如何通過array_agg遍歷迭代列？

Question

使用PostgreSQL 8.4，我已經成功地能夠使用ARRAY_AGG（）取多個訂單的情況下，讓每個客戶單行：

從這：

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 

結果：

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

的d我可以抓住陣列片創建新列，如果我硬代碼每個迭代：

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 

結果：

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

然而，想什麼，我做的，在兩個步驟：

1. For循環我的方式通過記錄，以便我不必創建該字段的每個迭代。好消息是上面的結構沒有錯誤，只是傳遞NULL，但是如果我遇到了一些瘋狂的記錄，我不必在我的聲明中手動創建order_55，order_56等。

2. 更妙的是最終沒有通過其特定的領域，有它在所有的領域進行迭代（除非該CUSTOMER_ID），給我的每場的迭代，該效果：

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

   等等基本上加入父表（客戶）的子（訂單），但有多個子記錄跨越一行而不是創建倍數。

   （是的，我知道這與您爲什麼首先執行父/子表的基本概念相反，不過，我將此傳遞給客戶端，這將使該過程變得更加容易。）

更新：我已經得到了與拆解功能更接近......我第一次稱呼它，它創建的列和填充的客戶之一。但由於某些原因，我的IF NOT EXISTS在單獨運行時工作，但不在函數內：我得到「column order_id1 exists」錯誤。我也想最終修改這個，以便特定的字段不被硬編碼;而不是customer_id我想要做一些事情，比如傳遞父表，子表和連接ID，並且用這種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; 

我爲是所有在地圖上道歉，因爲我一直在嘗試了一次，我不能完全得到解決這個幾件事情。任何幫助表示讚賞，謝謝！

Answer 1

您是否試圖獲取每個客戶訂單的序號？你可以用Postgres 8.4中的row_number（）函數來做到這一點。爲每個訂單號創建單獨的列在SQL中不可持續或高效。

喜歡的東西：

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 

給你三排：

"{"(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\")"}" 

這看起來非常接近，你說什麼是 「更好」 ：

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

唯一的區別是：一切都包含在一個列中。當然，這個單列是一個數組，每個元素都是一個複合類型，如果你將這個結構扁平化，你就得到了你要求的東西。當然，這取決於你是否有手段進行扁平化。