问题描述:
需要查询出学生在不同流程状态对应的活动个数,流程状态分为五个状态audit、havePass、notPass、reject、giveup;活动有社会活动(xg_credit_comm_activity_apply)、比赛活动(xg_credit_innov_match_apply)、培训活动(xg_credit_innov_train_apply)。
需要得到的效果:
SQL拼接步骤:
Step1: 统计学生在所有活动中不同流程状态对应的活动个数
SELECT student_id,`status`,COUNT(*) as counts from (
SELECT student_id, `status` from xg_credit_comm_activity_apply
UNION ALL
SELECT student_id, `status` from xg_credit_innov_match_apply
UNION ALL
SELECT student_id,`status` from xg_credit_innov_train_apply
)as t GROUP BY t.student_id, t.`status`
Step2:
SELECT t1.student_id as stuId,
CASE when t1.`status`='AUDIT' then t1.counts ELSE 0 END as audit,
CASE when t1.`status`='HAVEPASS' then t1.counts ELSE 0 END as havePass,
CASE when t1.`status`='NOTPASS' then t1.counts else 0 END as notPass,
CASE when t1.`status`='REJECT' then t1.counts else 0 END as reject,
CASE when t1.`status`='GIVEUP' then t1.counts else 0 END as giveup
from (
SELECT student_id,`status`,COUNT(*) as counts from (
SELECT student_id, `status` from xg_credit_comm_activity_apply
UNION ALL
SELECT student_id, `status` from xg_credit_innov_match_apply
UNION ALL
SELECT student_id,`status` from xg_credit_innov_train_apply
)as t GROUP BY t.student_id, t.`status`
)as t1
Step3:
SELECT stuId, SUM(audit) as audit,SUM(havePass) as havePass, SUM(notPass) as notPass, SUM(reject) as reject, SUM(giveup) as giveup from (
SELECT t1.student_id as stuId,
CASE when t1.`status`='AUDIT' then t1.counts ELSE 0 END as audit,
CASE when t1.`status`='HAVEPASS' then t1.counts ELSE 0 END as havePass,
CASE when t1.`status`='NOTPASS' then t1.counts else 0 END as notPass,
CASE when t1.`status`='REJECT' then t1.counts else 0 END as reject,
CASE when t1.`status`='GIVEUP' then t1.counts else 0 END as giveup
from (
SELECT student_id,`status`,COUNT(*) as counts from (
SELECT student_id, `status` from xg_credit_comm_activity_apply
UNION ALL
SELECT student_id, `status` from xg_credit_innov_match_apply
UNION ALL
SELECT student_id,`status` from xg_credit_innov_train_apply
)as t GROUP BY t.student_id, t.`status`
)as t1
)as t2 GROUP BY t2.stuId