Question

我最近遇到了一个Postgres SQL select测试问题，这个问题我没有看到，也很难找到正确的答案。一个表基本上包含一本书的页码和页面标题，并且对象是创建一个SELECT来对本书进行布局，如下所示：

左页，右页

其中left_page包含所有偶数页号，right_page包含所有奇数页号，但是其中0页显示空值，而任何非连续页都将具有空值。可以有空白页，因此页码可以跳过。

我的猜测是使用滞后或超前功能按顺序查找下一页，然后查看它是否连续，但是遇到问题。

这是一个例子：

create table mysharkbook (page_no integer not null, title varchar(30) not null, unique(page_no));
insert into mysharkbook (page_no,title) VALUES (1,'Hammerhead');
insert into mysharkbook (page_no,title) VALUES (2,'Great White');
insert into mysharkbook (page_no,title) VALUES (3,'Blue');
insert into mysharkbook (page_no,title) VALUES (4,'Tiger');
insert into mysharkbook (page_no,title) VALUES (6,'Blacktip');
select * from mysharkbook;

结果

1   "Hammerhead"
2   "Great White"
3   "Blue"
4   "Tiger"
6   "Blacktip"

现在，尝试使SELECT这样列出：

left_page, right_page
null        | Hammerhead
Great White | Blue
Tiger       | null
Blacktip    | null

这是我（可怜的）尝试：

SELECT CASE WHEN prev_page IS NULL OR prev_page <> page_no - 1 OR (page_no % 2) = 0 THEN NULL ELSE title END As left_title,
CASE WHEN (page_no % 2) = 1 THEN title ELSE NULL END As right_title
FROM
(
SELECT LAG (page_no, 1) OVER (
 ORDER BY
 page_no
 ) AS prev_page,
title,page_no
from mysharkbook
order by page_no
) d

有什么最好的想法吗？

Answer 1

完整的子查询联接可能会获得奇数页和偶数页。对于奇数页，我们还附加了generate_series()来获取可能丢失的页面。

SELECT even.title left_page,
       odd.title right_page
       FROM (SELECT *
                    FROM mysharkbook b
                    WHERE b.page_no % 2 = 0) even
            FULL JOIN (SELECT gs.pn page_no,
                              b.title
                              FROM generate_series(1,
                                                   (SELECT max(b.page_no)
                                                           FROM mysharkbook b),
                                                   2) gs (pn)
                                   LEFT JOIN mysharkbook b
                                             ON b.page_no = gs.pn) odd
                      ON odd.page_no = even.page_no + 1
       ORDER BY odd.page_no;

db<>fiddle

（P.S .：我喜欢以鲨鱼为样本的想法！不是一次又一次地使用水果。）

Answer 2

我相信这就是您想要的；

WITH cte
AS (
    SELECT generate_series(0, (
                SELECT max(page_no) max_pages
                FROM mysharkbook
                )) AS page
    )
SELECT cte.page
    ,l.title l_title
    ,r.title r_title
FROM cte
LEFT OUTER JOIN mysharkbook l ON l.page_no = cte.page
LEFT OUTER JOIN mysharkbook r ON r.page_no = cte.page + 1
WHERE mod(page, 2) = 0
ORDER BY page

也许有更好的方法。

-HTH

Answer 3

另一个：

select distinct on (page.no / 2)
    msb.title, lead(msb.title) over(order by page.no)
from generate_series(0, (select max(page_no) from  mysharkbook)) page(no) 
left join mysharkbook msb on page.no = msb.page_no
order by page.no / 2

输出：

Answer 4

另一种方法是将页面标题放入两个元素的数组中

select 
    page_titles[1] as left_page, page_titles[2] as right_page
from
(
    select 
        page.no / 2 as even_pages, array_agg(title order by page.no) as page_titles
    from generate_series(0, (select max(page_no) from  mysharkbook)) page(no) 
    left join mysharkbook msb on page.no = msb.page_no
    group by even_pages
) t

Answer 5

单次加入和分组依据的另一种方法。

使用最小/最大无关紧要。该技术是使用模2取偶数和奇数页。如果我没记错的话，这模拟了Oracle的MIN + KEEP DENSE_RANK组合。

select 
    min(title) filter(where page.no % 2 = 0) as left_page,
    min(title) filter(where page.no % 2 = 1) as right_page,
    max(title) filter(where page.no % 2 = 0) as left_page_x,
    max(title) filter(where page.no % 2 = 1) as right_page_x                           
from generate_series(0, (select max(page_no) from  mysharkbook)) page(no) 
left join mysharkbook msb on page.no = msb.page_no
group by page.no / 2
order by page.no / 2

说明：

Answer 6

由于页码可以跳过，因此生成从0到max（page_no）的序列将产生完全空白的结果（左侧为null，右侧为null）。

如何将记录集分为偶数和奇数堆，并按顺序将2个记录集连接在一起->左页+ 1 =右页。

-- Test accepted solution by an additional row
insert into mysharkbook (page_no,title) VALUES (10,'Baby shark');

SELECT ls.title, rs.title
FROM
(SELECT * FROM  mysharkbook WHERE mod(page_no,2) = 0) as ls
FULL OUTER JOIN
(SELECT * FROM  mysharkbook WHERE mod(page_no,2) <> 0) as rs
ON ls.page_no + 1 = rs.page_no
ORDER BY COALESCE(ls.page_no, rs.page_no);

但这确实需要在同一张桌子上进行2次传球-也许不是那么出色。

SQL选择书籍格式

6 个答案: