Início / dba / Perguntas / 45158
Accepted
cubetwo1729
Asked: 2013-06-25 05:50:45 +0800 CST

Como escrever uma consulta que encontra todas as referências circulares quando uma tabela faz referência a si mesma?

  • 772

Tenho o seguinte esquema (nomes alterados), que não posso alterar:

CREATE TABLE MyTable (
    Id INT NOT NULL PRIMARY KEY,
    ParentId INT NOT NULL
);

ALTER TABLE MyTable ADD FOREIGN KEY (ParentId) REFERENCES MyTable(Id);

Ou seja, cada registro é filho de outro registro. Se um registro ParentIdfor igual ao seu Id, o registro será considerado um nó raiz.

Eu quero executar a consulta que encontrará todas as referências circulares. Por exemplo, com os dados

INSERT INTO MyTable (Id, ParentId) VALUES
    (0, 0),
    (1, 0),
    (2, 4),
    (3, 2),
    (4, 3);

a consulta deve retornar

Id | Cycle
2  | 2 < 4 < 3 < 2
3  | 3 < 2 < 4 < 3
4  | 4 < 3 < 2 < 4

Escrevi a seguinte consulta para o SQL Server 2008 R2 e gostaria de saber se essa consulta pode ser melhorada:

IF OBJECT_ID(N'tempdb..#Results') IS NOT NULL DROP TABLE #Results;
CREATE TABLE #Results (Id INT, HasParentalCycle BIT, Cycle VARCHAR(MAX));

DECLARE @i INT,
    @j INT,
    @flag BIT,
    @isRoot BIT,
    @ids VARCHAR(MAX);

DECLARE MyCursor CURSOR FAST_FORWARD FOR
    SELECT Id
    FROM MyTable;

OPEN MyCursor;
FETCH NEXT FROM MyCursor INTO @i;
WHILE @@FETCH_STATUS = 0
BEGIN
    IF OBJECT_ID(N'tempdb..#Parents') IS NOT NULL DROP TABLE #Parents;
    CREATE TABLE #Parents (Id INT);

    SET @ids = NULL;
    SET @isRoot = 0;
    SET @flag = 0;
    SET @j = @i;
    INSERT INTO #Parents (Id) VALUES (@j);

    WHILE (1=1)
    BEGIN
        SELECT
            @j = ParentId,
            @isRoot = CASE WHEN ParentId = Id THEN 1 ELSE 0 END
        FROM MyTable
        WHERE Id = @j;

        IF (@isRoot = 1)
        BEGIN
            SET @flag = 0;
            BREAK;
        END        

        IF EXISTS (SELECT 1 FROM #Parents WHERE Id = @j)
        BEGIN
            INSERT INTO #Parents (Id) VALUES (@j);
            SET @flag = 1;
            SELECT @ids = COALESCE(@ids + ' < ', '') + CAST(Id AS VARCHAR) FROM #Parents;
            BREAK;
        END
        ELSE
        BEGIN
            INSERT INTO #Parents (Id) VALUES (@j);
        END        
    END

    INSERT INTO #Results (Id, HasParentalCycle, Cycle) VALUES (@i, @flag, @ids);

    FETCH NEXT FROM MyCursor INTO @i;
END
CLOSE MyCursor;
DEALLOCATE MyCursor;

SELECT Id, Cycle
FROM #Results
WHERE HasParentalCycle = 1;
sql-server sql-server-2008-r2

2 respostas

  1. Best Answer

    Isso exige um CTE recursivo:

    WITH FindRoot AS
(
    SELECT Id,ParentId, CAST(Id AS NVARCHAR(MAX)) Path
    FROM dbo.MyTable

    UNION ALL

    SELECT C.Id, P.ParentId, C.Path + N' > ' + CAST(P.Id AS NVARCHAR(MAX))
    FROM dbo.MyTable P
    JOIN FindRoot C
    ON C.ParentId = P.Id AND P.ParentId <> P.Id AND C.ParentId <> C.Id
 )
SELECT *
FROM FindRoot R
WHERE R.Id = R.ParentId 
  AND R.ParentId <> 0;

    Veja-o em ação aqui: SQL Fiddle

    Atualizar:

    Adicionada distância para poder excluir todos os ciclos próprios (veja o comentário do ypercube):

    WITH FindRoot AS
(
    SELECT Id,ParentId, CAST(Id AS NVARCHAR(MAX)) Path, 0 Distance
    FROM dbo.MyTable

    UNION ALL

    SELECT C.Id, P.ParentId, C.Path + N' > ' + CAST(P.Id AS NVARCHAR(MAX)), C.Distance + 1
    FROM dbo.MyTable P
    JOIN FindRoot C
    ON C.ParentId = P.Id AND P.ParentId <> P.Id AND C.ParentId <> C.Id
 )
SELECT *
FROM FindRoot R
WHERE R.Id = R.ParentId 
  AND R.ParentId <> 0
  AND R.Distance > 0;

    SQL Fiddle

    Qual deles você deve usar depende de sua exigência.

    • 35 
  2. SELECT RC.CONSTRAINT_NAME FK_Name
, KF.TABLE_SCHEMA FK_Schema
, KF.TABLE_NAME FK_Table
, KF.COLUMN_NAME FK_Column
, RC.UNIQUE_CONSTRAINT_NAME PK_Name
, KP.TABLE_SCHEMA PK_Schema
, KP.TABLE_NAME PK_Table
, KP.COLUMN_NAME PK_Column
, RC.MATCH_OPTION MatchOption
, RC.UPDATE_RULE UpdateRule
, RC.DELETE_RULE DeleteRule
FROM INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS RC
JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE KF ON RC.CONSTRAINT_NAME = KF.CONSTRAINT_NAME
JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE KP ON RC.UNIQUE_CONSTRAINT_NAME = KP.CONSTRAINT_NAME
WHERE KF.TABLE_NAME = KP.TABLE_NAME
    • 2

relate perguntas