Cómo puedo encontrar claves externas no indexadas en SQL Server

Question

Tengo una base de datos SQL Server 2000 con aproximadamente 220 tablas. Estas tablas tienen un número de relaciones de clave externa entre ellos. A través del análisis de rendimiento, descubrimos que a varias de estas relaciones de clave foránea les faltan índices. En lugar de ser reactivo con los problemas de rendimiento, me gustaría ser proactivo y encontrar todas las claves externas que faltan índices.

¿Cómo puedo determinar mediante programación qué clave externa faltan los índices?

Answer 1

Aquí es una respuesta que trabaja para SQL Server 2000 escrito por un compañero de trabajo:

/* 
Description: 
    This script outputs a table with all the current database un-indexed foreign keys. 

    The table has three columns (TableName , ColumnName, ForeignKeyName) 
    TableName: The table containing the un-indexed foreign key 
    ColumnName: The foreign key column that’s not indexed 
    ForeignKeyName: Name of foreign key witch column doesn’t have an index 
    */ 
DECLARE 
    @TableName varchar(255), 
    @ColumnName varchar(255), 
    @ForeignKeyName sysname 

SET NOCOUNT ON 
SET TRANSACTION ISOLATION LEVEL READ UNCOMMITTED 

DECLARE FKColumns_cursor CURSOR Fast_Forward FOR 
SELECT cu.TABLE_NAME, cu.COLUMN_NAME, cu.CONSTRAINT_NAME 
FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS ic 
    INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE cu ON ic.CONSTRAINT_NAME = cu.CONSTRAINT_NAME 
WHERE ic.CONSTRAINT_TYPE = 'FOREIGN KEY' 

CREATE TABLE #temp1( 
    TableName varchar(255), 
    ColumnName varchar(255), 
    ForeignKeyName sysname 
) 

OPEN FKColumns_cursor 
FETCH NEXT FROM FKColumns_cursor INTO @TableName, @ColumnName, @ForeignKeyName 

WHILE @@FETCH_STATUS = 0 
BEGIN 

    IF (SELECT COUNT(*) 
    FROM sysobjects o  
     INNER JOIN sysindexes x ON x.id = o.id 
     INNER JOIN syscolumns c ON o.id = c.id 
     INNER JOIN sysindexkeys xk ON c.colid = xk.colid AND o.id = xk.id AND x.indid = xk.indid 
    WHERE o.type in ('U') 
     AND xk.keyno <= x.keycnt 
     AND permissions(o.id, c.name) <> 0 
     AND (x.status&32) = 0 
     AND o.name = @TableName 
     AND c.name = @ColumnName 
    ) = 0 
    BEGIN 
     INSERT INTO #temp1 SELECT @TableName, @ColumnName, @ForeignKeyName 
    END 


    FETCH NEXT FROM FKColumns_cursor INTO @TableName, @ColumnName, @ForeignKeyName 
END 
CLOSE FKColumns_cursor 
DEALLOCATE FKColumns_cursor 

SELECT * FROM #temp1 ORDER BY TableName 

Answer 2

En primer lugar: enumere las columnas con una restricción de clave externa. Esto ayudará a:

Query to get all foreign key constraints in SQL Server 2000

Cruz-Comparar con sysindexes y syscolumns tablas; el campo keys en sysindexes tiene una lista de todas las claves en el índice.

Answer 3

SELECT * 
FROM sys.foreign_keys fk 
WHERE EXISTS 
     (
     SELECT * 
     FROM sys.foreign_key_columns fkc 
     WHERE fkc.constraint_object_id = fk.object_id 
       AND NOT EXISTS 
       (
       SELECT * 
       FROM sys.index_columns ic 
       WHERE ic.object_id = fkc.parent_object_id 
         AND ic.column_id = fkc.parent_column_id 
         AND ic.index_column_id = fkc.constraint_column_id 
       ) 
     ) 

no tengo una copia de SQL Server 2000 útil, pero puede que tenga que cambiar sys.foreign_key-sysforeignkeys etc., como se describe here.

Esta consulta selecciona todas las claves foráneas que no tienen un índice que cubra todas las columnas que componen la clave.

Esto también admite claves externas de varias columnas.

Esto, sin embargo, devolverá un falso positivo si hay un índice compuesto que cubre todas las columnas, pero no son las columnas más a la izquierda en este índice.

Me gusta, si hay un FOREIGN KEY (col2, col3) y un índice en (col1, col2, col3), esto devolverá que hay un índice a pesar de que este índice no se puede usar para esta clave externa.

Answer 4

incorporado en el código anterior para eliminar la tabla temporal y obtener secuencias de comandos para crear los índices.

/* 
Description: 

    */ 
DECLARE 
    @SchemaName varchar(255), 
    @TableName varchar(255), 
    @ColumnName varchar(255), 
    @ForeignKeyName sysname 

SET NOCOUNT ON 
SET TRANSACTION ISOLATION LEVEL READ UNCOMMITTED 

DECLARE FKColumns_cursor CURSOR Fast_Forward FOR 
SELECT cu.TABLE_SCHEMA, cu.TABLE_NAME, cu.COLUMN_NAME, cu.CONSTRAINT_NAME 
FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS ic 
    INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE cu ON ic.CONSTRAINT_NAME = cu.CONSTRAINT_NAME 
WHERE ic.CONSTRAINT_TYPE = 'FOREIGN KEY' 

CREATE TABLE #temp1( 
    SchemaName varchar(255), 
    TableName varchar(255), 
    ColumnName varchar(255), 
    ForeignKeyName sysname 
) 

OPEN FKColumns_cursor 
FETCH NEXT FROM FKColumns_cursor INTO @SchemaName,@TableName, @ColumnName, @ForeignKeyName 

WHILE @@FETCH_STATUS = 0 
BEGIN 

    IF (SELECT COUNT(*) 
    FROM  sysobjects o  
     INNER JOIN sysindexes x ON x.id = o.id 
     INNER JOIN syscolumns c ON o.id = c.id 
     INNER JOIN sysindexkeys xk ON c.colid = xk.colid AND o.id = xk.id AND x.indid = xk.indid 
    WHERE  o.type in ('U') 
     AND xk.keyno <= x.keycnt 
     AND permissions(o.id, c.name) <> 0 
     AND (x.status&32) = 0 
     AND o.name = @TableName 
     AND c.name = @ColumnName 
    ) = 0 
    BEGIN 
     INSERT INTO #temp1 SELECT @SchemaName, @TableName, @ColumnName, @ForeignKeyName 
    END 


    FETCH NEXT FROM FKColumns_cursor INTO @SchemaName,@TableName, @ColumnName, @ForeignKeyName 
END 
CLOSE FKColumns_cursor 
DEALLOCATE FKColumns_cursor 

SELECT 'CREATE INDEX IDX_' + ForeignKeyName + ' ON ' + SchemaName + '.' + TableName + '(' + ColumnName +')' 
FROM #temp1 
ORDER BY TableName 

drop table #temp1 

Answer 5

En mi post "SQL Script to create indexes for Foreign keys" He puesto enlaces a 2 implementaciones: paul_nielsen 's y tklimczak's (login to sqlservercentral se requiere)

Answer 6

Nota: Esto es para SQL Server 2005 + pero esta fue la única pregunta que encontré sobre este tema.

--Finds foreign keys without indexes 
--How to interpret: 
--When we delete frpm PkTable, it checks FkColumn for the PkId we are deleting. 
--So if FkTable doesn't have an index on FkColumn, then we cannot delete a row from PkTable because it is too slow. 
SELECT rt.name as PkTableName, rc.name as PkColumnName, 
fk.name FkName, t.name as FkTableName, c.name as FkColumnName, ddps.row_count, i.name as IndexName 
FROM sys.foreign_key_columns fkc 
inner join sys.foreign_keys fk on fkc.constraint_object_id = fk.object_id 
inner join sys.tables t on fkc.parent_object_id = t.object_id 
inner join sys.columns c on fkc.parent_object_id = c.object_id and fkc.parent_column_id = c.column_id 
inner join sys.tables rt on fkc.referenced_object_id = rt.object_id 
inner join sys.columns rc on fkc.referenced_object_id = rc.object_id and fkc.referenced_column_id = rc.column_id 
inner join sys.indexes ri on t.object_id = ri.object_id 
inner JOIN sys.dm_db_partition_stats AS ddps ON ri.OBJECT_ID = ddps.OBJECT_ID AND ri.index_id = ddps.index_id 
left join sys.index_columns ic on ic.object_id = t.object_id and ic.column_id = c.column_id 
left join sys.indexes i on ic.object_id = i.object_id and i.index_id = ic.index_id 
where ri.index_id < 2 and i.index_id is null and ddps.row_count > 0 
order by 
--PkTableName, 
ddps.row_count desc