mysql 有语句不返回行

Question

我尝试根据 fifo_in_date 和 order_date 之间的差异为给定订单分配正确的 fifo_cost:与 order_date 和 fifo_date_in 之间的最小差异关联的 fifo_cost 应分配给该订单。

以下 mysql 片段不会返回任何记录。我希望它返回一条 fifo_date_in 最接近 order_date 的记录，但显然我遗漏了一些东西。

drop table if exists tmp;

create table tmp (
order_sequence int,
order_number int,
order_date date,
fifo_date_in date,
fifo_cost float);

INSERT INTO tmp (order_sequence, order_number, order_date, fifo_date_in, fifo_cost) VALUES (5613, 561, '2013-01-02','2009-01-01',1.55);
INSERT INTO tmp (order_sequence, order_number, order_date, fifo_date_in, fifo_cost) VALUES (5613, 561, '2013-01-02','2009-02-01',2.55);
INSERT INTO tmp (order_sequence, order_number, order_date, fifo_date_in, fifo_cost) VALUES (5613, 561, '2013-01-02','2009-03-01',3.55);
INSERT INTO tmp (order_sequence, order_number, order_date, fifo_date_in, fifo_cost) VALUES (5613, 561, '2013-01-02','2009-04-01',4.55);
INSERT INTO tmp (order_sequence, order_number, order_date, fifo_date_in, fifo_cost) VALUES (5613, 561, '2013-01-02','2009-05-01',5.55);

SELECT
  order_sequence, order_number, order_date, fifo_date_in, fifo_cost, datediff(order_date,fifo_date_in) as ddiff
FROM tmp
GROUP BY order_sequence, order_number, order_date
HAVING datediff(order_date,fifo_date_in) = min(datediff(order_date,fifo_date_in))

Answer 1

如果你想获得成本，我认为你必须找到最小值并连接回基表:

SELECT t.order_sequence, t.order_number, t.order_date, t.fifo_date_in, t.fifo_cost
  FROM tmp t
  INNER JOIN ( SELECT order_sequence, order_number, order_date
                     ,MIN(datediff(order_date,fifo_date_in)) as ddiff
                 FROM tmp
                 GROUP BY order_sequence, order_number, order_date
              ) m
         ON (m.order_sequence = t.order_sequence
             AND m.order_number = t.order_number
             AND m.order_date = t.order_date
             AND datediff(t.order_date, t.fifo_date_in) = m.ddiff)

此外，如果最接近可以表示之前或之后，您可能必须考虑绝对值。

这是SQLFiddle