PHP和SQL不起作用？

$sql = mysqli_query($db, "SELECT COUNT(ID) from bruker");

if (!$sql) {
    while ($row1 = mysqli_fetch_array($sql, MYSQL_NUM)); {
        echo $row1;
    }
} else {
    echo "Funker ikke";
}

I cant get it to work, all I get is the: else "Funker ikke".

Your code are having two error there.
First, it should be true condition only can get into the display loop. But you put the condition as !$sql means false only can get into the loop. So, you should change !$sql to $sql.

Besides, after the While loop should not direct put the semicolon.

You try see the following code

$sql = mysqli_query($db, "SELECT COUNT(ID) from bruker");

if ($sql) {
    while ($row1 = mysqli_fetch_array($sql, MYSQL_NUM))
    {
        echo $row1[0]; 
    }
} 
else {
    echo "Funker ikke";
}

1) Using wrong if condition. If query return true then insert into if condition

2) Wrong echo data using MYSQL_NUM. It return numeric array

$sql = mysqli_query($db, "SELECT COUNT(ID) from bruker");      

if ($sql) {// if true
    while ($row1 = mysqli_fetch_array($sql, MYSQL_NUM)) {
        echo $row1['0'];//numeric array
    }
} else {
    echo "Funker ikke";
}    

UPDATED as per below comments

$sql = mysqli_query($db, "SELECT COUNT(ID) from bruker");
if ($sql) {
    $row1 = mysqli_fetch_array($sql, MYSQL_NUM);
    echo $row1['0']; //numeric array
} else {
    echo "Funker ikke";
}

Read http://php.net/manual/en/mysqli-result.fetch-array.php

Your code is correct except the not sign before sql in if condition.According to you if your sql query is wrong then the $row1 is displayed else if it is correct then "funker ikke" gets printed.Just the opposite. This code works:

 $sql = mysqli_query($db, "SELECT COUNT(ID) from bruker");      

 if ($sql) {
   // as the query can only return 1 row the while look is unnecessary
   //while ($row1 = mysqli_fetch_array($sql, MYSQL_NUM)) {

   $row = mysqli_fetch_array($sql, MYSQL_NUM)
   echo $row[0]; 

 } else {
   echo "Funker ikke";
 }

There are a few errors in your code

First you do not need a while loop to process a single result, SELECT COUNT() can only return one row.

Second as mysqli_query() for a SELECT query will return either FALSE or a mysqli_result object it would be better to flip your if condition as below and test specifically for FALSE rather than TRUE.

Third, a mysqli_fetch_array($sql, MYSQL_NUM) returns an array, you have to use array notation to get the only returned value i.e. $row[0]

Also when any mysqli_ function fails it leaves some error information that is very useful, so show that info or send it to an application error file if you dont want users to see them when the code goes live.

$sql = mysqli_query($db, "SELECT COUNT(ID) from bruker");

if ($sql === FALSE) {
    echo "Funker ikke";
    echo mysqli_error($db);
    exit;
} else {
    $row = mysqli_fetch_array($sql, MYSQL_NUM);
    echo $row[0];
}