使用 entityframework 核心检索 json

由于 用于 json 路径.

你如何使用 entity-framework-core 来使用它们?

How do you consume them with entity-framework-core?

以下不起作用:

var foo = _db.Set<JObject>()
             .FromSql("dbo.Mine @customerid = {0}", _user.guid)
             .FirstOrDefault();

因为 JObject 类型不是模型的一部分:

Because JObject type is not part of the model:

InvalidOperationException: Cannot create a DbSet for 'JObject' because this type is not included in the model for the context.

但是我们应该如何使用 entity-framework-core 来做到这一点?

But how are we supposed to do that with entity-framework-core?

推荐答案

在谷歌上搜索了一段时间后,我了解到尚不支持.如果上下文中没有模型,则无法使用 entityframework 检索数据,请指向:https://docs.microsoft.com/en-us/ef/core/querying/raw-sql 和 https://github.com/aspnet/EntityFrameworkCore/issues/1862

After searching on google for a while I understood is not supported yet. If you don't have a model in the context you can't retrieve data with entityframework, point: https://docs.microsoft.com/en-us/ef/core/querying/raw-sql and https://github.com/aspnet/EntityFrameworkCore/issues/1862

我决定用旧的方式来做:

I resolved doing it the old way:

  var jsonResult = new System.Text.StringBuilder();
  /*"using" would be bad, we should leave the connection open*/
  var connection = _db.Database.GetDbConnection() as SqlConnection;
  {
    await connection.OpenAsync();

    using (SqlCommand cmd = new SqlCommand(
        "Mine",
        connection))
    {
      cmd.CommandType = CommandType.StoredProcedure;

      cmd.Parameters.Add("@customerid", SqlDbType.NVarChar).Value = _user.guid;

      using (SqlDataReader reader = await cmd.ExecuteReaderAsync())
      {
        if (!reader.HasRows)
        {
          jsonResult.Append("[]");
        }
        else
        {
          while (reader.Read())
          {
            jsonResult.Append(reader.GetValue(0).ToString());
          }
        }
      }
    }
  }
  var raw = JArray.Parse(jsonResult.ToString());

  var ret = raw.ToObject<List<SiteData>>();

我怀疑明确关闭连接是否更好.

I am in doubt if it's better to explicitly close the connection or not.

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