Question

有问题的表格（条纹不需要的cols）：

car_id INT(11) PK
url VARCHAR(100)
signup TIMESTAMP

默认列表查询如下所示：

SELECT car_id, url FROM car ORDER By signup DESC

现在我需要在结果集中找到汽车的位置，就像在列表查询中一样。和上一行和下一行的URL。

我尝试了很多不同的方法，我终于不确定它是否可行。（在一个查询中）。

对查询的任何帮助或指向正确方向的指针都将非常感激。

修改

期望的结果将是：

pos, nextid, nexturl, previd, prevurl

Answer 1

试试这个 -

+行位置

SELECT car_id, url, signup, CONCAT(pos1, '/', @p1) position FROM (
  SELECT
    c.*,
    @p1:=@p1+1 pos1,
    @p2:=IF(car_id = 3 AND @p2 IS NULL, @p1, @p2)
  FROM
    cars c,
    (SELECT @p1:=0, @p2:=NULL) t
  ORDER BY
    signup
) t
WHERE
  pos1 BETWEEN @p2 - 1 AND @p2 + 1

你写道： 期望的结果将是：pos，nextid，nexturl，previd，prevurl

尝试此查询：

SELECT
  @p2 pos,
  MAX(IF(pos1 > @p2, car_id, NULL)) nextid,
  MAX(IF(pos1 > @p2, url, NULL)) nexturl,
  MAX(IF(pos1 < @p2, car_id, NULL)) previd,
  MAX(IF(pos1 < @p2, url, NULL)) prevurl
FROM (
  SELECT
    c.*,
    @p1:=@p1+1 pos1,
    @p2:=IF(car_id = 3 AND @p2 IS NULL, @p1, @p2)
  FROM
    cars c,
    (SELECT @p1:=0, @p2:=NULL) t
  ORDER BY
    signup
) t
WHERE
  pos1 BETWEEN @p2 - 1 AND @p2 + 1

Answer 2

(SELECT car_id, url FROM car WHERE car_id = ?)
UNION
(SELECT car_id, url FROM car WHERE signup >  (SELECT signup FROM car WHERE car_id = ?) ORDER By signup ASC LIMIT 1)
UNION
(SELECT car_id, url FROM car WHERE signup <  (SELECT signup FROM car WHERE car_id = ?) ORDER By signup DESC LIMIT 1)

Answer 3

SELECT
  @previd AS previousid,
  @prevurl AS previousurl,
  @previd:=@id AS currentid,
  @prevurl:=@url AS currenturl,
  @id:=car_id AS nextid,
  @url:=url AS nexturl
FROM
  (SELECT @previd:=0) AS initprevid,
  (SELECT @prevurl:='') AS initprevurl,
  (SELECT @id:=0) AS initid,
  (SELECT @url:='') AS initurl,
  car
WHERE currentid=<whatever>
ORDER BY
  car.signup DESC

您可以过滤掉spurious currentid = 0

找到给定PK和下一行和上一行的位置作为一个结果行

3 个答案: