答えて

12
select distinct 
    object_name(fic.[object_id]) table_name, 
    [name] column_name 
from 
    sys.fulltext_index_columns fic 
    inner join sys.columns c 
     on c.[object_id] = fic.[object_id] 
     and c.[column_id] = fic.[column_id] 
6

この1つは、私は少しだけでなく、データの列に関する情報が含まれるようにクエリを変更し、あなたに多くの情報

SELECT 
    t.name AS TableName, 
    c.name AS FTCatalogName , 
    i.name AS UniqueIdxName, 
    cl.name AS ColumnName, 
    cdt.name AS DataTypeColumnName 
FROM 
    sys.tables t 
INNER JOIN 
    sys.fulltext_indexes fi 
ON 
    t.[object_id] = fi.[object_id] 
INNER JOIN 
    sys.fulltext_index_columns ic 
ON 
    ic.[object_id] = t.[object_id] 
INNER JOIN 
    sys.columns cl 
ON 
    ic.column_id = cl.column_id 
    AND ic.[object_id] = cl.[object_id] 
INNER JOIN 
    sys.fulltext_catalogs c 
ON 
    fi.fulltext_catalog_id = c.fulltext_catalog_id 
INNER JOIN 
    sys.indexes i 
ON 
    fi.unique_index_id = i.index_id 
    AND fi.[object_id] = i.[object_id] 
LEFT JOIN 
    sys.columns cdt 
ON 
    ic.type_column_id = cdt.column_id 
    AND fi.object_id = cdt.object_id; 
+0

を与えます。このクエリ、おかげで多くの仕事を私に保存! –

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