我刚刚做了marc_s“presentation ready”:

SELECT 
    c.name 'Column Name',
    t.name 'Data type',
    IIF(t.name = 'nvarchar', c.max_length / 2, c.max_length) 'Max Length',
    c.precision 'Precision',
    c.scale 'Scale',
    IIF(c.is_nullable = 0, 'No', 'Yes') 'Nullable',
    IIF(ISNULL(i.is_primary_key, 0) = 0, 'No', 'Yes') 'Primary Key'
FROM    
    sys.columns c
INNER JOIN 
    sys.types t ON c.user_type_id = t.user_type_id
LEFT OUTER JOIN 
    sys.index_columns ic ON ic.object_id = c.object_id AND ic.column_id = c.column_id
LEFT OUTER JOIN 
    sys.indexes i ON ic.object_id = i.object_id AND ic.index_id = i.index_id
WHERE
    c.object_id = OBJECT_ID('YourTableName')

试试这个:

select COLUMN_NAME, DATA_TYPE, CHARACTER_MAXIMUM_LENGTH, IS_NULLABLE 
from INFORMATION_SCHEMA.COLUMNS IC
where TABLE_NAME = 'tablename' and COLUMN_NAME = 'columnname'

为了避免某些列的重复行，请使用user_type_id而不是system_type_id。

SELECT 
    c.name 'Column Name',
    t.Name 'Data type',
    c.max_length 'Max Length',
    c.precision ,
    c.scale ,
    c.is_nullable,
    ISNULL(i.is_primary_key, 0) 'Primary Key'
FROM    
    sys.columns c
INNER JOIN 
    sys.types t ON c.user_type_id = t.user_type_id
LEFT OUTER JOIN 
    sys.index_columns ic ON ic.object_id = c.object_id AND ic.column_id = c.column_id
LEFT OUTER JOIN 
    sys.indexes i ON ic.object_id = i.object_id AND ic.index_id = i.index_id
WHERE
    c.object_id = OBJECT_ID('YourTableName')

只需将YourTableName替换为实际的表名-适用于SQL Server 2005及更高版本。

如果您正在使用模式，请将YourTableName替换为YourSchemaName。YourTableName，其中YourSchemaName是实际的模式名，YourTableName是实际的表名。

我有点惊讶居然没人提

sp_help 'mytable'

IF EXISTS (SELECT 1 FROM INFORMATION_SCHEMA.TABLES 
     WHERE TABLE_TYPE = 'BASE TABLE' AND TABLE_NAME = 'Table')
      BEGIN
        SELECT COLS.COLUMN_NAME, COLS.DATA_TYPE, COLS.CHARACTER_MAXIMUM_LENGTH, 
              (SELECT 'Yes' FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS TC JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE KCU
                              ON COLS.TABLE_NAME = TC.TABLE_NAME 
                             AND TC.CONSTRAINT_TYPE = 'PRIMARY KEY'
                             AND KCU.TABLE_NAME = TC.TABLE_NAME
                             AND KCU.CONSTRAINT_NAME = TC.CONSTRAINT_NAME
                             AND KCU.COLUMN_NAME = COLS.COLUMN_NAME) AS KeyX
        FROM INFORMATION_SCHEMA.COLUMNS COLS WHERE TABLE_NAME = 'Table' ORDER BY KeyX DESC, COLUMN_NAME
      END

查询:执行sp_description _first_result_set并选择年收入 [DB_NAME]。[DBO]。(员工)”

注意:在某些ide中选择n之前是工作的，或者在某些ide中没有选择n是工作的