之前，选择最接近的值后

Question

我有这样

设A表说，表是SHOW_DATE

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

我要选择2个值，第一个是第一的负值，第二个是第一个正面价值。在这个示例中的 将导致-1（2011-04-19）和4（2011-04-24）

我尝试按abs（不同）顺序排列，但它给出2最近的-1和 - 2

Answer 1

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 2

(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 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) 

的另一种方法 - 上述方案对于大表获得问题：

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