Question

使用PostgreSQL 8.4，我已经成功地使用array_agg（）来处理多个订单并为每个客户创建一行：

由此：

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}"

如果我对每次迭代进行硬编码，我可以抓取数组的一部分来创建新列：

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，我想做的是分两步：

循环浏览记录，这样我就不必创建字段的每次迭代。好消息是上面的结构没有错误，只是传递NULL，但如果我得到一些疯狂的记录数，我不必在我的声明中手动创建order_55，order_56等。
更好的是最终不会将其传递给特定字段并让它遍历所有字段（除了customer_id）并给我每个字段的迭代，以达到以下效果：
```
customer|order_id1|order_date1|order_desc1|order_id2|order_date2|order_desc2| ... 
```
等等。基本上将父表（客户）连接到子（订单），但是有多个子记录跨越单行而不是创建倍数。

（是的，我理解这与你为什么首先做父/子表的基本概念背道而驰。但是，我将它传递给客户端，这将使这个过程变得更加容易。）

更新：我已经接近了相似的功能......第一次调用它时，它会创建列并填充其中一个客户。但由于某种原因，我的IF NOT EXISTS在单独运行时起作用，但不在函数内：我得到“列order_id1存在”错误。我也想最终修改它，因此特定字段不是硬编码的;而不是customer_id我想做一些事情，比如传递父表，子表和连接ID，并让它以这种交叉方式完全附加子表。

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;

我为在地图上的所有人道歉，因为我一直试图解决有关此问题的一些事情，我无法理解。感谢任何帮助，谢谢！

Answer 1

您是否尝试获取每个客户订单的序号？您可以使用Postgres 8.4中的row_number（）函数执行此操作。在SQL中，为每个订单号创建单独的列是不可持续或高效的。

类似的东西：

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

给你三行：

"{"(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所说的＃34;甚至更好＆＃34;：

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

唯一的区别是：所有内容都包含在一个列中。当然，单个列是一个数组，每个元素都是一个复合类型，如果你展平那个结构，你就得到了你所要求的。如果当然这取决于你是否有办法进行压扁。

如何将array_agg作为迭代列循环？

2 个答案: