Mysql 涉及带条件的部分组的SQL查询？_Mysql_Sql_Database_Postgresql

Mysql 涉及带条件的部分组的SQL查询？

mysql sql database postgresql

Mysql 涉及带条件的部分组的SQL查询？,mysql,sql,database,postgresql,Mysql,Sql,Database,Postgresql,以下是我无法理解正确方法的SQL查询问题：数据库表： Employee: emp_id, emp_name Credit: credit_id, emp_id, credit_date, credit_amount debit: debit_id, emp_id, debit_date, debit_amount 在这里，每个人都可以有多重收入和支出查询要求：每天结束时，每个员工都会有一些资产（“至今贷记”-“至今贷记”）。我们需要在最大资产方面找到前五名员工，以及他们拥有最大资产的日期

以下是我无法理解正确方法的SQL查询问题：

数据库表：

Employee: emp_id, emp_name
Credit: credit_id, emp_id, credit_date, credit_amount
debit: debit_id, emp_id, debit_date, debit_amount

在这里，每个人都可以有多重收入和支出

查询要求：每天结束时，每个员工都会有一些资产（“至今贷记”-“至今贷记”）。我们需要在最大资产方面找到前五名员工，以及他们拥有最大资产的日期

我尝试了以下查询，但似乎遗漏了一些内容：

select Credit.emp_id, Credit.date, (Credit.income_amount - Debit.credit_amount) from 
(select emp_id, sum(amount) as credit_amount 
from credit) Credit
LEFT JOIN LATERAL (
     select emp_id, sum(amount) as debit_amount
     from debits
     where debits.emp_id = Credit.emp_id and Credit.date >= debits.date
     group by debits.emp_id
    ) Debit
ON true

Group by和sum允许您将每个人的总积分记录到一个记录中。您可以在子查询中执行类似的操作来减去借方

Select top 5 emp_id, credit_date, (sum(credit_amount) - 
(select sum(debit_amount) from debit d 
where c.emp_id = d.emp_id and c.credit_date = d.debit_date)
) as total
from Credit c group by emp_id, credit_date order by total

Group by和sum允许您将每个人的总积分记录到一个记录中。您可以在子查询中执行类似的操作来减去借方

Select top 5 emp_id, credit_date, (sum(credit_amount) - 
(select sum(debit_amount) from debit d 
where c.emp_id = d.emp_id and c.credit_date = d.debit_date)
) as total
from Credit c group by emp_id, credit_date order by total

在这里，我将打破查询，使其更具可读性

首先，我们需要得到贷方和借方在一天级别上的总金额，这样我们就可以使用相同的emp_id在一天级别上加入贷方和借方表

with 
credit as(
    select emp_id,credit_date date,sum(credit_amount) as amount
    from credit  
    group by 1,2),

debit as(
    select emp_id,debit_date,sum(debit_amount) as amount
    from expenses 
    group by 1,2),

现在我们需要完全外部连接“credit”和“debit”子查询

payments as (
    select distinct
    case when c.emp_id is null then d.person_id else c.emp_id  end as emp_id ,
    case when c.emp_id is null then d.date else c.date end as date,
    case when c.emp_id is null then 0 else i.amount end as credit ,
    case when d.emp_id is null then 0 else d.amount end as debit 
    from credit c
    full outer join debit d on d.emp_id=c.emp_id and d.date=c.date
    ),

现在，我们将对贷方、借方和总余额进行逐日累计，如下所示

total_balance as(
    SELECT emp_id, date, 
    sum(credit) OVER (PARTITION BY emp_id ORDER BY date asc) AS total_credit,
    sum(debit) OVER (PARTITION BY emp_id ORDER BY date asc) AS total_debit,
    (sum(income) OVER (PARTITION BY person_id ORDER BY date asc) - 
    sum(expense) OVER (PARTITION BY person_id ORDER BY date asc)) as total_balance
    FROM group_payment
    ORDER BY person_id, date),

ranks as (select emp_id,date,total_balance,
rank() over (partition by emp_id order by total_balance desc) as rank
from total_balance ),

现在，我们需要使用rank（）函数根据emp_id的总余额（desc）分配排名（即，对于特定emp_id，rank=1将分配给一天中最大的总余额）。查询如下所示

total_balance as(
    SELECT emp_id, date, 
    sum(credit) OVER (PARTITION BY emp_id ORDER BY date asc) AS total_credit,
    sum(debit) OVER (PARTITION BY emp_id ORDER BY date asc) AS total_debit,
    (sum(income) OVER (PARTITION BY person_id ORDER BY date asc) - 
    sum(expense) OVER (PARTITION BY person_id ORDER BY date asc)) as total_balance
    FROM group_payment
    ORDER BY person_id, date),

ranks as (select emp_id,date,total_balance,
rank() over (partition by emp_id order by total_balance desc) as rank
from total_balance ),

现在选择秩=1的行（即emp_id的一天总余额的最大值和它最大值的日期）。按总余额降序排列，并选择前5行

emp_order as (select emp_id,date,total_balance 
from ranks
where rank=1
order by 3 desc
limit 5)

现在从employee表中选择名称

select emp_id,name, date, total_balance as balance
from emp_order eo
join Employee e on e.emp_id = eo.emp_id
order by 4 desc