 1

Single equal signs assign, not compare.
if(($loggedin == "1") and ($user == $uplayername)){
...
And since you really only have two output states, you shouldn't need 3 conditions; remove condition B.

Answer 2

The variable $loggedin isn't known inside your function showplusicon(). You will need to add it as a global along with global $uplayername.
function showplusicon(){
global $loggedin, $uplayername;
// etc
}
Since this was accepted but not totally complete, I'll just add that as others indicated, the == equality operator needs to be used instead of the = assignment operator.
if(($loggedin == "1")
^^^^

Answer 3

$loggedin = "1"
Surely this should be:
$loggedin == "1"
Otherwise I would echo $user and $uplayername to see if these differ.

Answer 4

First thing's first:
$loggedin = "1" is a bad idea, as you're actually giving $loggedin the value "1" instead of comparing. Use == or even === if you're sure about the datatype.
Further on, the $loggedin isn't available in the scope of showplusicon(), as you haven't declared it as a global like you did with $uplayername.
Fix the listed issues above and it should be working a bit better.

Answer 5

If you've got problems to understand your own code's logic, a simple way is to assign the conditions to self speaking variables to get used to it:
$userIsLoggedIn = $loggedin == "1";
$userIsPlayer = $user == $uplayername;
The variables make it easy to debug your code at the very beginning
var_dump($userIsLoggedIn, $userIsPlayer);
so to locate the actual errors:
The variable $loggedin is undefined
The if clauses are setting a value (=), not comparing it (== or ===).
You can then use additionally a more readable code-flow to make your decision more visible:
if ($userIsLoggedIn)
{ // user is logged in
if ($userIsPlayer)
{ // user is player
...
}
else
{ // user is not player
...
}
}
else
{ // user is not logged in
...
}
Depending of what you want to output, this can be simplified even:
if ($userIsLoggedIn && $userIsPlayer)
{
echo 'yes';
} else
{
echo 'no';
}
Hope this is helpful for you.

Answer 6

Your main problem is todo with global scope of your variables:
<?php
//Get cookie info
$cookie['user'] = $_COOKIE["user"];
$cookie['loggedin'] = (isset($_COOKIE['loggedin'])&&$_COOKIE['loggedin']=='1')?TRUE:FALSE;
//Set user array
$user['user'] = strtoupper($cookie['user']);
$user['loggedin'] = $cookie['loggedin'];
$user['player'] = $_GET['player'];
$user['uplayername']=strtoupper($user['player']);
function showplusicon(){
//Made $user array available within function
global $user;
if($user['loggedin'] === TRUE && $user['user'] == $user['uplayername']){
echo "yes";
}else{
echo "no";
}
}
?>