Question

我正在尝试在sql下面运行：但它没有取得结果，运行时间超过5小时。

SELECT TD.TIME_KEY, CUST.SITE_ID, CUST.NODE, 
       COUNT(CUST.ACCOUNT_NUMBER) AS CUSTOMER_CNT
FROM
TIME_DIM TD INNER JOIN
OSP_ACTIVE_SUB_STAGE CUST ON TIME_KEY BETWEEN CUST.CONNECT_DATE AND (CASE WHEN CUST.CUSTOMER_STATUS_CODE = 'A' THEN SYSDATE ELSE CUST.STATUS_DATE END)
WHERE TD.TIME_KEY >= '01-JAN-14'
GROUP BY TD.TIME_KEY, CUST.SITE_ID, CUST.NODE

OSP_ACTIVE_SUB_STAGE的表结构：

Name                  Null Type        
--------------------- ---- ----------- 
SITE_ID                    NUMBER(3)   
NODE                       VARCHAR2(5) 
HOUSE_NUMBER               NUMBER(10)  
HOUSE_RESIDENT_NUMBER      NUMBER(10)  
ACCOUNT_NUMBER             NUMBER(10)  
CONNECT_DATE               DATE        
STATUS_DATE                DATE        
CUSTOMER_STATUS_CODE       VARCHAR2(1)

Time_dim是正常的时间维度表。

如何改善和增加此查询的运行时间或重写此查询？

由于

Answer 1

这是您的查询：

SELECT TD.TIME_KEY, CUST.SITE_ID, CUST.NODE, COUNT(CUST.ACCOUNT_NUMBER) AS CUSTOMER_CNT
FROM TIME_DIM TD INNER JOIN
     OSP_ACTIVE_SUB_STAGE CUST
     ON TIME_KEY BETWEEN CUST.CONNECT_DATE AND 
        (CASE WHEN CUST.CUSTOMER_STATUS_CODE = 'A' THEN SYSDATE ELSE CUST.STATUS_DATE END)
WHERE TD.TIME_KEY >= '01-JAN-14';
GROUP BY TD.TIME_KEY, CUST.SITE_ID, CUST.NODE

Oy公司！那里有很多计算。你需要一个不同的方法。而不是在此期间的每一天加入，创建一个连接和结束的表，然后使用累积总和。

假设所有站点/节点组合每天都有连接或结束，您可以这样做：

  select site_id, node, dte, sum(inc) as ActivesOnDate
  from ((select oass.site_id, oass.node, oass.connect_date as dte, 1 as inc
         from osp_active_sub_stage oass
        ) union all
        (select oass.site_id, oass.node,
                (case when oass.customer_status_code = 'A'
                      then trunc(sysdate)
                      else oass.status_date + 1
                 end) , -1 as inc
         from osp_active_sub_stage oass
        )
       )
  group by site_id, node, dte;

这应该比原始版本运行得快一点。

oracle查询获取速度很慢

1 个答案: