mysql学习教程参考：https://www.w3school.com.cn/sql/index.asp
where 与 having by实例运用参考：https://www.2cto.com/database/201712/706595.html
--可选认证中至少有一项通过的用户占比，直接计算
select
count(distinct case when c.auth_type=‘可选‘ and c.auth_status in (‘已认证‘, ‘已过期‘) then c.user_id end) as pass_cnt
,count(distinct c.user_id) as tot_cnt
,count(distinct case when c.auth_type=‘可选‘ and c.auth_status in (‘已认证‘, ‘已过期‘) then c.user_id end)/count(distinct c.user_id) per_cnt
from(
select a.*,ath.auth_item,ath.auth_type,ath.auth_status,ath.auth_suc_time
from(
select laf.user_id,laf.main_order_id,laf.loan_apply_submit_time
from pl_performance_loan_apply_fact laf
left join pl_customer_basic_info usb on laf.user_id=usb.user_id
where laf.reborrow_status = ‘首借‘ and usb.merchant_no=‘DuitSayang‘
and date(laf.loan_apply_submit_time)>=‘2019-08-03‘ and date(laf.loan_apply_submit_time)<=‘2019-08-05‘
) a
left join pl_customer_auth_info ath on a.user_id = ath.user_id
) c
;
--找出未通过认证的用户，巧用having by
select a.user_id,a.phone
from(
select laf.user_id,laf.main_order_id,laf.loan_apply_submit_time,usb.phone
from pl_performance_loan_apply_fact laf
left join pl_customer_basic_info usb on laf.user_id=usb.user_id
where laf.reborrow_status = ‘首借‘ and usb.merchant_no=‘DuitSayang‘
and date(laf.loan_apply_submit_time)>=‘2019-08-03‘ and date(laf.loan_apply_submit_time)<=‘2019-08-05‘
) a
left join pl_customer_auth_info ath on a.user_id = ath.user_id
group by a.user_id,a.phone
having max( case when ath.auth_type=‘可选‘ and ath.auth_status in (‘已认证‘, ‘已过期‘) then 1 else 0 end) = 0
巧用having by
原文：https://www.cnblogs.com/dlp-527/p/11304503.html
