迭代案例类数据成员 [英] iterate over case class data members

问题描述

我正在用 mongodb 编写 play2.1 应用程序，我的模型对象有点广泛.在更新数据库中的条目时，我需要将来自表单的临时对象与数据库中的内容进行比较，以便我可以构建更新查询(并记录更改).

I am writing a play2.1 application with mongodb, and my model object is a bit extensive. when updating an entry in the DB, i need to compare the temp object coming from the form with what's in the DB, so i can build the update query (and log the changes).

我正在寻找一种方法来一般地采用 2 个实例并获取它们的差异.迭代每个数据成员很长，硬编码并且容易出错(如果 a.firstName.equalsIgnoreCase(b.firstName))，所以我正在寻找一种方法来迭代所有数据成员并水平比较它们(名称映射 -> value 可以，或者我可以信任的列表每次都以相同的顺序枚举数据成员).

i am looking for a way to generically take 2 instances and get a diff of them. iterating over each data member is long, hard-coded and error prone (if a.firstName.equalsIgnoreCase(b.firstName)) so i am looking for a way to iterate over all data members and compare them horizontally (a map of name -> value will do, or a list i can trust to enumerate the data members in the same order every time).

有什么想法吗?

case class Customer(
  id: Option[BSONObjectID] = Some(BSONObjectID.generate),
  firstName: String,
  middleName: String,
  lastName: String,
  address: List[Address],
  phoneNumbers: List[PhoneNumber],
  email: String,
  creationTime: Option[DateTime] = Some(DateTime.now()),
  lastUpdateTime: Option[DateTime] = Some(DateTime.now())
)

以下所有三个解决方案都很棒，但我仍然无法获得该字段的名称，对吗?这意味着我可以记录更改，但不能记录它影响的字段...

all three solutions below are great, but i still cannot get the field's name, right? that means i can log the change, but not what field it affected...

推荐答案

扩展@Malte_Schwerhoff 的答案，您可能会创建一个递归 diff 方法，该方法不仅生成差异的索引，而且将它们映射到该索引处的新值- 或者在嵌套产品类型的情况下，子产品差异的映射:

Expanding on @Malte_Schwerhoff's answer, you could potentially create a recursive diff method that not only generated the indexes of differences, but mapped them to the new value at that index - or in the case of nested Product types, a map of the sub-Product differences:

def diff(orig: Product, update: Product): Map[Int, Any] = {
  assert(orig != null && update != null, "Both products must be non-null")
  assert(orig.getClass == update.getClass, "Both products must be of the same class")

  val diffs = for (ix <- 0 until orig.productArity) yield {
    (orig.productElement(ix), update.productElement(ix)) match {
      case (s1: String, s2: String) if (!s1.equalsIgnoreCase(s2)) => Some((ix -> s2))
      case (s1: String, s2: String) => None
      case (p1: Product, p2: Product) if (p1 != p2) => Some((ix -> diff(p1, p2)))
      case (x, y) if (x != y) => Some((ix -> y))
      case _ => None
    }
  }

  diffs.flatten.toMap
}

从该答案扩展用例:

case class A(x: Int, y: String)
case class B(a: A, b: AnyRef, c: Any)

val a1 = A(4, "four")
val a2 = A(4, "Four")
val a3 = A(4, "quatre")
val a4 = A(5, "five")
val b1 = B(a1, null, 6)
val b2 = B(a1, null, 7)
val b3 = B(a2, a2, a2)
val b4 = B(a4, null, 8)

println(diff(a1, a2)) // Map()
println(diff(a1, a3)) // Map(0 -> 5)
println(diff(a1, a4)) // Map(0 -> 5, 1 -> five)

println(diff(b1, b2)) // Map(2 -> 7)
println(diff(b1, b3)) // Map(1 -> A(4,four), 2 -> A(4,four))
println(diff(b1, b4)) // Map(0 -> Map(0 -> 5, 1 -> five), 2 -> 8l

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