Scala和HTML解析

Question

如何將HTML DOM文檔加載到Scala中？嘗試加載xmlns標記時，XML單例錯誤。

import java.net._ 
import java.io._ 
import scala.xml._ 

object NetParse { 

    import java.net.{URLConnection, URL} 
    import scala.xml._ 

    def netParse(sUrl: String): Elem = { 
     var url = new URL(sUrl) 
     var connect = url.openConnection 

     XML.load(connect.getInputStream) 
    } 
} 

---

最後，我找到了解決辦法！ - 需要斯卡拉2.7.7或更高的工作（2.7.0有一個致命的錯誤）： How-to-use-TagSoup-with-Scala-XML

Answer 1

這可能會幫助你 Processing real world HTML as if it were XML in scala

Answer 2

嘗試使用scala.xml.parsing.XhtmlParser代替。

Answer 3

/* 
Copyright (c) 2008 Florian Hars, BIK Aschpurwis+Behrens GmbH, Hamburg 
Copyright (c) 2002-2008 EPFL, Lausanne, unless otherwise specified. 
All rights reserved. 

This software was developed by the Programming Methods Laboratory of the 
Swiss Federal Institute of Technology (EPFL), Lausanne, Switzerland. 

Permission to use, copy, modify, and distribute this software in source 
or binary form for any purpose with or without fee is hereby granted, 
provided that the following conditions are met: 

1. Redistributions of source code must retain the above copyright 
    notice, this list of conditions and the following disclaimer. 

2. Redistributions in binary form must reproduce the above copyright 
    notice, this list of conditions and the following disclaimer in the 
    documentation and/or other materials provided with the distribution. 

3. Neither the name of the EPFL nor the names of its contributors 
    may be used to endorse or promote products derived from this 
    software without specific prior written permission. 


THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 
ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 
IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 
ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 
FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 
DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR 
SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER 
CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 
LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 
OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 
SUCH DAMAGE. 
*/ 

package tagsoup 

import org.xml.sax.InputSource 
import javax.xml.parsers.SAXParser 
import org.ccil.cowan.tagsoup.jaxp.SAXFactoryImpl 
import scala.xml.parsing.FactoryAdapter 
import scala.xml._ 

class TagSoupFactoryAdapter extends FactoryAdapter { 

    val parserFactory = new SAXFactoryImpl 
    parserFactory.setNamespaceAware(false) 

    val emptyElements = Set("area", "base", "br", "col", "hr", "img", 
         "input", "link", "meta", "param") 

    /** Tests if an XML element contains text. 
    * @return true if element named <code>localName</code> contains text. 
    */ 
    def nodeContainsText(localName: String) = !(emptyElements contains localName) 

    /** creates a node. 
    */ 
    def createNode(pre:String, label: String, attrs: MetaData, 
      scpe: NamespaceBinding, children: List[Node]): Elem = { 
    Elem(pre, label, attrs, scpe, children:_*); 
    } 

    /** creates a text node 
    */ 
    def createText(text:String) = 
    Text(text); 

    /** Ignore Processing Instructions 
    */ 
    def createProcInstr(target: String, data: String) = Nil 

    /** load XML document 
    * @param source 
    * @return a new XML document object 
    */ 
    override def loadXML(source: InputSource) = { 
    val parser: SAXParser = parserFactory.newSAXParser() 

    scopeStack.push(TopScope) 
    parser.parse(source, this) 
    scopeStack.pop 
    rootElem 
    } 

} 

How-to-use-TagSoup-with-Scala-XML

Answer 4

我剛纔想用這個答案與斯卡拉2.8.1和最終使用從工作：

http://www.hars.de/2009/01/html-as-xml-in-scala.html

，我需要有趣的1位是：

val parserFactory = new org.ccil.cowan.tagsoup.jaxp.SAXFactoryImpl 
val parser = parserFactory.newSAXParser() 
val source = new org.xml.sax.InputSource("http://www.scala-lang.org") 
val adapter = new scala.xml.parsing.NoBindingFactoryAdapter 
adapter.loadXML(source, parser) 

Answer 5

Scala Scraper

我建議Scala Scraper，讓您解析HTML優雅是這樣的：

// Parse elements from files, URLs or plain strings 
val browser = JsoupBrowser() 
val doc = browser.parseFile("core/src/test/resources/example.html") 
val doc2 = browser.get("http://example.com") 
val doc3 = browser.parseString("<html><h1>parse me</h1></html>") 

// Extract the text inside the element with id "header" 
doc >> text("#header") 

// Extract the <span> elements inside #menu 
val items = doc >> elementList("#menu span") 

// From each item, extract all the text inside their <a> elements 
items.map(_ >> allText("a")) 

例子從斯卡拉刮板的readme拍攝。