源对象(JSON,如果重要的话,使用JSON.NET):
{
"conum" : 1001,
"name" : "CLN Industries Corporation",
"agencyName" : "Murphy, Holmes & Associates, LLC",
"sAA" : [{
"code" : 247,
"description" : "Mechanic\u0027s lien - Bond to Discharge - Fixed penalty - where principal has posted Performance and Pa"
}, {
"code" : 277,
"description" : "Mechanic\u0027s lien - Bond to Discharge - Open Penalty - where principal has posted Performance and Paym"
}, {
"code" : 505,
"description" : "Indemnity Bonds - Contractor\u0027s Indemnity Against Damages where there is a performance bond and addit"
}
]
}
目标对象(C#):
public class CorporateRatesInfo
{
public string Conum { get; set; }
public string Name { get; set; }
public string AgencyName { get; set; }
public List<SaaCode> SaaCodes { get; set; }
}
public class SaaCode
{
public string Code { get; set; }
public string Description { get; set; }
}
Automapper配置和映射:
var config = new MapperConfiguration(cfg =>
{
cfg.CreateMap<JObject, CorporateRatesInfo>();
cfg.AddProfile<CorporateRatesProfile>();
});
//config.AssertConfigurationIsValid();
_mapper = config.CreateMapper();
public class CorporateRatesProfile : Profile
{
protected override void Configure()
{
CreateMap<JObject, CorporateRatesInfo>()
.ForMember("SaaCodes", cfg => { cfg.MapFrom(jo => jo["sAA"]); })
.ForMember("Conum", cfg => { cfg.MapFrom(jo => jo["conum"]); })
.ForMember("Name", cfg => { cfg.MapFrom(jo => jo["name"]); })
.ForMember("AgencyName", cfg => { cfg.MapFrom(jo => jo["agencyName"]); });
}
}
除了SaaCodes转换之外,其他一切都有效,其中Automapper将每个条目转换为空的SaaCode对象(所有属性都设置为null)。
我如何/在何处告诉automapper如何将该JSON字段中的项目转换为其目标类型?
答案 0 :(得分:5)
试试这个 -
public class CorporateRatesInfo
{
public string Conum { get; set; }
public string Name { get; set; }
public string AgencyName { get; set; }
public List<SaaCode> SaaCodes { get; set; }
}
public class SaaCode
{
public string Code { get; set; }
public string Description { get; set; }
}
public class CorporateRatesProfile : Profile
{
protected override void Configure()
{
CreateMap<JObject, SaaCode>()
.ForMember("Code", cfg => { cfg.MapFrom(jo => jo["code"]); })
.ForMember("Description", cfg => { cfg.MapFrom(jo => jo["description"]); });
CreateMap<JObject, CorporateRatesInfo>()
.ForMember("SaaCodes", cfg => { cfg.MapFrom(jo => jo["sAA"]); })
.ForMember("Conum", cfg => { cfg.MapFrom(jo => jo["conum"]); })
.ForMember("Name", cfg => { cfg.MapFrom(jo => jo["name"]); })
.ForMember("AgencyName", cfg => { cfg.MapFrom(jo => jo["agencyName"]); });
}
}
class Program
{
static void Main(string[] args)
{
var config = new MapperConfiguration(cfg =>
{
cfg.CreateMap<JObject, CorporateRatesInfo>();
cfg.AddProfile<CorporateRatesProfile>();
});
//config.AssertConfigurationIsValid();
var mapper = config.CreateMapper();
var jsonText = @"
{
""conum"" : 1001,
""name"" : ""CLN Industries Corporation"",
""agencyName"" : ""Murphy, Holmes & Associates, LLC"",
""sAA"" : [{
""code"" : 247,
""description"" : ""Mechanic\u0027s lien - Bond to Discharge - Fixed penalty - where principal has posted Performance and Pa""
}, {
""code"" : 277,
""description"" : ""Mechanic\u0027s lien - Bond to Discharge - Open Penalty - where principal has posted Performance and Paym""
}, {
""code"" : 505,
""description"" : ""Indemnity Bonds - Contractor\u0027s Indemnity Against Damages where there is a performance bond and addit""
}
]
}
";
var jsonoObj = JObject.Parse(jsonText);
CorporateRatesInfo dto = mapper.Map<CorporateRatesInfo>(jsonoObj);
}
}
答案 1 :(得分:0)
不幸的是,我在新版本的AutoMapper中关注了Bookamp's question,并努力使它工作。
在以下source
中找到了我发现其他问题的其他方法示例数组
{
'Items': [{
'Id': '1',
'EmployeeNo': '1001',
'FirstName': 'Rajat',
'LastName': 'Kumar',
'Gender': {
'Id': '1',
'ShortName': 'M',
'FullName': 'Male'
}
}, {
'Id': '2',
'EmployeeNo': '1003',
'FirstName': 'Lisa',
'LastName': 'David',
'Gender': {
'Id': '2',
'ShortName': 'F',
'FullName': 'Female'
}
}]
}
配置文件配置示例
CreateMap<JObject, List<User>>().ConvertUsing<JObjectToUserListConverter>();
var employeeMap = CreateMap<JToken, User>();
employeeMap.ForMember(x => x.Id, y => y.MapFrom(j => j.SelectToken(".Id")));
employeeMap.ForMember(x => x.DisplayName, y => y.MapFrom(j => j.SelectToken(".LastName").ToString() + ", " + j.SelectToken(".FirstName").ToString()));
employeeMap.ForMember(x => x.Gender, y => y.MapFrom(j => j.SelectToken(".Gender.FullName")));
employeeMap.ForMember(x => x.Login, y => y.MapFrom(j => j.SelectToken(".Login")));
TypeConverter
public class JObjectToUserListConverter : ITypeConverter<JObject, List<User>>
{
public List<User> Convert(JObject source, List<User> destination, ResolutionContext context)
{
var userList = new List<User>();
var tokenCountItems = source.SelectTokens("Items").Children().Count();
for (int i = 0; i < tokenCountItems; i++)
{
var token = source.SelectToken($"Items[{i}]");
var result = new User();
if (token != null)
{
result = context.Mapper.Map<JToken, User>(token);
}
userList.Add(result);
}
}
return userList;
}
}
答案 2 :(得分:0)
一份报告,我写了一个简单的解决方案,如果字段名称匹配,则可以在mapper扩展帮助器方法中使用Newtonsoft的反序列化器。我写了一个通用的ResolveJson扩展方法。
public static class MapperExtensions
{
public static T ResolveJson<T>(this JObject jobj, string target)
{
return JsonConvert.DeserializeObject<T>(jobj.SelectToken(target).ToString());
}
}
您可以使用以下
进行调用CreateMap<JObject, CorporateRatesInfo>()
.ForMember(x => x.SaaCodes,m => { m.MapFrom(s => s.ResolveJson<SaaCode>("sAA"));});