如何通过array_agg遍历迭代列？

Question

使用PostgreSQL 8.4，我已经成功地能够使用ARRAY_AGG（）取多个订单的情况下，让每个客户单行：

从这：

order_id|customer_id|order_date |order_desc 
1  |1   |"2010-01-01"|"Tom's First" 
2  |1   |"2010-04-01"|"Tom's Second" 
7  |1   |"2010-04-13"|"Tom's Third" 
8  |1   |"2011-04-13"|"Tom's Last" 
5  |1   |"2011-06-20"|"Tom's Really Last." 
3  |2   |"2010-07-07"|"Dick's First" 
6  |2   |"2011-07-07"|"Dick's Other" 
4  |3   |"2011-04-04"|"Harry's Only" 

使用此：

select cu.customer, array_agg(ord.order_id) as orders from test_order ord 
inner join test_customer cu 
on ord.customer_id = cu.customer_id 
group by cu.customer 

结果：

customer |orders 
"Tom"  |"{1,2,7,8,5}" 
"Dick"  |"{3,6}" 
"Harry" |"{4}" 

的d我可以抓住阵列片创建新列，如果我硬代码每个迭代：

select cu.customer, 
(array_agg(ord.order_id))[1] as order_1, 
(array_agg(ord.order_id))[2] as order_2, 
(array_agg(ord.order_id))[3] as order_3, 
(array_agg(ord.order_id))[4] as order_4, 
(array_agg(ord.order_id))[5] as order_5 
from test_order ord 
inner join test_customer cu 
on ord.customer_id = cu.customer_id 
group by cu.customer 

结果：

customer|order_1|order_2|order_3|order_4|order_5 
"Dick" |3  |6  |  |  | 
"Harry" |4  |  |  |  | 
"Tom" |8  |1  |5  |2  |7 

然而，想什么，我做的，在两个步骤：

1. For循环我的方式通过记录，以便我不必创建该字段的每个迭代。好消息是上面的结构没有错误，只是传递NULL，但是如果我遇到了一些疯狂的记录，我不必在我的声明中手动创建order_55，order_56等。

2. 更妙的是最终没有通过其特定的领域，有它在所有的领域进行迭代（除非该CUSTOMER_ID），给我的每场的迭代，该效果：

   customer|order_id1|order_date1|order_desc1|order_id2|order_date2|order_desc2| ... 
   

   等等基本上加入父表（客户）的子（订单），但有多个子记录跨越一行而不是创建倍数。

   （是的，我知道这与您为什么首先执行父/子表的基本概念相反，不过，我将此传递给客户端，这将使该过程变得更加容易。）

更新：我已经得到了与拆解功能更接近......我第一次称呼它，它创建的列和填充的客户之一。但由于某些原因，我的IF NOT EXISTS在单独运行时工作，但不在函数内：我得到“column order_id1 exists”错误。我也想最终修改这个，以便特定的字段不被硬编码;而不是customer_id我想要做一些事情，比如传递父表，子表和连接ID，并且用这种交叉表方式完全追加子表。

CREATE FUNCTION loop_test(integer) RETURNS integer AS $$ 

DECLARE 
rOrder RECORD; 
loop_counter INT := 1; 
target_customer_id ALIAS FOR $1;   
BEGIN 

FOR rOrder IN SELECT * 
    FROM vdad_data.test_order 
    WHERE customer_id = target_customer_id 
    ORDER BY order_id LOOP 

    IF NOT EXISTS 
     (
     SELECT * FROM information_schema.COLUMNS 
     WHERE COLUMN_NAME= 'order_id' || loop_counter 
     AND TABLE_NAME='test_customer' 
     AND TABLE_SCHEMA='vdad_data' 
     ) 
     THEN 

     EXECUTE 'ALTER TABLE vdad_data.test_customer 
     ADD COLUMN order_id' || loop_counter || ' integer'; 
    END IF; 

    IF NOT EXISTS 
     (
     SELECT * FROM information_schema.COLUMNS 
     WHERE COLUMN_NAME= 'order_date' || loop_counter 
     AND TABLE_NAME='test_customer' 
     AND TABLE_SCHEMA='vdad_data' 
     ) 
     THEN 

     EXECUTE 'ALTER TABLE vdad_data.test_customer 
     ADD COLUMN order_date' || loop_counter || ' date'; 
    END IF; 


    IF NOT EXISTS 
     (
     SELECT * FROM information_schema.COLUMNS 
     WHERE COLUMN_NAME= 'order_desc' || loop_counter 
     AND TABLE_NAME='test_customer' 
     AND TABLE_SCHEMA='vdad_data' 
     ) 
     THEN 

     EXECUTE 'ALTER TABLE vdad_data.test_customer 
     ADD COLUMN order_desc' || loop_counter || ' character varying'; 
    END IF; 

EXECUTE 'UPDATE vdad_data.test_customer 
     SET order_id' || loop_counter || ' = ' || rOrder.order_id ||', 
     order_date' || loop_counter || ' = ' || quote_literal(to_char(rOrder.order_date,'yyyy-mm-dd')) ||',   
     order_desc' || loop_counter || ' = ' || quote_literal(rOrder.order_desc) ||' 
     WHERE customer_id = ' ||rOrder.customer_id; 

loop_counter = loop_counter + 1; 
END LOOP; 

RETURN 1; 
END; 
$$ LANGUAGE plpgsql; 

我为是所有在地图上道歉，因为我一直在尝试了一次，我不能完全得到解决这个几件事情。任何帮助表示赞赏，谢谢！

Answer 1

您是否试图获取每个客户订单的序号？你可以用Postgres 8.4中的row_number（）函数来做到这一点。为每个订单号创建单独的列在SQL中不可持续或高效。

喜欢的东西：

select cu.customer, 
     row_number() OVER(PARTITION BY cu.customer ORDER BY ord.order_date) 
from test_order ord inner join test_customer cu 
    on ord.customer_id = cu.customer_id 
group by cu.customer 

Answer 2

此：

select array_agg(row(order_id,order_date,order_desc)) 
from (
select 1 order_id,1 customer_id,'2010-01-01' order_date,'Tom''s First' order_desc union 
select 2 order_id,1 customer_id,'2010-04-01' order_date,'Tom''s Second' order_desc union 
select 7 order_id,1 customer_id,'2010-04-13' order_date,'Tom''s Third' order_desc union 
select 8 order_id,1 customer_id,'2011-04-13' order_date,'Tom''s Last' order_desc union 
select 5 order_id,1 customer_id,'2011-06-20' order_date,'Tom''s Really Last.' order_desc union 
select 3 order_id,2 customer_id,'2010-07-07' order_date,'Dick''s First' order_desc union 
select 6 order_id,2 customer_id,'2011-07-07' order_date,'Dick''s Other' order_desc union 
select 4 order_id,3 customer_id,'2011-04-04' order_date,'Harry''s Only' order_desc 
) orders 
group by orders.customer_id 

给你三排：

"{"(2,2010-04-01,\"Tom's Second\")","(1,2010-01-01,\"Tom's First\")","(7,2010-04-13,\"Tom's Third\")","(5,2011-06-20,\"Tom's Really Last.\")","(8,2011-04-13,\"Tom's Last\")"}" 

"{"(3,2010-07-07,\"Dick's First\")","(6,2011-07-07,\"Dick's Other\")"}" 

"{"(4,2011-04-04,\"Harry's Only\")"}" 

这看起来非常接近，你说什么是 “更好” ：

customer | order_id1 | order_date1 | order_desc1 | order_id2 | order_date2 | order_desc2 | ...

唯一的区别是：一切都包含在一个列中。当然，这个单列是一个数组，每个元素都是一个复合类型，如果你将这个结构扁平化，你就得到了你要求的东西。当然，这取决于你是否有手段进行扁平化。