Question

我有两个Map[String, String]，我需要创建具有更新值的新地图。如果第二个Map具有新的键或非空值或大于第一个Map的值，则完成更新。

例如：

更新方案

如果FirstMap("key1" -> "123")和SecondMap("key1" -> "456")则ResultMap("key1" -> "456")
如果FirstMap("key1" -> null)和SecondMap("key1" -> "value1")则ResultMap("key1" -> "value1")
如果FirstMap()和SecondMap("key1" -> "value1")则ResultMap("key1" -> "value1")

非更新方案

如果FirstMap("key1" -> "value1")和SecondMap()那么 ResultMap("key1" -> "value1")
如果FirstMap("key1" -> "value1")和SecondMap("key1" -> null)则ResultMap("key1" -> "value1")
如果FirstMap("key1" -> "789")和SecondMap("key1" -> "456")则ResultMap("key1" -> "789")

到目前为止，我的代码如下：

val currentFieldValue = currentDetail.get(field).get
val newFieldValue = newDetail.get(field).get

currentFieldValue match {
  case null if newFieldValue != null => currentDetail ++ Map(s"$field" -> s"$newFieldValue")
  case _ if newFieldValue != null && newFieldValue.toInt > currentFieldValue.toInt => currentDetail ++ Map(s"$field" -> s"$newFieldValue")
  case _  => currentDetail
}

关于如何容纳None状态的任何建议，即关键本身不存在？

Answer 1

map.get(key)在选项上返回Option模式匹配，并根据需要创建任意数量的相关案例

(currentDetail.get(field), newDetail.get(field)) match {
  case (Some(null), Some(null)) =>
  case (Some(null), Some(value)) =>
  case (Some(value), Some(null)) => 
  case (Some(value), None) =>
  case (None, Some(value)) =>
  case (None, None) =>
  case (_, _) =>
}

Answer 2

这里的替代方法是完整的比较功能（带调试信息）：

def compare(currentDetail: Map[String, String], newDetail: Map[String, String]): Map[String, String] = {
  var resMap = Map[String, String]()
  currentDetail.keys.foreach { field =>

    val currentField = currentDetail.get(field)
    val newField = newDetail.get(field)

    resMap = resMap + {
      (currentField, newField) match {
        case (Some(null), Some(newFieldValue: String)) =>
          println("null and not null")
          (s"${field}" -> newFieldValue)
        case (Some(currentFieldValue), Some(null)) =>
          println("not null and null")
          (s"${field}" -> currentFieldValue)
        case (None, Some(newFieldValue)) =>
          println("none and not null")
          (s"${field}" -> newFieldValue)
        case (Some(currentFieldValue), None) =>
          println("not null and none")
          (s"${field}" -> currentFieldValue)
        case (Some(currentFieldValue), Some(newFieldValue)) if newFieldValue.toInt > currentFieldValue.toInt =>
          println("not null not null if new bigger then current")
          (s"${field}" -> newFieldValue)
        case (Some(currentFieldValue), Some(newFieldValue)) if newFieldValue.toInt <= currentFieldValue.toInt =>
          println("not null not null if new less or equal to current")
          (s"${field}" -> currentFieldValue)
      }
    }
  }

  resMap
}

Answer 3

def combine(m1: Map[String,String], m2: Map[String,String]): Map[String,String] =
  (m1.keys ++ m2.keys).toSet.map{ k: String =>
    k -> Seq(Option(m1.getOrElse(k,"")), Option(m2.getOrElse(k,"")), Some(""))
         .flatten.sorted.reverse.head
  }.toMap

"456" > "123"对String和Int都适用。如果您有两个字符串"99"和"101"怎么办？如果您需要进行数字比较，则需要更多代码才能在String - ＆gt; Int - ＆gt; String之间传输。

将两个map键与可能为None进行比较并在scala中{null

3 个答案: