如果你有一个动态变量，例如DapperRow，你可以先构建一个ExpandoObject，然后将Expando转换为一个IDictionary<string对象>。从那时起，可以通过属性名获取值。

帮助方法ToExpandoObject:

public static ExpandoObject ToExpandoObject(object value)
    {
        IDictionary<string, object> dapperRowProperties = value as IDictionary<string, object>;
        IDictionary<string, object> expando = new ExpandoObject();
        if (dapperRowProperties == null)
        {
            return expando as ExpandoObject;
        }
        foreach (KeyValuePair<string, object> property in dapperRowProperties)
        {
            if (!expando.ContainsKey(property.Key))
            {
                expando.Add(property.Key, property.Value);
            }
            else
            {
                //prefix the colliding key with a random guid suffixed 
                expando.Add(property.Key + Guid.NewGuid().ToString("N"), property.Value);
            } 
        }
        return expando as ExpandoObject;
    }

示例用法，我用粗体标记了赋予我们访问权限的类型转换，我用**字母标记了重要的位:

  using (var transactionScope = new TransactionScope(TransactionScopeAsyncFlowOption.Enabled))
        {
            foreach (var dynamicParametersForItem in dynamicParametersForItems)
            {
                var idsAfterInsertion = (await connection.QueryAsync<object>(sql, dynamicParametersForItem)).ToList();
                if (idsAfterInsertion != null && idsAfterInsertion.Any())
                {
                    **var idAfterInsertionDict = (IDictionary<string, object>) ToExpandoObject(idsAfterInsertion.First());**
                    string firstColumnKey = columnKeys.Select(c => c.Key).First();
                    **object idAfterInsertionValue = idAfterInsertionDict[firstColumnKey];**
                    addedIds.Add(idAfterInsertionValue); //we do not support compound keys, only items with one key column. Perhaps later versions will return multiple ids per inserted row for compound keys, this must be tested.
                }
            }
        }

在我的示例中，我在动态对象DapperRow中查找一个属性值，并首先将Dapper行转换为ExpandoObject，并将其转换为一个字典属性包，如本文的其他回答中所示和提到的那样。

我的示例代码是我正在工作的Dapper扩展的InsertMany方法，我想在批量插入后抓住这里的多个id。

为一个适用于任何类型的属性(包括dynamic和ExpandoObject)获取setter和getter的最简单方法是使用FastMember，这也是最快的方法(它使用Emit)。

您既可以获得基于给定类型的TypeAccessor，也可以获得基于给定类型实例的ObjectAccessor。

例子:

var staticData = new Test { Id = 1, Name = "France" };
var objAccessor = ObjectAccessor.Create(staticData);
objAccessor["Id"].Should().Be(1);
objAccessor["Name"].Should().Be("France");

var anonymous = new { Id = 2, Name = "Hilton" };
objAccessor = ObjectAccessor.Create(anonymous);
objAccessor["Id"].Should().Be(2);
objAccessor["Name"].Should().Be("Hilton");

dynamic expando = new ExpandoObject();
expando.Id = 3;
expando.Name = "Monica";
objAccessor = ObjectAccessor.Create(expando);
objAccessor["Id"].Should().Be(3);
objAccessor["Name"].Should().Be("Monica");

var typeAccessor = TypeAccessor.Create(staticData.GetType());
typeAccessor[staticData, "Id"].Should().Be(1);
typeAccessor[staticData, "Name"].Should().Be("France");

typeAccessor = TypeAccessor.Create(anonymous.GetType());
typeAccessor[anonymous, "Id"].Should().Be(2);
typeAccessor[anonymous, "Name"].Should().Be("Hilton");

typeAccessor = TypeAccessor.Create(expando.GetType());
((int)typeAccessor[expando, "Id"]).Should().Be(3);
((string)typeAccessor[expando, "Name"]).Should().Be("Monica");

如果你有一个动态变量，例如DapperRow，你可以先构建一个ExpandoObject，然后将Expando转换为一个IDictionary<string对象>。从那时起，可以通过属性名获取值。

帮助方法ToExpandoObject:

public static ExpandoObject ToExpandoObject(object value)
    {
        IDictionary<string, object> dapperRowProperties = value as IDictionary<string, object>;
        IDictionary<string, object> expando = new ExpandoObject();
        if (dapperRowProperties == null)
        {
            return expando as ExpandoObject;
        }
        foreach (KeyValuePair<string, object> property in dapperRowProperties)
        {
            if (!expando.ContainsKey(property.Key))
            {
                expando.Add(property.Key, property.Value);
            }
            else
            {
                //prefix the colliding key with a random guid suffixed 
                expando.Add(property.Key + Guid.NewGuid().ToString("N"), property.Value);
            } 
        }
        return expando as ExpandoObject;
    }

示例用法，我用粗体标记了赋予我们访问权限的类型转换，我用**字母标记了重要的位:

  using (var transactionScope = new TransactionScope(TransactionScopeAsyncFlowOption.Enabled))
        {
            foreach (var dynamicParametersForItem in dynamicParametersForItems)
            {
                var idsAfterInsertion = (await connection.QueryAsync<object>(sql, dynamicParametersForItem)).ToList();
                if (idsAfterInsertion != null && idsAfterInsertion.Any())
                {
                    **var idAfterInsertionDict = (IDictionary<string, object>) ToExpandoObject(idsAfterInsertion.First());**
                    string firstColumnKey = columnKeys.Select(c => c.Key).First();
                    **object idAfterInsertionValue = idAfterInsertionDict[firstColumnKey];**
                    addedIds.Add(idAfterInsertionValue); //we do not support compound keys, only items with one key column. Perhaps later versions will return multiple ids per inserted row for compound keys, this must be tested.
                }
            }
        }

在我的示例中，我在动态对象DapperRow中查找一个属性值，并首先将Dapper行转换为ExpandoObject，并将其转换为一个字典属性包，如本文的其他回答中所示和提到的那样。

我的示例代码是我正在工作的Dapper扩展的InsertMany方法，我想在批量插入后抓住这里的多个id。

一些解决方案不是工作的valuekind对象，我从一个json字符串获得的，也许是因为我没有一个具体的类型在我的代码，类似于对象，我将从json字符串获得，所以我是如何去做的

JsonElement myObject = System.Text.Json.JsonSerializer.Deserialize<JsonElement>(jsonStringRepresentationOfMyObject);

/*In this case I know that there is a property with 
the name Code, otherwise use TryGetProperty. This will 
still return a JsonElement*/

JsonElement propertyCode = myObject.GetProperty("Code");
        
/*Now with the JsonElement that represents the property, 
you can use several methods to retrieve the actual value, 
in this case I know that the value in the property is a string, 
so I use the GetString method on the object. If I knew the value 
was a double, then I would use the GetDouble() method on the object*/

string code = propertyCode.GetString();

这对我很有用

GetProperty/GetValue对Json数据不起作用，它总是生成一个空异常，但是，你可以尝试这种方法:

使用JsonConvert序列化你的对象:

var z = Newtonsoft.Json.JsonConvert.DeserializeObject(Convert.ToString(request));

然后直接访问它，将其转换回string:

var pn = (string)z["DynamicFieldName"];

它可以直接应用Convert.ToString(request)["DynamicFieldName"]，但我还没有测试。

一旦你有了你的PropertyInfo(从GetProperty)，你需要调用GetValue并传递你想要从中获取值的实例。在你的情况下:

d.GetType().GetProperty("value2").GetValue(d, null);