昨天(星期五)下班,19:00左右回到家,洗个澡,然后20:30左右开始写代码,写完代码之后,上床看了《生活大爆炸10季》17、18两集,发现没有更新到19集,瞄了一眼手机,竟然已经是凌晨02:00多了,关掉电视睡觉,10:30左右被老婆电话吵醒,洗漱完毕,去麦当劳吃了一个早餐,然后屁颠屁颠地坐地铁到很远的地方去爬山。爬山回来之后,闲来无事,写篇文章记录一下昨晚所花的几个小时干的事情——使用EntityFrameworkCore实现Repository
, UnitOfWork
,支持MySQL分库分表。
由于是使用业余时间写来玩的,时间也有限,所以,全部代码做了一个基本假设:Repository
, UnitOfWork
只支持同一个IP上的MySQL分库分表,不同IP上的MySQL分库分表,需要使用不同的Repository
, UnitOfWork
对象。以下示例代码,假设数据库是按年分库按月分表。
EntityFrameworkCore
默认并不支持分库分表,我们看一眼EntityFrameworkCore
默认生成的SQL:
Executed DbCommand [Parameters=[@p2='?', @p4='?' (Size = 8000), @p6='?' (Size = 8000)], CommandType='Text', CommandTimeout='0']INSERT INTO `t_user_201703` (`Fis_deleted`, `Fpassword`, `Fname`)VALUES (@p2, @p4, @p6);SELECT LAST_INSERT_ID();
默认生成的SQL并没有带上库名,而想要让EntityFrameworkCore
支持MySQL分库分表,首要条件是必须能做到可以动态地改变库名表名。软件界有一句老话叫:凡是做不到的就多抽象一层,所以,想要让EntityFrameworkCore
支持MySQL分库分表,我抽象了以下两个接口, IRepository
和IUnitOfWork
很多人都自己动手实现过Repository
和UnitOfWork
,虽然各自实现不尽相同,但是其实现本身并没有难度,但在这里,我们需要特别关注两个方法:void ChangeTable(string table)
和void ChangeDatabase(string database)
/// /// Changes the table name. This require the tables in the same database. /// /// /// /// This only been used for supporting multiple tables in the same model. This require the tables in the same database. ///
void ChangeTable(string table);
/// /// Changes the database name. This require the databases in the same machine. /// /// The database name. /// /// This only been used for supporting multiple databases in the same model. This require the databases in the same machine. ///
void ChangeDatabase(string database);
怎么实现这两个方法,就需要一定的技术功底了,我以前在一家创业公司的时候,因为看不惯架构师自以为是的样子,自己动手写了一个轻量级的ORM框架,如果以后有时间,我打算写一篇《如何基于Dapper实现一个轻量级的ORM框架》的文章。ORM框架背后的动机很单纯,就是数据库与Domain之间的一种双向映射,真正把这种单纯的动机搞复杂是的那些性能优化,各种缓存实现。而从Domain到数据库这一单方向上的映射,在.NET领域借助了一种代码即数据的思想,再细化到C#语言代码即数据就是表达式树。所以,我们有理由相信:SQL是根据表达式树生成的。现在我们已经找准了方向,那么我们看看EntityFrameworkCore
在什么地方生成表名的,也就是说,我们只需要修改一下生成表名的代码,就可以做到动态生成database
.table
SQL。EntityFrameworkCore
是通过TableExpression
来生成表名的:
public class TableExpression{
public virtual string Table { get; }
public virtual string Schema { get; }}
如果你MySQL知识至少跟我一样的水平的话,看到TableExpression
表达式有一个Schema
是不是立即就可以想到:哈哈,太好了,我压根就不用修改EntityFrameworkCore
本身的代码就可以实现。为什么呢?好吧,看看MySQL官网怎么说Schema的:
In MySQL, physically, a schema is synonymous with a database. You can substitute the keyword SCHEMA instead of DATABASE in MySQL SQL syntax, for example using CREATE SCHEMA instead of CREATE DATABASE. Some other database products draw a distinction. For example, in the Oracle Database product, a schema represents only a part of a database: the tables and other objects owned by a single user.
好吧,Schema就是Database,那么我们就用Schema
.Table
来表示database
.table
。现在事情就变得简单了,变成了我们如何动态地改变Schema
和Table
了,以下是我提供的简化实现:
/// /// Changes the database name. This require the databases in the same machine. /// /// The database name. /// /// This only been used for supporting multiple databases in the same model. This require the databases in the same machine. /// public void ChangeDatabase(string database){
if (_context.Model.Relational() is RelationalModelAnnotations relational)
{
relational.DatabaseName = database;
}
var connection = _context.Database.GetDbConnection();
if (connection.State.HasFlag(ConnectionState.Open))
{
connection.ChangeDatabase(database);
}
var items = _context.Model.GetEntityTypes();
foreach (var item in items)
{
if (item.Relational() is RelationalEntityTypeAnnotations extensions)
{
extensions.Schema = database;
}
}}/// /// Changes the table name. This require the tables in the same database. /// /// /// /// This only been used for supporting multiple tables in the same model. This require the tables in the same database. /// public void ChangeTable(string table){
if (_dbContext.Model.FindEntityType(typeof(TEntity)).Relational() is RelationalEntityTypeAnnotations relational)
{
relational.TableName = table;
}}
OK, 虽然有点low,但是毕竟支持了MySQL分库分表,看看怎么用:
namespace QuickStart.Controllers{
[Route("api/[controller]")]
public class UserController : ApiController
{
private readonly IUnitOfWork _unitOfWork;
// 1. IRepositoryFactory used for readonly scenario; // 2. IUnitOfWork used for read/write scenario; // 3. IUnitOfWork used for multiple databases scenario;
public UserController(IUnitOfWork unitOfWork)
{
_unitOfWork = unitOfWork;
unitOfWork.ChangeDatabase($"rigofunc_{DateTime.Now.Year}");
var userRepo = unitOfWork.GetRepository<User>();
var postRepo = unitOfWork.GetRepository<Post>();
var ym = DateTime.Now.ToString("yyyyMM");
userRepo.ChangeTable($"t_user_{ym}");
postRepo.ChangeTable($"t_post_{ym}");
var user = new User
{
//UserId = 123, UserName = "rigofunc",
Password = "password"
};
userRepo.Insert(user);
var post = new Post
{
//PostId = 123, UserId = user.UserId,
Content = "What a piece of junk!"
};
postRepo.Insert(post);
unitOfWork.SaveChanges();
var find = userRepo.Find(user.UserId);
find.Password = "p@ssword";
unitOfWork.SaveChanges();
}
[HttpGet]
public IPagedList<User> Get()
{
_unitOfWork.ChangeDatabase($"rigofunc_2018");
var userRepo = _unitOfWork.GetRepository<User>();
return userRepo.Query(u => true).OrderBy(u => u.UserId).ToPagedList(0, 20);
}
}}
以下是生成的SQL:
Executed DbCommand [Parameters=[@p2='?', @p4='?' (Size = 8000), @p6='?' (Size = 8000)], CommandType='Text', CommandTimeout='0']
INSERT INTO `rigofunc_2017`.`t_user_201703` (`Fis_deleted`, `Fpassword`, `Fname`)
VALUES (@p2, @p4, @p6);
SELECT LAST_INSERT_ID();
Executed DbCommand [Parameters=[@p10='?' (Size = 8000), @p12='?', @p14='?'], CommandType='Text', CommandTimeout='0']
INSERT INTO `rigofunc_2017`.`t_post_201703` (`Fcontent`, `Fis_deleted`, `Fuser_id`)
VALUES (@p10, @p12, @p14);
SELECT LAST_INSERT_ID();
Executed DbCommand [Parameters=[@p0='?', @p3='?', @p4='?' (Size = 8000)], CommandType='Text', CommandTimeout='0']
UPDATE `rigofunc_2017`.`t_user_201703` SET `Fpassword` = @p4
WHERE `Fid` = @p0 AND `Fis_deleted` = @p3;
SELECT ROW_COUNT();
Executed DbCommand (1ms) [Parameters=[], CommandType='Text', CommandTimeout='0']
SELECT `u`.`Fid`, `u`.`Fis_deleted`, `u`.`Fpassword`, `u`.`Fname`
FROM `rigofunc_2017`.`t_user_201703` AS `u`
ORDER BY `u`.`Fid`
Executed DbCommand (2ms) [Parameters=[], CommandType='Text', CommandTimeout='0']
SELECT `u`.`Fid`, `u`.`Fis_deleted`, `u`.`Fpassword`, `u`.`Fname`
FROM `rigofunc_2018`.`t_user_201703` AS `u`
ORDER BY `u`.`Fid`
ABOUT ME:
I’m a software architect, particularly love .NET Core, but I also embrace all the new stuff. I’m on GitHub with xyting, and my packages publish on NuGet with rigofunc
原文地址:http://www.xyting.org/2017/03/25/EFCore-Support-MySQL-Multiple-Databases-And-Tables-sharding.html
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