显示同一个表中的行之间的差异

Question

我有一个名为 Employee_audit 的表，具有以下架构，

emp_audit_id	eid	name	salary
1	1	Daniel	1000
2	1	Dani	1000
3	1	Danny	3000

我的目标是编写一个 SQL 查询，它将以以下格式返回，同时考虑到第一行也从 null 更改为值。

columnName	oldValue	newValue
name	null	Daniel
salary	null	1000
name	Daniel	Dani
name	Dani	Danny
salary	1000	3000

我已经编写了下面的 SQL 查询，

WITH cte  AS 
(
  SELECT empid,
         name,
         salary, 
         rn=ROW_NUMBER()OVER(PARTITION BY empid ORDER BY emp_audit_id)
  FROM   Employee_audit 
) 
SELECT oldname=CASE WHEN c1.Name=c2.Name THEN '' ELSE C1.Name END,
       newname=CASE WHEN c1.Name=c2.Name THEN '' ELSE C2.Name END,
       oldsalary=CASE WHEN c1.salary=c2.salary THEN NULL ELSE C1.salary END,
       newsalary=CASE WHEN c1.salary=c2.salary THEN NULL ELSE C2.salary END
FROM cte c1 INNER JOIN cte c2 
ON c1.empid=c2.empid AND c2.RN=c1.RN + 1

但它以以下格式给出结果

oldname	newname	oldsalary	newsalary
Daniel	Dani	null	null
Dani	Danny	1000	3000

你能回答我吗，我怎样才能得到所需的结果。

Answer 1

如果您在 CTE 中为每一行指定一个行号，然后将自己连接到下一行，您可以比较旧值和新值。合并 2 个不同的列名有点笨拙，但是如果您需要更强大的解决方案，您可能会考虑对数据进行透视。

您显然还必须将所有值转换为通用数据类型，例如一个字符串。

declare @Test table (emp_audit_id int, eid int, [name] varchar(32), salary money);

insert into @Test (emp_audit_id, eid, [name], salary)
values
(1, 1, 'Daniel', 1000),
(2, 1, 'Dani', 1000),
(3, 1, 'Danny', 3000);

with cte as (
    select emp_audit_id, eid, [name], salary
      , row_number() over (partition by eid order by emp_audit_id) rn
    from @Test
)
select C.emp_audit_id, 'name' columnName, P.[Name] oldValue, C.[name] newValue
from cte C
left join cte P on P.eid = C.eid and P.rn + 1 = C.rn
where coalesce(C.[name],'') != coalesce(P.[Name],'')
union all
select C.emp_audit_id, 'salary' columnName, convert(varchar(21),P.salary), convert(varchar(21),C.salary)
from cte C
left join cte P on P.eid = C.eid and P.rn + 1 = C.rn
where coalesce(C.salary,0) != coalesce(P.salary,0)
order by C.emp_audit_id, columnName;

返回：

emp_audit_id	columnName	oldValue	newValue
1	name	NULL	Daniel
1	salary	NULL	1000.00
2	name	Daniel	Dani
3	name	Dani	Danny
3	salary	1000.00	3000.00

我强烈建议您将 DDL+DML（如上所示）添加到您未来的所有问题中，因为它使人们更容易提供帮助。

Answer 2

超前和滞后功能可以帮助您。

“差异”计算您需要查找差异的每一列的差异

with diffs as (
    select 'name' colName, emp_audit_id, eid, lag(name, 1, null) over (partition by eid order by emp_audit_id) oldValue, name newValue
    from some_table
    union all
    select 'salary', emp_audit_id, eid, cast(lag(salary, 1, null) over (partition by eid order by emp_audit_id) as varchar), cast(salary as varchar) newValue
    from some_table
)
select * 
from diffs 
where oldValue <> newValue or oldValue is null 
order by emp_audit_id, eid