Question

我有像这样的NodeSeq

val article_template =
 <div>
      <div class="title"></div>
      <div class="content"></div>
      <!-- some other markups don't know -->
 </div>

我想为List中的每篇文章创建此结构（定义为val articleList），结果如下：

 <div>
      <div class="title">title a</div>
      <div class="content">content a</div>
      <!-- some other markups don't know -->
 </div>
 <div>
      <div class="title">title b</div>
      <div class="content">content b</div>
      <!-- some other markups don't know -->
 </div>
...

已更新：

请注意部分，所以我不能使用XML文字语法来构造它。我需要一个修改后的article_template列表以及相应的文章信息。

Answer 1

请参阅scala.xml.transform._以及相应的Stack Overflow问题。

它可能看起来像这样：

import scala.xml._
import transform._

case class Article( title: String, content: String )

val articleList = List(
    Article("title 1","content 1"),
    Article("title 2","content 2")
)

class TransformArticle(title: String, content: String) extends RewriteRule {
  override def transform(n: Node): Seq[Node] = n match {
    case div @ <div/> => div match {
        case elem: Elem if elem \ "@class" contains Text("title") =>
            elem copy (child = Text(title))
        case elem: Elem if elem \ "@class" contains Text("content") =>
            elem copy (child = Text(content))
        case other => other
    }
    case other => other
  }
}

val article_template =
 <div>
      <div class="title"></div>
      <div class="content"></div>
      <!-- some other markups don't know -->
 </div>

Group(
    articleList.flatMap( article => {
        new RuleTransformer(new TransformArticle(article.title, article.content)) transform article_template
    })
)

Answer 2

这是lift对

非常有用的东西

import net.liftweb.util.Helpers._ 

case class Article( title: String, content: String )

val articles = List(
    Article("title a","content a"),
    Article("title b","content b")
)  

val snipet =
"article *" #> articles.map { article =>
    "header h1 *" #> article.title &
    ".content *" #> article.content
}

val article_template =
<article>
    <header>
        <h1>Title</h1>
    </header>
    <section class="content">
        Content
    </section>
</article>

snipet( article_template )

Answer 3

您可以使用Group对XML元素进行分组。在XML文字中，它将如下所示：

val xml =
  <xml:group>
    <div>
      <div class="title">title a</div>
      <div class="content">content a</div>
    </div>
    <div>
      <div class="title">title b</div>
      <div class="content">content b</div>
    </div>
  </xml:group>

更新

您可以像这样动态创建XML：

case class Article(title: String, content: String)
val articles = List(Article("A1", "A1 content"), Article("A2", "A2 content"))

val xml =
  <xml:group>{articles map { a =>
    <div>
      <div class="title">{a title}</div>
      <div class="content">{a content}</div>
    </div>
  }}</xml:group>

更新1

这是使用模板的另一种变体。模板只是一个返回XML的函数。在这种情况下，我使用currying来生成每篇文章的模板列表（仍然需要提供一些东西）：

case class Article(title: String, content: String)
val articles = List(Article("A1", "A1 content"), Article("A2", "A2 content"))

def articleTemplate[T](article: Article)(stuff: T) =
  <div>
    <div class="title">{article title}</div>
    <div class="content">{article content}</div>

    <div class="stuff">{stuff}</div>
  </div>

val allArticles = articles map articleTemplate

val allArticlesWithStuff: NodeSeq =
  allArticles.zipWithIndex map {case (f, i) => f(i)}

Answer 4

假设Article的定义类似于：

case class Article(title: String, content: String)

和articleList as：

val articleList = List(
  Article("title a","content a"),
  Article("title b", "content b"))

您可以使用以下命令生成XML片段：

val xml: scala.xml.NodeSeq = articleList map { article => 
  <div>
    <div class="title">{article.title}</div>
    <div class="content">{article.content}</div>
  </div>
}

Answer 5

使用flatmap将它们添加到一起：

val xml: NodeSeq = List("Test", "Something", "Yo").flatMap(s => <a>{s}</a>)

使用文章而不仅仅是字符串

更新：适用于匹配

val xml: NodeSeq = List("Test", "Something", "Yo", 1).flatMap{
  case s: String => <string>{s}</string>
  case i: Int => <int>{i}</int>
}

如何以编程方式构造NodeSeq？

5 个答案:

更新

更新1