select COUNT(up.*) as "Total Upsell Accounts in October" from
list_all_upsells('2015-10-01 00:00:00'::timestamp,'2015-10-31 23:59:59'::timestamp) as up
where up.user_id not in
(select distinct user_id from paid_users_no_more
where concat(extract(month from payment_stop_date),'-',extract(year from payment_stop_date))<>
concat(extract(month from payment_start_date),extract(year from payment_start_date)));
list_all_upsells过程如下所示：
DECLARE payor_email_2 text; BEGIN FOR payor_email_2 in select distinct payor_email from paid_users LOOP return query execute 'select paid_users.* from paid_users,( select payment_start_date as first_time from paid_users where payor_email = $3 order by payment_start_date limit 1 ) as dummy where payor_email = $3 and payment_start_date > first_time and payment_start_date between $1 and $2 and first_time < $1' using a,b,payor_email_2; END LOOP; return; END
我想能够在我们有记录的所有月份运行这个并在一个表中查询数据,如下所示：
Month | Total Upselled Accounts --------------------------------- 08/2014 | 23 09/2014 | 35 ETC... 10/2015 | 56
我有一个问题是抓住每个月的第一个月和每个月的最后一个月我们经营的月份：
select distinct date_trunc('month',payment_start_date)::date as startmonth from paid_users ORDER BY startmonth;
上个月：
SELECT distinct (date_trunc('MONTH',payment_start_date) + INTERVAL '1 MONTH - 1 day')::date as endmonth from paid_users ORDER BY endmonth;
现在,我将如何创建一个循环list_all_upsells的函数并获取这几个月中的每一个的计数？即startmonth的第一个查询给了我2014-03-01,2014-04-01,…到2015-10-01,而endmonth的第二个查询给了我2014-03-31,2014-04-30,.. .to 2015-10-31.我想在这几个月的每个月运行list_all_sells,这样我就可以每个月得到一个总计数我们拥有多少外包帐户
我的paid_users表看起来像这样：
CREATE TABLE paid_users ( user_id integer,user_email character varying(255),payor_id integer,payor_email character varying(255),payment_start_date timestamp without time zone DEFAULT Now() )
paid_users_no_more：
CREATE TABLE paid_users_no_more ( user_id integer,payment_stop_date timestamp without time zone DEFAULT Now() )
解决方法
从逻辑开始：显然,由他的电子邮件地址标识的用户从某个payment_start_date取消订阅,直到某个payment_stop_date,并且可以多次执行此操作.您正在寻找那些在相关月份之前购买了第一个订阅的用户,以及在相关月份开始新订阅而不是第一个订阅的用户.在这种情况下,过滤器payment_start_date> first_time没用,因为您已经过滤了有关月份之前的第一个订阅(first_time< $1)和新订阅(payment_start_date BETWEEN $1和$2). 当重写函数内的查询时,Points(1),(2)和(3)实际上只会变得明显：
CREATE FUNCTION list_all_upsells(timestamp) RETURNS SetoF paid_users AS $$ SELECT paid_users.* FROM paid_users JOIN ( -- This JOIN keeps only those rows where the payor_email has a prior subscription SELECT disTINCT payor_email,first_value(payment_start_date) OVER (PARTITION BY payor_email ORDER BY payment_start_date) AS dummy FROM paid_users WHERE payment_start_date < date_trunc('month',$1) ) dummy USING (payor_email) -- This filter keeps only those rows with new subscriptions in the month WHERE date_trunc('month',payment_start_date) = date_trunc('month',$1) $$LANGUAGE sql STRICT;
由于函数体已缩减为单个sql语句,因此该函数现在是一个sql语言函数,它比plpgsql更有效.您现在只提供一个参数,可以是您想要数据的月份中的任何时刻,因此list_all_upsells(LOCALTIMESTAMP)将为您提供当月的结果.就您发布的查询而言,它将是：
SELECT count(up.*) AS "Total Upsell Accounts in October" FROM list_all_upsells(LOCALTIMESTAMP) up WHERE up.user_id NOT IN (SELECT disTINCT user_id FROM paid_users_no_more WHERE date_trunc('month',payment_stop_date) <> date_trunc('month',up.payment_start_date) );
顺便说一下,这确实引起了为什么你有表pay_users_no_more的问题.为什么不简单地将一列payment_stop_date添加到表paid_users？如果该列为NULL,则仍然订阅用户.但整个查询相当奇怪,因为list_all_upsells()在月内返回新的订阅,那么为什么还要在其他时间打扰取消的订阅呢？
现在回答你真正的问题：
SELECT months.m "Month",coalesce(count(up.*),0) "Total Upselled Accounts"
FROM generate_series('2014-08-01'::timestamp,date_trunc('month',LOCALTIMESTAMP),'1 month') AS months(m)
LEFT JOIN list_all_upsells(months.m) AS up ON date_trunc('month',payment_start_date) = m
GROUP BY 1
ORDER BY 1;
Generate a series of months从某个起始月到当月,然后计算每个月的新订阅,可能为0.
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