mysqli_fetch_array如何工作？

A while loop usually has code that tries to avoid an infinite loop. I don't understand how this works while($row = mysqli_fetch_array($result_set)? From the php manual, it says that mysqli_fetch_array returns an array per result row. So we are assigning here an array to the variable $row. Well I tried to replicate this:

$result_set = $database->query($sql);
while($row = mysqli_fetch_array($result_set)){
      some code
    }

with this:

$a = ["a","b","c"];
$num = 1;

while($row = $a){
 echo $num;
  $num++;
}

and I get an infinite loop. What am I missing?

In order to replicate the behavior of mysqli_fetch_array() with an array you create consider the following: first create an array of arrays:

$a = ["a"=>array(1,2,3),"b"=>array(4,5,6),"c"=>array(7,8,9)];
$num = 1;

One way (there is more than one way to skin this proverbial cat) is to use a foreach()loop to get and manipulate (echo) each row:

foreach($a AS $row){
 echo $num ."
";
 $num++;
 echo $row[0] . " " . $row[1] . " " .$row[2] . "
";
}

This returns just what you would expect, your number followed by the data in each row of $a example:

1
1 2 3
2
4 5 6
3
7 8 9

Once the loop has reached the end of the array it will exit, same as mysqli_fetch_array() which essentially says, "foreach result_set as row".

You can try this with while loop which is like while($row = mysqli_fetch_array($result_set))

 $a = ["a","b","c"];
    $num = 0;

    while($num < count($a)){         
        echo $a[$num]."
";
        $num++;       
    }