我试图在下面的示例中排除_passthroughFields属性。当我使用调试器时,看起来我的PropertyFilter
从未使用过。我做错了什么?
import java.util
import com.fasterxml.jackson.databind.ser.impl.SimpleBeanPropertyFilter.SerializeExceptFilter
import com.fasterxml.jackson.databind.ser.impl.SimpleFilterProvider
import com.fasterxml.jackson.databind.{DeserializationFeature, ObjectMapper, ObjectWriter}
import com.fasterxml.jackson.module.scala.DefaultScalaModule
import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper
import org.scalatest.{Matchers, WordSpec}
import scala.collection.immutable.Map
class PassthroughFieldsSpec extends WordSpec with Matchers {
"JacksonParser" when {
"given an Object and undesired fields" should {
"not include those fields in the json response" in {
trait Foo {
def id: String
def _passthroughFields: Map[String, String] = Map.empty
}
class Bar(val id: String, override val _passthroughFields: Map[String, String]) extends Foo
val item = new Bar("abcd", Map.empty)
val mapper = new ObjectMapper() with ScalaObjectMapper
mapper.registerModule(DefaultScalaModule)
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
val excludes = new util.HashSet[String](1)
excludes.add("_passthroughFields")
excludes.add("get_passthroughFields")
val filters = new SimpleFilterProvider()
.addFilter("filter properties by name", new SerializeExceptFilter(excludes))
val writer: ObjectWriter = mapper.writer(filters)
val json = writer.writeValueAsString(item)
json.contains("_passthroughFields") shouldBe false
}
}
}
}
答案 0 :(得分:1)
我认为你可以使用像@JsonIgnore这样的东西来排除它,或者使场成为瞬态。
否则,如果您需要在案例类代码之外定义它(如您的示例中所示),则可以使用Genson执行此操作。
import com.owlike.genson._
import com.owlike.genson.reflect.VisibilityFilter
// Note that if you use case classes you don't need the visibility filter stuff
// it is used to indicate that private fields should be ser/de
val genson = new GensonBuilder()
.withBundle(ScalaBundle())
.useFields(true, VisibilityFilter.PRIVATE)
.exclude("_passthroughFields")
.create()
genson.toJson(item)
免责声明:我是Gensons的作者。