選擇進入而不在PostgreSQL中創建新表

Question

我有很多這樣的查詢，它們的結果在最后結合在一起，因此我可以計算兩個不同結果集之間的聯接數，然后使用聯合查詢生成完整的結果表。 這里的問題是，使用select into為每個結果創建一個新表，而不是將其存儲在內存中。 使用PostgreSQL時，如何將每個結果存儲在其自己的變量中，而不在數據庫中創建一大堆偽造的表（此后我將刪除這些表）？

select distinct cust_num into t1 
from all_visits_x_cust 
where date > '06/02/2015'
and date <= '07/02/2015' 
and cust_num > 9999;

select distinct cust_num into t1_2 
from cc_calls_x_cust 
where date = '06/02/2015' 
and cust_num > 9999);

select count(*) as post_call_visitors 
into v1 
from t1_2 join t1 
on t1.cust_num = t1_2.cust_num);

Answer 1

Postgres中沒有全局變量。 您可以將with query一起使用來運行這些查詢：

with t1 as (
    select distinct cust_num
    from all_visits_x_cust 
    where date > '06/02/2015'
    and date <= '07/02/2015' 
    and cust_num > 9999
    ),
t1_2 as (
    select distinct cust_num into t1_2 
    from cc_calls_x_cust 
    where date = '06/02/2015' 
    and cust_num > 9999
    )
select count(*) as post_call_visitors 
from t1_2 
join t1 
on t1.cust_num = t1_2.cust_num;

您可以在PL / pgSQL函數或DO語句中使用SELECT ... INTO variable 。

Answer 2

您可以只使用前兩個查詢作為最終查詢中的子查詢：

select count(*) as post_call_visitors 
into temp v1
from (
        select distinct cust_num
        from all_visits_x_cust 
        where date > '06/02/2015'
        and date <= '07/02/2015' 
        and cust_num > 9999
    ) as t1
join (
        select distinct cust_num 
        from cc_calls_x_cust 
        where date = '06/02/2015' 
        and cust_num > 9999
     ) as t1_2
on t1.cust_num = t1_2.cust_num; 

確保刪除into那些子查詢的條款。 最后，他們將自己的名字as一部分。

通過在第二行使用temp關鍵字，將在特殊模式中創建臨時表。 在會話結束時將自動刪除它。

附帶說明，下面的較短查詢應返回相同的結果：

select count(distinct cust_num) as post_call_visitors 
from cc_calls_x_cust 
where date = '06/02/2015' 
and cust_num > 9999
and cust_num in (
        select cust_num 
        from all_visits_x_cust 
        where date > '06/02/2015'
        and date <= '07/02/2015' 
    );