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提取内部元素而不循环(Extract inner element without looping)

 如果我循环遍历所有，则从以下示例HTML代码中提取href值是直截了当的  并在第一个之后立即中断：  
  <li class="parts partname parts_first">
    <div id="dpdn10" uri="/public/page/part1" class="partype partstate">
      <div class="ptctainer">
        <div class="ptitle">
          <p class="ptypead">
            <span class="rtext"><a href="http://www.example.com/page/ptname.html?dv=rfirst" class="mnLabel">First</a></span>
            <span class="ndx">
              <a href="#" dndx="dpdn10" class="xpnd _t" style="opacity:1">Details: </a>
            </span>
          </p>
        </div>
      </div>

      <div id="dpdn10_content" class="xpns">
        <div class="ptctainer">
          <div class="ptitle">
            <p class="ptypead">
              <span class="rtext"><a href="http://www.example.com/page/ptname.html?dv=rfirst" class="mnLabel">First</a></span>
              <span class="ndx"><a href="#" class="xpnd">Details: </a></span>
            </p>
          </div>
        </div>    
      </div>
    </div>
  </li>
 
 当我可以假设两个实例的href值相同时，我当然可以这样做  如上例所示。  
 但是，如果它们不相同并且我想提取特定的（第一个或第二个），则此方法会失败。  
 这让我在Jsoup中搜索允许“嵌套选择”的机制：到目前为止，我已熟悉单级选择，如：  
Elements links = doc.select("a[href]"); // a with href
Elements pngs = doc.select("img[src$=.png]");  // img with src ending .png
Element masthead = doc.select("div.masthead").first();  // div with class=masthead
 
 但我找不到文档或多级选择的例子，例如  
Element link= doc.select("div.xpns.div.ptctainer.div.ptitle.p.ptypead.span.rtext");
 
 当然，上面的内容仅用于说明而不是真正的语法。 我不知道Jsoup中是否还有这样的东西。  
 Jsoup中是否存在这种“嵌套选择”？ 

Extracting the href value from the following sample HTML code is straight forward if I loop through all  and break immediately after the first one: 
  <li class="parts partname parts_first">
    <div id="dpdn10" uri="/public/page/part1" class="partype partstate">
      <div class="ptctainer">
        <div class="ptitle">
          <p class="ptypead">
            <span class="rtext"><a href="http://www.example.com/page/ptname.html?dv=rfirst" class="mnLabel">First</a></span>
            <span class="ndx">
              <a href="#" dndx="dpdn10" class="xpnd _t" style="opacity:1">Details: </a>
            </span>
          </p>
        </div>
      </div>

      <div id="dpdn10_content" class="xpns">
        <div class="ptctainer">
          <div class="ptitle">
            <p class="ptypead">
              <span class="rtext"><a href="http://www.example.com/page/ptname.html?dv=rfirst" class="mnLabel">First</a></span>
              <span class="ndx"><a href="#" class="xpnd">Details: </a></span>
            </p>
          </div>
        </div>    
      </div>
    </div>
  </li>
 
I can certainly do that when I can assume the href value is identical for both instances of  as in the example above. 
However, this approach fails if they are not identical and I want to extract a specific one (either the first or the second). 
Which brings me to searching for a mechanism in Jsoup that allows "nested selection": Up until now I have been familiar with single-level selection as in: 
Elements links = doc.select("a[href]"); // a with href
Elements pngs = doc.select("img[src$=.png]");  // img with src ending .png
Element masthead = doc.select("div.masthead").first();  // div with class=masthead
 
But I can't find documentation or an example for multi-level selection, e.g. 
Element link= doc.select("div.xpns.div.ptctainer.div.ptitle.p.ptypead.span.rtext");
 
The above is for illustration and not real syntax, of course. I don't know if something like this is possible (yet) in Jsoup. 
Does such "nested selection" exist in Jsoup?

原文：https://stackoverflow.com/questions/5771462

更新时间：2022-05-21 11:05

最满意答案

 发生这种情况是因为您传递的是“4 2 1 -5”而不是作为参数，因此它们会进入main方法的args参数。 要将它们作为输入传递，请等待应用程序启动并输入它们或使用输入流重定向  
 更新 （参数处理代码）  
 
  我应该如何在java中编写它以将输入作为参数呢？ 有一个特定的简单方法吗？  
 
 我会说使用非固定数量的输入参数的参数并不是一个好主意。 我仍然说使用输入流是要走的路 - 这就是它的设计目的！ 您的解析要求也有点不合理。 我的尝试还在这里  
public class Main
{
    static OptionalInt parseOnlyInt(String s)
    {
        String digitsOnly = s.replaceAll("[^0-9]", "");
        if (digitsOnly.length() == 0)
            return OptionalInt.empty();
        else
            return OptionalInt.of(Integer.parseInt(digitsOnly));
    }

    public static void main(String[] args)
    {
        int sum = Arrays.stream(args)
                .map(Main::parseOnlyInt)
                // probably in Java 9 OptionalInt::stream would be better than next 2 lines
                // see https://bugs.openjdk.java.net/browse/JDK-8050820
                .filter(OptionalInt::isPresent)
                .mapToInt(OptionalInt::getAsInt)
                .sum(); 

        System.out.println(Arrays.toString(args));
        System.out.println(sum);    
    }
}
 
 或者，如果您更喜欢旧的（非Java-8）方式，那么它就是  
public static void main(String[] args)
{
    int sum = 0;
    for (int i = 0; i < args.length; i++)
    {
        String digitsOnly = args[i].replaceAll("[^0-9]", "");
        if (digitsOnly.length() != 0)
            sum += Integer.parseInt(digitsOnly);
    }
    System.out.println(Arrays.toString(args));
    System.out.println(sum);
}

This happens because you pass "4 2 1 -5" not as input but as arguments so they come into args parameter of the main method. To pass them as input wait till the app starts and enter them or use input stream redirection  
Update (code for arguments processing) 
 
 How should I write it in java to take input as arguments then? Is there a specific easy way? 
 
I'd say that using arguments for non-fixed number of input params is not so good idea. I'd still say that using input stream is the way to go - this is what it was designed for! Also your parsing requirements are a bit umbigious. Still here is my attempt 
public class Main
{
    static OptionalInt parseOnlyInt(String s)
    {
        String digitsOnly = s.replaceAll("[^0-9]", "");
        if (digitsOnly.length() == 0)
            return OptionalInt.empty();
        else
            return OptionalInt.of(Integer.parseInt(digitsOnly));
    }

    public static void main(String[] args)
    {
        int sum = Arrays.stream(args)
                .map(Main::parseOnlyInt)
                // probably in Java 9 OptionalInt::stream would be better than next 2 lines
                // see https://bugs.openjdk.java.net/browse/JDK-8050820
                .filter(OptionalInt::isPresent)
                .mapToInt(OptionalInt::getAsInt)
                .sum(); 

        System.out.println(Arrays.toString(args));
        System.out.println(sum);    
    }
}
 
Or if you prefer the old (non-Java-8) way here it is 
public static void main(String[] args)
{
    int sum = 0;
    for (int i = 0; i < args.length; i++)
    {
        String digitsOnly = args[i].replaceAll("[^0-9]", "");
        if (digitsOnly.length() != 0)
            sum += Integer.parseInt(digitsOnly);
    }
    System.out.println(Arrays.toString(args));
    System.out.println(sum);
}

提取内部元素而不循环(Extract inner element without looping)

最满意答案

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