sql - SQL查询查找缺少的连续数字

Question

我在 SQL 表中有以下数据：

我需要查询数据，以便获得每个员工缺少的“ familyid ”列表。

例如，对于序列中缺少的 Employee 1021，我应该得到 ID：2 和 5，对于 Employee 1027，应该得到缺少的数字 1 和 6。

关于如何查询的任何线索？

感谢任何帮助。

Answer 1

找到第一个缺失值

我会使用ROW_NUMBER 窗口函数来分配“正确的”序列 ID 号。假设每次员工 ID 更改时序列 ID 都会重新启动：

SELECT
  e.id,
  e.name,
  e.employee_number,
  e.relation,
  e.familyid,
  ROW_NUMBER() OVER(PARTITION BY e.employeeid ORDER BY familyid) - 1 AS sequenceid
FROM employee_members e

然后，我将过滤结果集以仅包含序列 ID 不匹配的行：

SELECT *
FROM (
  SELECT
    e.id,
    e.name,
    e.employee_number,
    e.relation,
    e.familyid,
    ROW_NUMBER() OVER(PARTITION BY e.employeeid ORDER BY familyid) - 1 AS sequenceid
  FROM employee_members e
) a
WHERE a.familyid <> a.sequenceid

再说一次，您应该轻松地分组employee_number并找到每个员工的第一个缺失的序列 ID：

SELECT
  a.employee_number,
  MIN(a.sequence_id) AS first_missing
FROM (
  SELECT
    e.id,
    e.name,
    e.employee_number,
    e.relation,
    e.familyid,
    ROW_NUMBER() OVER(PARTITION BY e.employeeid ORDER BY familyid) - 1 AS sequenceid
  FROM employee_members e
) a
WHERE a.familyid <> a.sequenceid
GROUP BY a.employee_number

查找所有缺失值

扩展之前的查询，我们可以在每次familyid和之间的差异sequenceid发生变化时检测到一个缺失值：

-- Warning: this is totally untested :-/
SELECT
  b.employee_number,
  MIN(b.sequence_id) AS missing
FROM (
  SELECT
    a.*,
    a.familyid - a.sequenceid AS displacement
    SELECT
      e.*,
      ROW_NUMBER() OVER(PARTITION BY e.employeeid ORDER BY familyid) - 1 AS sequenceid
    FROM employee_members e
  ) a
) b
WHERE b.displacement <> 0
GROUP BY
  b.employee_number,
  b.displacement

Answer 2

这是一种方法。计算每个员工的最大家庭 ID。然后将其加入到最大家庭 ID 的数字列表中。结果对于每个员工和预期的家庭 ID 都有一行。

做一个left outer join从这个回到原来的数据，就familyid和个数。在没有匹配项的地方，这些是缺失值：

with nums as (
      select 1 as n
      union all
      select n+1
      from nums
      where n < 20
     )
select en.employee, n.n as MissingFamilyId
from (select employee, min(familyid) as minfi, max(familyid) as maxfi
      from t
      group by employee
     ) en join
     nums n
     on n.n <= maxfi left outer join
     t
     on t.employee = en.employee and
        t.familyid = n.n
where t.employee_number is null;

familyid请注意，当缺少序列中的最后一个数字时，这将不起作用。但这可能是您可以使用数据结构做的最好的事情。

此外，上述查询假设最多有 20 个家庭成员。

Answer 3

这将起作用，您选择所有“依赖项”并在前一行留下连接。如果该行不存在，那么您将显示结果：

SELECT 'Missing Prior', t1.*
FROM employee_members t1
LEFT JOIN employee_members t2 ON t1.employee_number = t2.employee_number 
                  AND (t1.familyid-1) = t2.familyid 
WHERE t2.employee_number is null and t1.relation == 'Dependent'

---

另一个向您显示丢失号码的版本：

SELECT t1.employee_number, t1.familyid-1 as Missing_Member
FROM employee_members t1
LEFT JOIN employee_members t2 ON t1.employee_number = t2.employee_number 
                  AND (t1.familyid-1) = t2.familyid 
WHERE t2.employee_number is null and t1.relation == 'Dependent'

Answer 4

另一种解决方案：建立一个包含序列中所有可能值的表（可以为此使用身份）。然后在源表为空的表上左连接。

DECLARE @Seq TABLE (id INT IDENTITY(1, 1))
DECLARE @iter INT = 1

WHILE @iter <= (
        SELECT MAX([your ID column])
        FROM [Offending Table]
        )
BEGIN
    INSERT @Seq DEFAULT
    VALUES

    SET @iter = @iter + 1
END

SELECT id
FROM @seq s
LEFT JOIN [Offending Table] ot ON s.id = ot.[your ID column]
WHERE ot.[your ID column] IS NULL

Answer 5

此选择将使用 CTE 方法检索每个员工缺少的“familyid”列表。

询问 ：

   WITH emp_grp (
        EmployeeID
        ,MaxFamilyID
        )
    AS (
        SELECT e2.EmployeeID
            ,MAX(e2.FamilyID) MaxFamilyID
        FROM employee_number e2
        GROUP BY e2.EmployeeID
        )
        ,emp_mem
    AS (
        SELECT EmployeeID
            ,0 AS FamilyID
            ,MaxFamilyID
        FROM emp_grp

        UNION ALL

        SELECT EmployeeID
            ,FamilyID + 1 AS FamilyID
            ,MaxFamilyID
        FROM emp_mem
        WHERE emp_mem.FamilyID < MaxFamilyID
        )

    SELECT emp_mem.EmployeeID
        ,emp_mem.FamilyID
    FROM emp_mem
    LEFT JOIN employee_number emp_num ON emp_mem.EmployeeID = emp_num.EmployeeID
        AND emp_mem.FamilyID = emp_num.FamilyID
    WHERE emp_num.EmployeeID IS NULL
    ORDER BY emp_mem.EmployeeID
        ,emp_mem.FamilyID

OPTION ( MAXRECURSION 32767)

输出 ：

EmployeeID  FamilyID
----------- -----------
1021        2
1021        5
1027        1
1027        6