Scala 解析 JSON 在一个函数中表现不同

Question

我正在尝试创建一个函数来解析我的记录，当我调用一个函数与对它进行硬编码时，我得到了两种不同的行为：

我正在使用：

import org.json4s.JsonAST.{JString, JField, JObject, JArray}
import org.json4s.jackson.JsonMethods._

val parsed = parse("""{"timestamp":"2016-06-02 13:40:16,772","tableName":"stg_mde_campaign_master","dbName":"stg_bankrtl_mde","owner":"hive","location":"null"}""")
      val output = for {
        JObject(child) <- parsed
        JField("timestamp", JString(subject1)) <- child
        JField("tableName", JString(obj1)) <- child 
        } yield (subject1,obj1)

会输出（我想要的）：

output: List[(String, String)] = List((2016-06-02 13:40:16,772,stg_mde_campaign_master)

但是当我将它转移到一个函数时，我得到：

    def getSubOb(record: String, subject:String, obj:String): List[(String, String)] = {
      val parsed = parse(record)
      val output: List[(String, String)] = for {
        JObject(child) <- parsed
        JField(subject, JString(subject1)) <- child
        JField(obj, JString(obj1)) <- child
    } yield (subject1, obj1)
      output
  }
val something = getSubOb("""{"timestamp":"2016-06-02 13:40:16,772","tableName":"stg_mde_campaign_master","dbName":"stg_bankrtl_mde","owner":"hive","location":"null"}""", "timestamp", "tableName")

输出行为很奇怪：

something: List[(String, String)] = List((2016-06-02 13:40:16,772,2016-06-02 13:40:16,772), (2016-06-02 13:40:16,772,stg_mde_campaign_master), (2016-06-02 13:40:16,772,stg_bankrtl_mde), (2016-06-02 13:40:16,772,hive), (2016-06-02 13:40:16,772,null), (stg_mde_campaign_master,2016-06-02 13:40:16,772), (stg_mde_campaign_master,stg_mde_campaign_master), (stg_mde_campaign_master,stg_bankrtl_mde), (stg_mde_campaign_master,hive), (stg_mde_campaign_master,null), (stg_bankrtl_mde,2016-06-02 13:40:16,772), (stg_bankrtl_mde,stg_mde_campaign_master), (stg_bankrtl_mde,stg_bankrtl_mde), (stg_bankrtl_mde,hive), (stg_bankrtl_mde,null), (hive,2016-06-02 13:40:16,772), (hive,stg_mde_campaign_master), (hive,stg_bankrtl_mde), (hive,hive), (hive,null), (null,2016-06-02 13:40:16,772), (null,stg_mde_campaign_...

Answer 1

您在 unapply 上有一个细微的错误。

模式匹配左侧的小写术语被视为变量。所以一切都匹配并绑定在那里。

您可以像在“变量名”中那样使用反引号来告诉 Scala 它不是要绑定的变量，而是匹配模式左侧要匹配的值。

参见：lowercased variables in pattern matching

这应该可以正常工作：

def getSubOb(record: String, subject:String, obj:String): List[(String, String)] = for {
    JObject(child) <-  parse(record)
    JField(`subject`, JString(subject1)) <- child
    JField(`obj`, JString(obj1)) <- child
} yield (subject1, obj1)

Scala 解析 JSON 在一个函数中表现不同

Scala Parsing JSON behaving differently in a function

scala

json4s