[英]Scala Salat Deserialization: how to get a Map[String, Number]?
My database looks like 我的数据库看起来像
[
{
name: "domenic",
records: {
today: 5,
yesterday: 1.5
}
},
{
name: "bob",
records: { ... }
}
]
When I try queries like 当我尝试查询
val result: Option[DBObject] = myCollection.findOne(
MongoDBObject("name" -> "domenic")
MongoDBObject("records" -> 1),
)
val records = result.get.getAs[BasicDBObject]("records").get
grater[Map[String, Number]].asObject(records)
it fails (at runtime!) with 它失败(在运行时!)与
GRATER GLITCH - unable to find or instantiate a grater using supplied path name
REASON: Class scala.collection.immutable.Map is an interface
Context: 'global'
Path from pickled Scala sig: 'scala.collection.immutable.Map'
I think I could make this work by creating a case class whose only field is a Map[String, Number]
and then getting its property. 我想我可以通过创建一个case类(其唯一的字段是一个
Map[String, Number]
,然后获取其属性来完成这项工作。 Is that really necessary? 那真的有必要吗?
grater
doesn't take a collection as a type argument, only a case class or a trait/abstract class whose concrete representations are case classes. grater
不会将集合作为类型参数,而只会将其具体表示形式为case类的case类或trait / abstract类作为参数。 Since you're just querying for a map, just extract the values you need out of the DBObject
using getAs[T]
. 由于您只是在查询地图,因此只需使用
getAs[T]
从DBObject
提取所需的值getAs[T]
。
Number
may not be a supported type in Salat - I've certainly never tried it. Number
可能无法在礼拜支持的类型-我当然从来没有尝试过。 If you need Number
you can write a custom transformer or send a pull request to add real support to Salat. 如果您需要
Number
,则可以编写自定义转换器或发送拉动请求以向Salat添加真正的支持。
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