Question

我有一个自定义类，我想转换为JSON，但我在这里发现了一个奇怪的错误：

Exception in thread "main" scala.MatchError: (23,com.xxx.dts.dq.common.utils.DQOpsStoreProfileStatus@5f275ae4) (of class scala.Tuple2)

代码在这里：

implicit val formats = org.json4s.DefaultFormats
val A = Serialization.write(resultsMap)
println(A)

现在，如果我做一个foreach：

 resultsMap.foreach(x => println(Serialization.write(x)))

我得到了一些结果，但看起来不正确：

{"_1":23,"_2":{}}
{"_1":32,"_2":{}}

元组缺少核心信息。我假设因为我们使用的自定义类导致某种问题？它有什么办法吗？

如果我要拉出地图的第二个元素并将其转换为JSON，它将如下所示：

{"errorCode":null,"id":null,"fieldType":"STRING","fieldIndex":0,"datasetFieldName":"RECORD_ID","datasetFieldSum":0.0,"datasetFieldMin":0.0,"datasetFieldMax":0.0,"datasetFieldMean":0.0,"datasetFieldSigma":0.0,"datasetFieldNullCount":0.0,"datasetFieldObsCount":0.0,"datasetFieldKurtosis":0.0,"datasetFieldSkewness":0.0,"frequencyDistribution":"(D,4488)","runStatusId":null,"lakeHdfsPath":"/user/jvy234/20140817_011500_zoot_kohls_offer_init.dat"}

另外一方面注意，这个类是用java编写的，如果那可能是罪魁祸首？

完整堆栈跟踪：

Exception in thread "main" scala.MatchError: (0,com.xxx.dts.dq.common.utils.DQOpsStoreProfileStatus@315a29f4) (of class scala.Tuple2)
    at org.json4s.Extraction$.internalDecomposeWithBuilder(Extraction.scala:132)
    at org.json4s.Extraction$.decomposeWithBuilder(Extraction.scala:67)
    at org.json4s.Extraction$.decompose(Extraction.scala:194)
    at org.json4s.jackson.Serialization$.write(Serialization.scala:22)
    at com.xxx.dts.toolset.jsonWrite$.jsonClob(jsonWrite.scala:16)
    at com.xxx.dts.dq.profiling.DQProfilingEngine.profile(DQProfilingEngine.scala:255)
    at com.xxx.dts.dq.profiling.Profiler$.main(DQProfilingEngine.scala:64)
    at com.xxx.dts.dq.profiling.Profiler.main(DQProfilingEngine.scala)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:606)
    at org.apache.spark.deploy.SparkSubmit$.org$apache$spark$deploy$SparkSubmit$$runMain(SparkSubmit.scala:569)
    at org.apache.spark.deploy.SparkSubmit$.doRunMain$1(SparkSubmit.scala:166)
    at org.apache.spark.deploy.SparkSubmit$.submit(SparkSubmit.scala:189)
    at org.apache.spark.deploy.SparkSubmit$.main(SparkSubmit.scala:110)
    at org.apache.spark.deploy.SparkSubmit.main(SparkSubmit.scala)

Answer 1

我认为你只有两种方式：

为tuple2编写序列化

或
将其转换为列表地图，例如：resultsMap.map(Map(_)).foreach(...)

<强>更新对于序列化，您可以使用以下内容：

class Tuple2Serializer extends CustomSerializer[(String, Int)](   format => (
    {
      case JObject(JField(k, JInt(v))) => (k, v)
    },
    {
      case (s: String, t: Int) => (s -> t)
    }   ) )

implicit val formats = org.json4s.DefaultFormats + new Tuple2Serializer

json4s scala.MatchError（类scala.Tuple2）

1 个答案: