MySQL php查询似乎返回错误的结果（可能很明显）

Question

MySql Appears to be returning the wrong result and I do not know why (new to mysql so its probably obvious but I couldn't find the answer)

home.php:

    $testUID = query("SELECT `uid` FROM `userid` WHERE `session_id` = '".$_COOKIE['phpAuth']."'",$link);
            echo $testUID;

echoes "1" when $_COOKIE['phpAuth'] is "ABCD"

But the MySQL userid table looks like this (test values obviously) so surely it should return 0?

    name  | pass     | csrf | uid  | session_id
    -------------------------------------------
    admin | password | ABCD | 0    | ABCD

Sorry I forgot completely that query() was my own function and ($link is the connection the the database)

function query($request,$link) {
    return mysqli_query($link,$request)or die(mysqli_error($link));
}

EDIT:

Simplified,

echo mysqli_fetch_assoc(mysqli_query(mysqli_connect("localhost", "root", "", "testdb"),"SELECT `uid` FROM `userid` WHERE `session_id` = 'ABCD'"))['uid'];

works perfectly but is in no way different to the current process as far as I can see so I'm even more confused.

echo mysqli_fetch_assoc($testuid)['uid'];

doesn't work with error

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, 
boolean given in C:\xampp\htdocs\Home.php on line 37

Answer 1

To anyone interested, changing

function query($request,$link) {
    return mysqli_query($link,$request)or die(mysqli_error($link));
}

to

function query($request,$link) {
    return mysqli_query($link,$request);// or die(mysqli_error($link));
}

resolves the error completely