How can I show the table structure in SQL Server query? - sql-server

Question

SELECT DateTime, Skill, Name, TimeZone, ID, User, Employee, Leader
FROM t_Agent_Skill_Group_Half_Hour AS t
I need to view the table structure in a query.

Answer 1

For SQL Server, if using a newer version, you can use
select *
from INFORMATION_SCHEMA.COLUMNS
where TABLE_NAME='tableName'
There are different ways to get the schema. Using ADO.NET, you can use the schema methods. Use the DbConnection's GetSchema method or the DataReader'sGetSchemaTable method.
Provided that you have a reader for the for the query, you can do something like this:
using(DbCommand cmd = ...)
using(var reader = cmd.ExecuteReader())
{
var schema = reader.GetSchemaTable();
foreach(DataRow row in schema.Rows)
{
Debug.WriteLine(row["ColumnName"] + " - " + row["DataTypeName"])
}
}
See this article for further details.

Answer 2

sp_help tablename in sql server
desc tablename in oracle

Answer 3

Try this query:
DECLARE #table_name SYSNAME
SELECT #table_name = 'dbo.test_table'
DECLARE
#object_name SYSNAME
, #object_id INT
SELECT
#object_name = '[' + s.name + '].[' + o.name + ']'
, #object_id = o.[object_id]
FROM sys.objects o WITH (NOWAIT)
JOIN sys.schemas s WITH (NOWAIT) ON o.[schema_id] = s.[schema_id]
WHERE s.name + '.' + o.name = #table_name
AND o.[type] = 'U'
AND o.is_ms_shipped = 0
DECLARE #SQL NVARCHAR(MAX) = ''
;WITH index_column AS
(
SELECT
ic.[object_id]
, ic.index_id
, ic.is_descending_key
, ic.is_included_column
, c.name
FROM sys.index_columns ic WITH (NOWAIT)
JOIN sys.columns c WITH (NOWAIT) ON ic.[object_id] = c.[object_id] AND ic.column_id = c.column_id
WHERE ic.[object_id] = #object_id
)
SELECT #SQL = 'CREATE TABLE ' + #object_name + CHAR(13) + '(' + CHAR(13) + STUFF((
SELECT CHAR(9) + ', [' + c.name + '] ' +
CASE WHEN c.is_computed = 1
THEN 'AS ' + cc.[definition]
ELSE UPPER(tp.name) +
CASE WHEN tp.name IN ('varchar', 'char', 'varbinary', 'binary', 'text')
THEN '(' + CASE WHEN c.max_length = -1 THEN 'MAX' ELSE CAST(c.max_length AS VARCHAR(5)) END + ')'
WHEN tp.name IN ('nvarchar', 'nchar', 'ntext')
THEN '(' + CASE WHEN c.max_length = -1 THEN 'MAX' ELSE CAST(c.max_length / 2 AS VARCHAR(5)) END + ')'
WHEN tp.name IN ('datetime2', 'time2', 'datetimeoffset')
THEN '(' + CAST(c.scale AS VARCHAR(5)) + ')'
WHEN tp.name = 'decimal'
THEN '(' + CAST(c.[precision] AS VARCHAR(5)) + ',' + CAST(c.scale AS VARCHAR(5)) + ')'
ELSE ''
END +
CASE WHEN c.collation_name IS NOT NULL THEN ' COLLATE ' + c.collation_name ELSE '' END +
CASE WHEN c.is_nullable = 1 THEN ' NULL' ELSE ' NOT NULL' END +
CASE WHEN dc.[definition] IS NOT NULL THEN ' DEFAULT' + dc.[definition] ELSE '' END +
CASE WHEN ic.is_identity = 1 THEN ' IDENTITY(' + CAST(ISNULL(ic.seed_value, '0') AS CHAR(1)) + ',' + CAST(ISNULL(ic.increment_value, '1') AS CHAR(1)) + ')' ELSE '' END
END + CHAR(13)
FROM sys.columns c WITH (NOWAIT)
JOIN sys.types tp WITH (NOWAIT) ON c.user_type_id = tp.user_type_id
LEFT JOIN sys.computed_columns cc WITH (NOWAIT) ON c.[object_id] = cc.[object_id] AND c.column_id = cc.column_id
LEFT JOIN sys.default_constraints dc WITH (NOWAIT) ON c.default_object_id != 0 AND c.[object_id] = dc.parent_object_id AND c.column_id = dc.parent_column_id
LEFT JOIN sys.identity_columns ic WITH (NOWAIT) ON c.is_identity = 1 AND c.[object_id] = ic.[object_id] AND c.column_id = ic.column_id
WHERE c.[object_id] = #object_id
ORDER BY c.column_id
FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'), 1, 2, CHAR(9) + ' ')
+ ISNULL((SELECT CHAR(9) + ', CONSTRAINT [' + k.name + '] PRIMARY KEY (' +
(SELECT STUFF((
SELECT ', [' + c.name + '] ' + CASE WHEN ic.is_descending_key = 1 THEN 'DESC' ELSE 'ASC' END
FROM sys.index_columns ic WITH (NOWAIT)
JOIN sys.columns c WITH (NOWAIT) ON c.[object_id] = ic.[object_id] AND c.column_id = ic.column_id
WHERE ic.is_included_column = 0
AND ic.[object_id] = k.parent_object_id
AND ic.index_id = k.unique_index_id
FOR XML PATH(N''), TYPE).value('.', 'NVARCHAR(MAX)'), 1, 2, ''))
+ ')' + CHAR(13)
FROM sys.key_constraints k WITH (NOWAIT)
WHERE k.parent_object_id = #object_id
AND k.[type] = 'PK'), '') + ')' + CHAR(13)
PRINT #SQL
Output:
CREATE TABLE [dbo].[test_table]
(
[WorkOutID] BIGINT NOT NULL IDENTITY(1,1)
, [DateOut] DATETIME NOT NULL
, [EmployeeID] INT NOT NULL
, [IsMainWorkPlace] BIT NOT NULL DEFAULT((1))
, [WorkPlaceUID] UNIQUEIDENTIFIER NULL
, [WorkShiftCD] NVARCHAR(10) COLLATE Cyrillic_General_CI_AS NULL
, [CategoryID] INT NULL
, CONSTRAINT [PK_WorkOut] PRIMARY KEY ([WorkOutID] ASC)
)
Also read this:
http://www.c-sharpcorner.com/UploadFile/67b45a/how-to-generate-a-create-table-script-for-an-existing-table/

Answer 4

On SQL Server 2012, you can use the following stored procedure:
sp_columns '<table name>'
For example, given a database table named users:
sp_columns 'users'

Answer 5

In SQL Server, you can use this query:
USE Database_name
SELECT *
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME='Table_Name';
And do not forget to replace Database_name and Table_name with the exact names of your database and table names.

Answer 6

For recent versions of SQL Server Management Studio Write the in a query editor and Do "Alt" + "F1"

Answer 7

Try this query:
select *
from (SELECT TABLE_SCHEMA
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_NAME='brands')[.brands];

Answer 8

To print a schema, I use jade and do an export to a file of the database then bring it into word to format and print

Answer 9

I was trying 'DESC table_name' but then this worked for me in psql:
select *
from INFORMATION_SCHEMA.COLUMNS
where TABLE_NAME='table_name';

Answer 10

Another way is,
mysql > SHOW CREATE TABLE my_db.my_table;
You should get the table name and create table sql