Question

我在SQL表中有以下数据：

enter image description here

我需要查询数据，以便每个员工都能找到一个缺少“ familyid ”的列表。

例如，我应该找到ID：2和5以及员工1027中缺少的员工1021，其中缺少数字1和6。

有关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 window function来指定“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，我会过滤结果集，只包含序列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

然后，您应该轻松按employee_number进行分组，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

扩展上一个查询，我们可以在每次familyid和sequenceid之间的差异发生变化时检测到缺失值：

-- 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。结果每个员工和预期的家庭ID都有一行。

从此left outer join返回原始数据，familyid和号码。如果没有匹配，那就是缺失值：

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;

请注意，当缺少的familyid是序列中的最后一个数字时，这将不起作用。但是，对数据结构可能是最好的。

此外，上述查询假设最多有20个家庭成员。

Answer 3

这将有效，您选择所有“Dependents”并在前一行左连接。如果那行不存在，则显示结果：

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'

另一个版本显示缺少的数字：

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

另一个解决方案：构建一个包含序列中所有可能值的表（可以为此使用标识）。然后在源表为null的表上保持连接。

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.[Offending Table]IS NULL

Answer 5

此选择将使用CTE方法检索每位员工缺少的“familyid”列表。

QUERY：

   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

SQL查询查找缺少的连续数字

5 个答案: