Where do I find Sql Server metadata for column datatypes?
You are close. You can look at sys.columns
to get the columns.
You can filter on a table with OBJECT_ID=OBJECT_ID('dbo.Foo')
.
You can get the length from sys.columns
. The data type is in the user_type_id
field. The keys for that field are in sys.types
.
In its entirety you can do:
select object_NAME(c.object_id), c.name, t.name, c.max_length
from sys.columns c
INNER JOIN sys.types t
ON t.user_type_id = c.user_type_id
As a side note, in SQL Server the system tables are deprecated (i.e. syscolumns
, sysobjects
) and it's recommended as a best practice to use the views instead, sys.columns
, sys.objects
, etc.
This will give you Table, column, data type, and maxlength for each one.
To build on the answers above, it's often useful to get the column data type in the same format that you need to declare columns.
For example, varchar(50)
, varchar(max)
, decimal(p, s)
.
This allows you to do that:
SELECT
[Name] = c.[name]
, [Type] =
CASE
WHEN tp.[name] IN ('varchar', 'char') THEN tp.[name] + '(' + IIF(c.max_length = -1, 'max', CAST(c.max_length AS VARCHAR(25))) + ')'
WHEN tp.[name] IN ('nvarchar','nchar') THEN tp.[name] + '(' + IIF(c.max_length = -1, 'max', CAST(c.max_length / 2 AS VARCHAR(25)))+ ')'
WHEN tp.[name] IN ('decimal', 'numeric') THEN tp.[name] + '(' + CAST(c.[precision] AS VARCHAR(25)) + ', ' + CAST(c.[scale] AS VARCHAR(25)) + ')'
WHEN tp.[name] IN ('datetime2') THEN tp.[name] + '(' + CAST(c.[scale] AS VARCHAR(25)) + ')'
ELSE tp.[name]
END
, [RawType] = tp.[name]
, [MaxLength] = c.max_length
, [Precision] = c.[precision]
, [Scale] = c.scale
FROM sys.tables t
JOIN sys.schemas s ON t.schema_id = s.schema_id
JOIN sys.columns c ON t.object_id = c.object_id
JOIN sys.types tp ON c.user_type_id = tp.user_type_id
WHERE s.[name] = 'dbo' AND t.[name] = 'MyTable'
The correct way to do this is to join to user_type_id in the sys.types table:
select object_NAME(c.object_id), c.name, t.name, c.max_length
from sys.columns c
INNER JOIN sys.types t
ON t.user_type_id = c.user_type_id
user_type_id
is identical to system_type_id for system types - see documentation: https://msdn.microsoft.com/en-gb/library/ms188021.aspx