“wie man Tabelle mit fremden Schlüsselbeschränkungen abschneidet” Code-Antworten

wie man Tabelle mit fremden Schlüsselbeschränkungen abschneidet

SET FOREIGN_KEY_CHECKS = 0;
TRUNCATE table1;
Ill Ibis

So schneiden die Tisch für die Einschränkung des Fremdschlüssels ab


DELETE FROM [TableName]
DBCC CHECKIDENT ([TableName], RESEED, 0)
Johan Leabirt

Tischtabelle SQL Server Fremdschlüssel

CREATE TABLE #x -- feel free to use a permanent table
(
  drop_script NVARCHAR(MAX),
  create_script NVARCHAR(MAX)
);

DECLARE @drop   NVARCHAR(MAX) = N'',
        @create NVARCHAR(MAX) = N'';

-- drop is easy, just build a simple concatenated list from sys.foreign_keys:
SELECT @drop += N'
ALTER TABLE ' + QUOTENAME(cs.name) + '.' + QUOTENAME(ct.name) 
    + ' DROP CONSTRAINT ' + QUOTENAME(fk.name) + ';'
FROM sys.foreign_keys AS fk
INNER JOIN sys.tables AS ct
  ON fk.parent_object_id = ct.[object_id]
INNER JOIN sys.schemas AS cs 
  ON ct.[schema_id] = cs.[schema_id];

INSERT #x(drop_script) SELECT @drop;

-- create is a little more complex. We need to generate the list of 
-- columns on both sides of the constraint, even though in most cases
-- there is only one column.
SELECT @create += N'
ALTER TABLE ' 
   + QUOTENAME(cs.name) + '.' + QUOTENAME(ct.name) 
   + ' ADD CONSTRAINT ' + QUOTENAME(fk.name) 
   + ' FOREIGN KEY (' + STUFF((SELECT ',' + QUOTENAME(c.name)
   -- get all the columns in the constraint table
    FROM sys.columns AS c 
    INNER JOIN sys.foreign_key_columns AS fkc 
    ON fkc.parent_column_id = c.column_id
    AND fkc.parent_object_id = c.[object_id]
    WHERE fkc.constraint_object_id = fk.[object_id]
    ORDER BY fkc.constraint_column_id 
    FOR XML PATH(N''), TYPE).value(N'.[1]', N'nvarchar(max)'), 1, 1, N'')
  + ') REFERENCES ' + QUOTENAME(rs.name) + '.' + QUOTENAME(rt.name)
  + '(' + STUFF((SELECT ',' + QUOTENAME(c.name)
   -- get all the referenced columns
    FROM sys.columns AS c 
    INNER JOIN sys.foreign_key_columns AS fkc 
    ON fkc.referenced_column_id = c.column_id
    AND fkc.referenced_object_id = c.[object_id]
    WHERE fkc.constraint_object_id = fk.[object_id]
    ORDER BY fkc.constraint_column_id 
    FOR XML PATH(N''), TYPE).value(N'.[1]', N'nvarchar(max)'), 1, 1, N'') + ');'
FROM sys.foreign_keys AS fk
INNER JOIN sys.tables AS rt -- referenced table
  ON fk.referenced_object_id = rt.[object_id]
INNER JOIN sys.schemas AS rs 
  ON rt.[schema_id] = rs.[schema_id]
INNER JOIN sys.tables AS ct -- constraint table
  ON fk.parent_object_id = ct.[object_id]
INNER JOIN sys.schemas AS cs 
  ON ct.[schema_id] = cs.[schema_id]
WHERE rt.is_ms_shipped = 0 AND ct.is_ms_shipped = 0;

UPDATE #x SET create_script = @create;

PRINT @drop;
PRINT @create;

/*
EXEC sp_executesql @drop
-- clear out data etc. here
EXEC sp_executesql @create;
*/
Wrong Willet

Ähnliche Antworten wie “wie man Tabelle mit fremden Schlüsselbeschränkungen abschneidet”

Fragen ähnlich wie “wie man Tabelle mit fremden Schlüsselbeschränkungen abschneidet”

Weitere verwandte Antworten zu “wie man Tabelle mit fremden Schlüsselbeschränkungen abschneidet” auf Sql

Durchsuchen Sie beliebte Code-Antworten nach Sprache

Durchsuchen Sie andere Codesprachen