Php 使用mysqli_fetch_assoc（）从查询中填充数组

我有这个，一切都很好（我有一个通用的表生成器，但现在我不得不偏离它）：

现在我有了类似的东西：

 $result = mysqli_query($con, 'SELECT r.id AS ID, CONCAT(g.fname, g.lname) AS Name, r.apple AS Apple, 
                            r.dog AS Dog, DATEDIFF(r.Dog, r.Apple) AS Days, 
                            r.total_price AS "Total Price", u.name AS Name, r.in AS "In",
                            r.out AS "Out", r.time_in AS "Time In", r.time_out AS "Time Out", 
                            CONCAT(c.fname,c.lname) AS Charlie, r.here AS "Apple",
                            r.leave AS "Dog"
                            FROM really r, georgia g, unit u, charlie c 
                            WHERE g.id = r.georgia AND r.unit = u.id AND r.charlie = c.id
                            HAVING r.in = TRUE AND r.out = FALSE');

    //fill fields array with fields from table in database
    while ($x = mysqli_fetch_assoc($result))
    {
        $fields[] = $x['Field'];
    }

---

我现在得到一个行

$fields[]=$x['Field']的错误由于单词
字段
。为什么？因为我现在有一个完整的查询？如何在不引用每个字段名的情况下修复此问题？
因为在查询结果中没有名为
字段的字段：

'SELECT r.id AS ID, CONCAT(g.fname, g.lname) AS Name, r.apple AS Apple, 
                        r.dog AS Dog, DATEDIFF(r.Dog, r.Apple) AS Days, 
                        r.total_price AS "Total Price", u.name AS Name, r.in AS "In",
                        r.out AS "Out", r.time_in AS "Time In", r.time_out AS "Time Out", 
                        CONCAT(c.fname,c.lname) AS Charlie, r.here AS "Apple",
                        r.leave AS "Dog"
                        FROM really r, georgia g, unit u, charlie c 
                        WHERE g.id = r.georgia AND r.unit = u.id AND r.charlie = c.id
                        HAVING r.in = TRUE AND r.out = FALSE'

查询结果中有一些字段：
ID
、
Name
、
Apple
，等等。。您可以尝试按如下方式获取这些字段，或更改查询命令

while ($x = mysqli_fetch_assoc($result))
{
    $fields[] = $x['ID'];
}

为什么不使用mysqli_fetch_数组
？
while ($x = mysqli_fetch_assoc($result))
{
    $fields[] = $x['ID'];
}