斯卡拉解析JSON在功能表现不同

我想创建一个函数来分析我的记录，我得到两个不同的行为时，我调用一个函数VS硬编码是：斯卡拉解析JSON在功能表现不同

我使用：

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)

将输出（我想要什么）：

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

，但是当我把它转移到一个功能我得到：

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_...

来源

2016-07-28 theMadKing

张贴一些可运行的代码。 – Falmarri

添加了导入语句，但其他所有内容都可以运行 – theMadKing

您对unapply一个subtele错误。

将模式匹配左侧的下限项视为变量。因此，所有的事情都是相匹配的，并且在那里受到约束。

您可以像`变量名称'中那样使用反引号来告诉Scala它不是一个要绑定的变量，而是一个匹配的模式左侧匹配的值。

看到：lowercased variables in pattern matching

如预期这应该工作：

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)

来源

2016-07-29 11:35:11

斯卡拉解析JSON在功能表现不同

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