mySQL - 使用返回多行的选择更新多列

我有一个邮政编码表，我想用它的 3 个最近邻居更新每个邮政编码。即填写此表中的空白：

postcode  nearestPostcode1  nearestPostcode2  nearestPostcode3
_______________________________________________________________

KY6 1DA      -                -                  -
KY6 1DG      -                -                  -
KY6 2DT      -                -                  -
KY6 1RG      -                -                  -
....

我已经想出了一个 SELECT 查询来查找最近的邮政编码，这是更新第一行的一种笨拙方法：

update table1 set 
nearestPostcode1 = (select query for returning the first nearest postcode),
nearestPostcode2 = (select query for returning the second nearest postcode),
nearestPostcode3 = (select query for returning the third nearest postcode)
where postcode = 'KY6 1DA';

然而，这将导致为每个行更新运行 3 个选择查询。如果有某种方法可以执行此伪代码所表达的操作，那么效率会更高：

update table1 set 
(nearestPostcode1, nearestPostcode2, nearestPostcode3) = 
(select query to return the 3 nearest postcodes)
where postcode = 'KY6 1DA';

上面的“选择查询”如下所示：

select postcode from postcodeTable 
order by <equation to calculate distance> ASC 
limit 3

无论如何，从选择返回的行是否可以放入可用于更新多个字段的表单中？ 谢谢。

Update Table1
    Cross Join  (
                Select Min( Case When Z1.Num = 1 Then Z1.postcode End ) As PostCode1
                    , Min( Case When Z1.Num = 2 Then Z1.postcode End ) As PostCode2
                    , Min( Case When Z1.Num = 3 Then Z1.postcode End ) As PostCode3
                From    (
                        Select postcode 
                            , @num := @num + 1 As Num
                        From postcodeTable 
                        Where postcode = 'KY6 IDA'
                        Order By <equation to calculate distance> ASC 
                        Limit 3
                        ) As Z1
                ) As Z
Set nearestPostCode1 = Z.PostCode1
    , nearestPostCode2 = Z.PostCode2
    , nearestPostCode3 = Z.PostCode3
Where Table1.postcode =  'KY6 IDA'