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Question

我正在尝试比较两个Seq字符串，这是我使用Apache和Play获得的！Scala的框架。几天后我就没有主意了。以下是开始：

@(比较:SeqCompare，compares1:SeqCompare，compareForm: Form(String，String))(隐式请求:RequestAny) @import helper._ @main(“管理工具”，request.uri){

我有两个插曲--怎么比较？

@for(compare <- compares) {
    <td>@for(group <- compare.memberOf) {
        <li style="list-style-type: none;">@group.replaceAll(",(.*)","").replaceAll("(.*)=","")</li>
    }</td>
}
@for(compare1 <- compares1) {
    <td>@for(group1 <- compare1.memberOf) {
        <li style="list-style-type: none;">@group1.replaceAll(",(.*)","").replaceAll("(.*)=","")</li>
    }</td>
}   

Answer 1

我知道这可能是一个迟来的答案，但当时我设法解决了这个问题。代码可能有点脏，但我希望它能派上用场。

模型部分：

def intersection(c1:Seq[String], c2:Seq[String]): Seq[String] = {
    (Set(c1: _*) & Set(c2: _*)).toSeq
}

控制器部分：

compareForm.bindFromRequest.fold(
hasErrors = {errors => BadRequest(views.html.compares.index(compareForm, username, User.find(username)))},
success = { ldapName => 
  val leftUser = Compare.findAll(ldapName._1)
  val rightUser = Compare.findAll(ldapName._2)
  val intersection = Compare.intersection(leftUser.memberOf,rightUser.memberOf) 
  val rightLacking: Seq[String] = (leftUser.memberOf.toSet -- intersection.toSet).toList.map(_.toString).sorted      
  val leftLacking: Seq[String] = (rightUser.memberOf.toSet -- intersection.toSet).toList.map(_.toString).sorted
  val bothOk: Seq[String] = (intersection.toSet).toList.sorted
  Ok(views.html.compares.list(leftUser, rightUser, rightLacking, leftLacking, bothOk, compareForm, username, User.find(username)))
}  )

致以问候！