【发布时间】:2016-01-05 15:41:50
【问题描述】:
我正在使用 Play JSON API(最新版本;Play 2.4),将传入的 JSON 读取到对象中。
编写 JSON 时,使用自定义对象列表绝对没有问题,只要我有implicit val writes = Json.writes[CustomType]。
但显然反之亦然,因为即使为顶级类型和列表项类型(使用Json.reads[Incoming] 和Json.reads[Item])生成了Reads,以下内容也不起作用。自定义Reads 实施是强制性的吗?还是我错过了一些明显的东西?完成这项工作的最简单方法是什么?
简化示例:
JSON:
{
"test": "...",
"items": [
{ "id": 44, "time": "2015-11-20T11:04:03.544" },
{ "id": 45, "time": "2015-11-20T11:10:10.101" }
]
}
与传入数据匹配的模型/DTO:
import play.api.libs.json.Json
case class Incoming(test: String, items: List[Item])
object Incoming {
implicit val reads = Json.reads[Incoming]
}
case class Item(id: Long, time: String)
object Item {
implicit val reads = Json.reads[Item]
}
控制器:
def test() = Action(parse.json) { request =>
request.body.validate[Incoming].map(incoming => {
// ... handle valid incoming data ...
}).getOrElse(BadRequest)
}
编译器有这样的说法:
No implicit format for List[models.Item] available.
[error] implicit val reads = Json.reads[Incoming]
^
No Json deserializer found for type models.Incoming.
Try to implement an implicit Reads or Format for this type.
[error] request.body.validate[Incoming].map(incoming => {
【问题讨论】:
标签: json scala playframework playframework-2.4