我处理作业:内加入在ON语句
比较日均收入最高msa_income商店和实体店的(正如我在Teradata的第5周 锻炼指南定义它) 与最低的中位数msa_income(根据msa_income字段)。 这两家商店在哪个城市和州,哪家商店的平均每日收入增加了 ?
...并且答案关键字在ON
声明中有一个内部联接,这使得我很困惑。我只学过FROM
加入。所以我在网上搜索了关于ON
声明中的一个内部联接,关于它的内容并不多。
我是一个新的学习者,所以这个问题可能是非常基础的。预先感谢您的耐心!
,我有一个问题,该生产线是我:ON m.store=t.store JOIN strinfo s
SELECT SUM(store_rev. tot_sales)
SUM(store_rev.numdays) AS daily_average,
store_rev.msa_income as med_income,
store_rev.city, store_rev.state
FROM (SELECT COUNT (DISTINCT t.saledate) as numdays,
EXTRACT(YEAR from t.saledate) as s_year,
EXTRACT(MONTH from t.saledate) as s_month, t.store,
sum(t.amt) as tot_sales,
CASE
when extract(year from t.saledate) = 2005 AND extract(month from t.saledate) = 8 then 'exclude'
END as exclude_flag, m.msa_income, s.city, s.state
FROM trnsact t JOIN store_msa m
ON m.store=t.store JOIN strinfo s
ON t.store=s.store
WHERE t.stype = 'P' AND exclude_flag IS NULL
GROUP BY s_year, s_month, t.store, m.msa_income, s.city, s.state
HAVING numdays >= 20) as store_rev
WHERE store_rev.msa_income IN ((SELECT MAX(msa_income)
FROM store_msa),(SELECT MIN(msa_income) FROM store_msa))
GROUP BY med_income, store_rev.city, store_rev.state;
感谢您的版本!现在阅读确实容易得多。 :) –