82 
SELECT DateTime, Skill, Name, TimeZone, ID, User, Employee, Leader 
FROM t_Agent_Skill_Group_Half_Hour AS t

我需要在查询中查看表结构。

4

9 回答 9

156

对于 SQL Server，如果使用较新的版本，您可以使用

select *
from INFORMATION_SCHEMA.COLUMNS
where TABLE_NAME='tableName'

有不同的方法来获取架构。使用 ADO.NET，您可以使用架构方法。使用DbConnection'GetSchema方法或DataReader'GetSchemaTable方法.

如果您有一个用于查询的阅读器，您可以执行以下操作：

using(DbCommand cmd = ...)
using(var reader = cmd.ExecuteReader())
{
    var schema = reader.GetSchemaTable();
    foreach(DataRow row in schema.Rows)
    {
        Debug.WriteLine(row["ColumnName"] + " - " + row["DataTypeName"])
    }
}

有关更多详细信息，请参阅本文。

于 2013-08-18T11:42:32.457 回答
55

sp_help tablename在 sql 服务器中

desc tablename在甲骨文中

于 2014-05-16T04:10:30.873 回答
17

试试这个查询：

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

输出：

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)
)

另请阅读：

http://www.c-sharpcorner.com/UploadFile/67b45a/how-to-generate-a-create-table-script-for-an-existing-table/

于 2013-08-19T05:13:06.733 回答
16

在 SQL Server 2012 上，您可以使用以下存储过程：

sp_columns '<table name>'

例如，给定一个名为 users 的数据库表：

sp_columns 'users'
于 2014-08-20T23:24:43.267 回答
4

对于 SQL Server Management Studio 的最新版本，在查询编辑器中编写并执行“Alt”+“F1”

于 2019-04-25T20:30:42.590 回答
1

在 SQL Server 中，您可以使用以下查询：

USE Database_name

SELECT *
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME='Table_Name';

并且不要忘记用您的数据库和表名的确切名称替换Database_name和。Table_name

于 2020-07-11T20:30:29.247 回答
1

另一种方式是，

mysql > SHOW CREATE TABLE my_db.my_table;

您应该获取表名并创建表 sql

于 2020-06-06T09:45:17.953 回答
0

I was trying 'DESC table_name' but then this worked for me in psql:

select * 
from INFORMATION_SCHEMA.COLUMNS
where TABLE_NAME='table_name';
于 2019-06-23T10:26:03.097 回答
0

要打印模式，我使用玉并导出到数据库的文件，然后将其转换为 word 以格式化和打印

于 2018-12-31T23:42:33.907 回答