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我想将字符串列表作为 json 存储到 mysql 表中。我在 pomelo entityframework 中看到了对此的支持。我跟着这个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_context.Database.EnsureDeleted();被调用时，它会给出以下错误

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、未密封的、具有虚拟导航属性并具有公共或受保护的构造函数。

但是我必须添加虚拟关键字不是导航属性，而是一列。不知道我在这里错过了什么。

4

1 回答 1

1

看看下面的示例代码，它取自我在我们的 GitHub 存储库上的帖子，并且没有问题：

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`

仔细看看这个IceCream.Comments属性，这正是你想要的。

在下面的同一个 GitHub 问题上，您可以找到我的另一篇文章，其中包含一个更复杂的示例。

此外，我们接下来将为 Pomelo 实现完整的 JSON 支持（可能在一周内）。

于 2020-06-10T14:11:46.290 回答