如何以编程方式构造NodeSeq？

Question

我有一个像这样的NodeSeq

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

我想为List中的每个文章（定义为val articleList ）创建此结构，结果是：

 <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>
...

更新时间 ：

请注意<!-- some other markups don't know -->部分，因此我不能使用XML文字语法来构造它。 我需要带有相应文章信息的经过修改的article_template的列表。

Answer 1

这真的是电梯的优点

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 2

请参阅scala.xml.transform._和相应的堆栈溢出问题。

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 3

您可以使用Group分组XML元素。 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> 

更新资料

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个变体。 模板只是一个返回XML的函数。 在这种情况下，我使用currying来为每篇文章生成模板列表（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

假设对Article的定义如下：

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

和articleList为：

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

您可以通过以下方式生成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将它们添加在一起，如下所示：

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

• 用文章而不是当然的字符串

更新 ：匹配也很好

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