Question

我有一张这样的表

假设该表是show_date

different | date
----------------------  
    -2    | 2011-04-18
----------------------
    -1    | 2011-04-19
----------------------
     4    | 2011-04-24
----------------------
     5    | 2011-04-25 
----------------------

我想选择2个值，第一个是第一个负值，第二个是第一个正值。在此示例中将得到-1（2011-04-19）和4（2011-04-24）

我尝试通过abs（不同）进行排序，但它给出最近的2，即-1和-2

Answer 1

(SELECT * FROM show_date WHERE different < 0 ORDER BY different DESC LIMIT 1)
UNION ALL
(SELECT * FROM show_date WHERE different >= 0 ORDER BY different LIMIT 1)

Answer 2

SELECT
  MIN(CASE WHEN different > 0 THEN different ELSE NULL END) AS MinPositive,
  MAX(CASE WHEN different < 0 THEN different ELSE NULL END) AS MaxNegative
FROM show_date

Answer 3

SELECT 
   min (a.nr), 
   max (b.nr) 
FROM 
  demo a, 
  demo b 
WHERE a.nr > 0 
  AND b.nr < 0;

 min | max
-----+----
   1 | -1
(1 Row)

另一种方法 - 上述解决方案对于大型表来说会产生问题：

SELECT min(date)
  FROM tbl 
  WHERE date > 0
UNION 
SELECT max(date) 
  FROM tbl 
  WHERE date < 0;

选择之前和之后的最近值

3 个答案: