如何从SQL Server表对象生成类?
我说的不是使用ORM。我只需要创建实体(简单类)。喜欢的东西:
public class Person
{
public string Name { get;set; }
public string Phone { get;set; }
}
给定一个表，比如:
+----+-------+----------------+
| ID | Name | Phone |
+----+-------+----------------+
| 1 | Alice | (555) 555-5550 |
| 2 | Bob | (555) 555-5551 |
| 3 | Cathy | (555) 555-5552 |
+----+-------+----------------+
当前回答
Visual Studio杂志发表了这篇文章:
为SQL查询结果生成。net POCO类
它有一个可下载的项目，你可以构建，给它你的SQL信息，它会为你制作出类。
现在，如果该工具刚刚为SELECT、INSERT和UPDATE创建了SQL命令....
其他回答
你刚刚做了，只要你的表包含两列，并被称为'tblPeople'之类的东西。
您总是可以编写自己的SQL包装器。实际上我更喜欢这样做，我讨厌生成代码，以任何方式。
也许创建一个DAL类，并有一个名为GetPerson(int id)的方法，用于查询数据库中的person，然后从结果集中创建person对象。
Oracle DB的另一个解决方案-> c#
单查询 无功能无步骤 Mulitple表
添加DataAnnotations
(例子) (需要) (表) (StringLength) (列) 可以为空
查询
https://gist.github.com/omansak/f19eefffd2d639ac72a1f4b506d8471a
输出
[Table("AGENTS")]
public class Agents
{
[Key]
[Required]
[Column("INT_ID", TypeName = "NUMBER(10,0)", Order = 1)]
public long IntId { get; set; }
[Key]
[Required]
[StringLength(15)]
[Column("REFERENCE_CODE", TypeName = "VARCHAR2(15)", Order = 2)]
public string ReferenceCode { get; set; }
[Required]
[Column("PARENT_INT_ID", TypeName = "NUMBER(10,0)", Order = 3)]
public long ParentIntId { get; set; }
[Required]
[StringLength(200)]
[Column("TITLE", TypeName = "VARCHAR2(200)", Order = 4)]
public string Title { get; set; }
[Required]
[Column("START_DATE", TypeName = "DATE", Order = 5)]
public DateTime StartDate { get; set; }
[Required]
[Column("END_DATE", TypeName = "DATE", Order = 6)]
public DateTime EndDate { get; set; }
[Required]
[StringLength(1)]
[Column("AGENT_TYPE", TypeName = "VARCHAR2(1)", Order = 7)]
public string AgentType { get; set; }
[Required]
[Column("CREATE_DATE", TypeName = "DATE", Order = 8)]
public DateTime CreateDate { get; set; }
[Required]
[StringLength(32)]
[Column("CREATE_USER", TypeName = "VARCHAR2(32)", Order = 9)]
public string CreateUser { get; set; }
[StringLength(200)]
[Column("RESPONSIBLE_CONTACT", TypeName = "VARCHAR2(200)", Order = 10)]
public string ResponsibleContact { get; set; }
[StringLength(100)]
[Column("RESPONSIBLE_TITLE", TypeName = "VARCHAR2(100)", Order = 11)]
public string ResponsibleTitle { get; set; }
[StringLength(100)]
[Column("AGENCY_PLATE_NO", TypeName = "VARCHAR2(100)", Order = 12)]
public string AgencyPlateNo { get; set; }
[Column("AGENCY_COVER_AMOUNT", TypeName = "NUMBER(24,2)", Order = 13)]
public double? AgencyCoverAmount { get; set; }
[StringLength(100)]
[Column("MERSIS_NO", TypeName = "VARCHAR2(100)", Order = 14)]
public string MersisNo { get; set; }
[StringLength(100)]
[Column("TECH_PERSONEL_NO", TypeName = "VARCHAR2(100)", Order = 15)]
public string TechPersonelNo { get; set; }
[StringLength(100)]
[Column("TECH_PERSONEL_NAME", TypeName = "VARCHAR2(100)", Order = 16)]
public string TechPersonelName { get; set; }
[Column("COVER_END_DATE", TypeName = "DATE", Order = 17)]
public DateTime? CoverEndDate { get; set; }
[Column("BRANCH_NUMBER", TypeName = "NUMBER(10,0)", Order = 18)]
public long? BranchNumber { get; set; }
[Column("ACTION_NUMBER", TypeName = "NUMBER(10,0)", Order = 19)]
public long? ActionNumber { get; set; }
[Column("CLUB_PARTICIPATION_COUNT", TypeName = "NUMBER(10,0)", Order = 20)]
public long? ClubParticipationCount { get; set; }
[Column("AGENCY_CONTRACT_DATE", TypeName = "DATE", Order = 21)]
public DateTime? AgencyContractDate { get; set; }
[StringLength(200)]
[Column("KEP_ADDRESS", TypeName = "VARCHAR2(200)", Order = 22)]
public string KepAddress { get; set; }
}
我将几个基于SQL的答案(主要是Alex Aza的根答案)打包到kassify中，这是一个控制台应用程序，可以一次性为指定的数据库生成所有类:
例如，给定一个Users表，它是这样的:
+----+------------------+-----------+---------------------+
| Id | Name | Username | Email |
+----+------------------+-----------+---------------------+
| 1 | Leanne Graham | Bret | Sincere@april.biz |
| 2 | Ervin Howell | Antonette | Shanna@melissa.tv |
| 3 | Clementine Bauch | Samantha | Nathan@yesenia.net |
+----+------------------+-----------+---------------------+
klassify将生成一个名为Users.cs的文件，看起来像这样:
public class User
{
public int Id {get; set; }
public string Name { get;set; }
public string Username { get; set; }
public string Email { get; set; }
}
它将为每个表输出一个文件。丢弃你不用的东西。
使用
--out, -o:
output directory << defaults to the current directory >>
--user, -u:
sql server user id << required >>
--password, -p:
sql server password << required >>
--server, -s:
sql server << defaults to localhost >>
--database, -d:
sql database << required >>
--timeout, -t:
connection timeout << defaults to 30 >>
--help, -h:
show help
因为之前没有人提到它，所以还有Scaffold-DbContext
在NuGet Package Manager控制台类型如下: Microsoft.EntityFrameworkCore.SqlServer -OutputDir "输出目录"
为了感谢Alex的解决方案和Guilherme的要求，我为MySQL生成c#类做了这个
set @schema := 'schema_name';
set @table := 'table_name';
SET group_concat_max_len = 2048;
SELECT
concat('public class ', @table, '\n{\n', GROUP_CONCAT(a.property_ SEPARATOR '\n'), '\n}') class_
FROM
(select
CONCAT(
'\tpublic ',
case
when DATA_TYPE = 'bigint' then 'long'
when DATA_TYPE = 'BINARY' then 'byte[]'
when DATA_TYPE = 'bit' then 'bool'
when DATA_TYPE = 'char' then 'string'
when DATA_TYPE = 'date' then 'DateTime'
when DATA_TYPE = 'datetime' then 'DateTime'
when DATA_TYPE = 'datetime2' then 'DateTime'
when DATA_TYPE = 'datetimeoffset' then 'DateTimeOffset'
when DATA_TYPE = 'decimal' then 'decimal'
when DATA_TYPE = 'double' then 'double'
when DATA_TYPE = 'float' then 'float'
when DATA_TYPE = 'image' then 'byte[]'
when DATA_TYPE = 'int' then 'int'
when DATA_TYPE = 'money' then 'decimal'
when DATA_TYPE = 'nchar' then 'char'
when DATA_TYPE = 'ntext' then 'string'
when DATA_TYPE = 'numeric' then 'decimal'
when DATA_TYPE = 'nvarchar' then 'string'
when DATA_TYPE = 'real' then 'double'
when DATA_TYPE = 'smalldatetime' then 'DateTime'
when DATA_TYPE = 'smallint' then 'short'
when DATA_TYPE = 'smallmoney' then 'decimal'
when DATA_TYPE = 'text' then 'string'
when DATA_TYPE = 'time' then 'TimeSpan'
when DATA_TYPE = 'timestamp' then 'DateTime'
when DATA_TYPE = 'tinyint' then 'byte'
when DATA_TYPE = 'uniqueidentifier' then 'Guid'
when DATA_TYPE = 'varbinary' then 'byte[]'
when DATA_TYPE = 'varchar' then 'string'
else '_UNKNOWN_'
end, ' ',
COLUMN_NAME, ' {get; set;}') as property_
FROM INFORMATION_SCHEMA.COLUMNS
WHERE table_name = @table AND table_schema = @schema) a
;
谢谢Alex和Guilherme!
