獲得最後一個節點使用CTE

Question

考慮下面的PostgreSQL表屬性給出的所有祖先的節點的完整路徑：

items 
    integer id 
    integer parent_id 
    string name 

unique key on [parent_id, name] 

parent_id is null for all root nodes 

目前，我手動生成SQL查詢，在做每一個路徑元素的加入。但對我來說似乎相當難看，當然這限制了可能的深度。

例子：

path: holiday,images,spain 
SELECT i3.* 
FROM items AS i1 
    , items AS i2 
    , items AS i3 
WHERE i1.parent_id IS NULL AND i1.name = 'holiday' 
    AND i2.parent_id=i1.id AND i2.name = 'images' 
    AND i3.parent_id=i2.id AND i3.name = 'spain' 

我不知道是否有更好的方法，可能使用CTE？

你可以看到我當前的代碼是如何工作的以及預期的輸出就是在這裏：
http://sqlfiddle.com/#!1/4537c/2

Answer 1

這應該表現非常好，因爲它消除了立即不可能路徑：

WITH RECURSIVE cte AS (
    SELECT id, parent_id, name 
     ,'{holiday,spain,2013}'::text[] AS path -- provide path as array here 
     ,2 AS lvl        -- next level 
    FROM items 
    WHERE parent_id IS NULL 
    AND name = 'holiday'      -- being path[1] 

    UNION ALL 
    SELECT i.id, i.parent_id, i.name 
     ,cte.path, cte.lvl + 1 
    FROM cte 
    JOIN items i ON i.parent_id = cte.id AND i.name = path[lvl] 
) 
SELECT id, parent_id, name 
FROM cte 
ORDER BY lvl DESC 
LIMIT 1; 

假設你提供的唯一路徑（只有1個結果）。

->SQLfiddle demo

Answer 2

UPDATE2這裏是一個函數，它peforms好，因爲搜索那張只能在路徑，從父母開始：

create or replace function get_item(path text[]) 
returns items 
as 
$$ 
    with recursive cte as (
     select i.id, i.name, i.parent_id, 1 as level 
     from items as i 
     where i.parent_id is null and i.name = $1[1] 

     union all 

     select i.id, i.name, i.parent_id, c.level + 1 
     from items as i 
      inner join cte as c on c.id = i.parent_id 
     where i.name = $1[level + 1] 
    ) 
    select c.id, c.parent_id, c.name 
    from cte as c 
    where c.level = array_length($1, 1) 
$$ 
language sql; 

sql fiddle demo

更新我認爲你可以做遞歸遍歷。我寫的這個版本的SQL，所以這是一個有點亂，因爲熱膨脹係數，但它是可以寫一個函數：

with recursive cte_path as (
    select array['holiday', 'spain', '2013'] as arr 
), cte as (
    select i.id, i.name, i.parent_id, 1 as level 
    from items as i 
     cross join cte_path as p 
    where i.parent_id is null and name = p.arr[1] 

    union all 

    select i.id, i.name, i.parent_id, c.level + 1 
    from items as i 
     inner join cte as c on c.id = i.parent_id 
     cross join cte_path as p 
    where i.name = p.arr[level + 1] 
) 
select c.* 
from cte as c 
    cross join cte_path as p 
where level = array_length(p.arr, 1) 

sql fiddle demo

，或者你可以爲所有使用遞歸的元素的構建路徑CTE爲和accumuate您的路徑到數組或字符串：

with recursive cte as (
    select i.id, i.name, i.parent_id, i.name::text as path 
    from items as i 
    where i.parent_id is null 

    union all 

    select i.id, i.name, i.parent_id, c.path || '->' || i.name::text as path 
    from items as i 
     inner join cte as c on c.id = i.parent_id 
) 
select * 
from cte 
where path = 'holiday->spain->2013'; 

或

with recursive cte as (
    select i.id, i.name, i.parent_id, array[i.name::text] as path 
    from items as i 
    where i.parent_id is null 

    union all 

    select i.id, i.name, i.parent_id, c.path || array[i.name::text] as path 
    from items as i 
     inner join cte as c on c.id = i.parent_id 
) 
select * 
from cte 
where path = array['holiday', 'spain', '2013'] 

sql fiddle demo

Answer 3

來不及發佈我的回答（很相當於羅馬的和歐文的），但在表定義的改善，而不是：

CREATE TABLE items 
     (id integer NOT NULL PRIMARY KEY 
     , parent_id integer REFERENCES items(id) 
     , name varchar 
     , UNIQUE (parent_id,name) -- I don't actually like this one 
     );      -- ; UNIQUE on a NULLable column ... 

INSERT INTO items (id, parent_id, name) values 
     (1, null, 'holiday') 
     , (2, 1, 'spain'), (3, 2, '2013') 
     , (4, 1, 'usa'), (5, 4, '2013') 
     ;