如何使用Pomelo.EntityFramework的JsonObject [英] How to use JsonObject of Pomelo.EntityFramework
问题描述
我想将字符串列表存储为mysql表作为json.我看到柚子实体框架对此提供了支持.我遵循了这个 https://libraries.io/github/tuanbs/Pomelo.EntityFrameworkCore.MySql
I want to store list of string into mysql table as json. I saw there is support for this in pomelo entityframework. I followed this https://libraries.io/github/tuanbs/Pomelo.EntityFrameworkCore.MySql
这是我的实体
public class Project
{
public int Id {get;set;}
public string Title {get;set;}
public JsonObject<List<string>> Tags {get;set;}
}
但是，当调用 _context.Database.EnsureDeleted(); 时，它给出以下错误
But when _context.Database.EnsureDeleted(); is called it gives below error
实体类型项目"上的导航属性标签"不是虚拟的.UseLazyLoadingProxies要求所有实体类型都是公共的，未密封，具有虚拟导航属性，并具有公共场所或受保护的构造函数.
Navigation property 'Tags' on entity type 'Project' is not virtual. UseLazyLoadingProxies requires all entity types to be public, unsealed, have virtual navigation properties, and have a public or protected constructor.
但这不是导航属性，我必须为其添加虚拟关键字，而是一列.不知道我在这里想念什么.
But it is not navigation property that I have to add virtual keyword with it but is a column. Don't know what am I missing here.
推荐答案
Take a look at the following sample code, that is taken from my post on our GitHub repository, and works without issues:
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using Microsoft.EntityFrameworkCore;
using Microsoft.Extensions.Logging;
using Pomelo.EntityFrameworkCore.MySql.Storage;
namespace IssueConsoleTemplate
{
public class IceCream
{
public int IceCreamId { get; set; }
public string Name { get; set; }
public JsonObject<Energy> Energy { get; set; }
public JsonObject<List<string>> Comments { get; set; }
}
public class Energy
{
public double Kilojoules { get; set; }
public double Kilocalories { get; set; }
}
public class Context : DbContext
{
public virtual DbSet<IceCream> IceCreams { get; set; }
protected override void OnConfiguring(DbContextOptionsBuilder optionsBuilder)
{
optionsBuilder
.UseMySql("server=127.0.0.1;port=3306;user=root;password=;database=So62301095",
b => b.ServerVersion(new ServerVersion("8.0.20-mysql")))
.UseLoggerFactory(LoggerFactory.Create(b => b
.AddConsole()
.AddFilter(level => level >= LogLevel.Information)))
.EnableSensitiveDataLogging()
.EnableDetailedErrors();
}
}
internal class Program
{
private static void Main()
{
using (var context = new Context())
{
context.Database.EnsureDeleted();
context.Database.EnsureCreated();
context.IceCreams.AddRange(
new IceCream
{
Name = "Vanilla",
Energy = new Energy
{
Kilojoules = 866.0,
Kilocalories = 207.0
},
Comments = new List<string>
{
"First!",
"Delicious!"
}
},
new IceCream
{
Name = "Chocolate",
Energy = new Energy
{
Kilojoules = 904.0,
Kilocalories = 216.0
},
Comments = new List<string>
{
"My husband likes this one a lot."
}
});
context.SaveChanges();
}
using (var context = new Context())
{
var result = context.IceCreams
.OrderBy(e => e.IceCreamId)
.ToList();
Debug.Assert(result.Count == 2);
Debug.Assert(result[0].Name == "Vanilla");
Debug.Assert(result[0].Energy.Object.Kilojoules == 866.0);
Debug.Assert(result[0].Comments.Object.Count == 2);
Debug.Assert(result[0].Comments.Object[0] == "First!");
}
}
}
}
它生成以下SQL:
info: Microsoft.EntityFrameworkCore.Infrastructure[10403]
Entity Framework Core 3.1.3 initialized 'Context' using provider 'Pomelo.EntityFrameworkCore.MySql' with options: ServerVersion 8.0.20 MySql SensitiveDataLoggingEnabled DetailedErrorsEnabled
info: Microsoft.EntityFrameworkCore.Database.Command[20101]
Executed DbCommand (81ms) [Parameters=[], CommandType='Text', CommandTimeout='30']
DROP DATABASE `So62301095`;
info: Microsoft.EntityFrameworkCore.Database.Command[20101]
Executed DbCommand (12ms) [Parameters=[], CommandType='Text', CommandTimeout='30']
CREATE DATABASE `So62301095`;
info: Microsoft.EntityFrameworkCore.Database.Command[20101]
Executed DbCommand (66ms) [Parameters=[], CommandType='Text', CommandTimeout='30']
CREATE TABLE `IceCreams` (
`IceCreamId` int NOT NULL AUTO_INCREMENT,
`Name` longtext CHARACTER SET utf8mb4 NULL,
`Energy` json NULL,
`Comments` json NULL,
CONSTRAINT `PK_IceCreams` PRIMARY KEY (`IceCreamId`)
);
info: Microsoft.EntityFrameworkCore.Database.Command[20101]
Executed DbCommand (15ms) [Parameters=[@p0='["First!","Delicious!"]', @p1='{"Kilojoules":866.0,"Kilocalories":207.0}', @p2='Vanilla' (Size = 4000)], CommandType='Text', CommandTimeout='30']
INSERT INTO `IceCreams` (`Comments`, `Energy`, `Name`)
VALUES (@p0, @p1, @p2);
SELECT `IceCreamId`
FROM `IceCreams`
WHERE ROW_COUNT() = 1 AND `IceCreamId` = LAST_INSERT_ID();
info: Microsoft.EntityFrameworkCore.Database.Command[20101]
Executed DbCommand (1ms) [Parameters=[@p0='["My husband likes this one a lot."]', @p1='{"Kilojoules":904.0,"Kilocalories":216.0}', @p2='Chocolate' (Size = 4000)], CommandType='Text', CommandTimeout='30']
INSERT INTO `IceCreams` (`Comments`, `Energy`, `Name`)
VALUES (@p0, @p1, @p2);
SELECT `IceCreamId`
FROM `IceCreams`
WHERE ROW_COUNT() = 1 AND `IceCreamId` = LAST_INSERT_ID();
info: Microsoft.EntityFrameworkCore.Database.Command[20101]
Executed DbCommand (1ms) [Parameters=[], CommandType='Text', CommandTimeout='30']
SELECT `i`.`IceCreamId`, `i`.`Comments`, `i`.`Energy`, `i`.`Name`
FROM `IceCreams` AS `i`
ORDER BY `i`.`IceCreamId`
仔细查看 IceCream.Comments 属性，该属性正是您想要的.
Take a close look at the IceCream.Comments property, that does exactly what you want.
在下面的同一GitHub问题上，您找到另一个由我发布，并提供了更复杂的示例.
On the same GitHub issue further below, you find another post by me, with a much more sophisticated example.
此外，接下来我们将为Pomelo实现完整的JSON支持(可能在一周之内).
Also, we are going to implement full JSON support next for Pomelo (probably within a week).
这篇关于如何使用Pomelo.EntityFramework的JsonObject的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
