如何使用Xpath检索XML树的一个接一个的节点? [英] How to retrieve node after node of XML tree using Xpath?

问题描述

首先，我必须说，我发现 Xpath 是一个非常不错的解析器，与其他解析器进行比较时，我认为它非常强大.

First, I must say that I find Xpath as a very nice parser , and I guess pretty powerful when comparing it to other parsers .

给出以下代码:

  DocumentBuilderFactory domFactory = 
  DocumentBuilderFactory.newInstance();
  domFactory.setNamespaceAware(true); 
  DocumentBuilder builder = domFactory.newDocumentBuilder();
  Document doc = builder.parse("input.xml");
  XPath xpath = XPathFactory.newInstance().newXPath();

如果我想找到第1轮&的 first 节点，1号门，在这里:

If I wanted to find the first node of Round 1 & Door 1 , here :

<Game>
    <Round>
        <roundNumber>1</roundNumber>
        <Door>
            <doorName>abd11</doorName>
            <Value>
                <xVal1>0</xVal1>
                <xVal2>25</xVal2>
                <pVal>0.31</pVal>
            </Value>
            <Value>
                <xVal1>25</xVal1>
                <xVal2>50</xVal2>
                <pVal>0.04</pVal>
            </Value>
            <Value>
                <xVal1>50</xVal1>
                <xVal2>75</xVal2>
                <pVal>0.19</pVal>
            </Value>
            <Value>
                <xVal1>75</xVal1>
                <xVal2>100</xVal2>
                <pVal>0.46</pVal>
            </Value>
        </Door>
        <Door>
            <doorName>vvv1133</doorName>
            <Value>
                <xVal1>60</xVal1>
                <xVal2>62</xVal2>
                <pVal>1.0</pVal>
            </Value>
        </Door>
    </Round>
    <Round>
        <roundNumber>2</roundNumber>
        <Door>
            <doorName>eee</doorName>
            <Value>
                <xVal1>0</xVal1>
                <xVal2>-25</xVal2>
                <pVal>0.31</pVal>
            </Value>
            <Value>
                <xVal1>-25</xVal1>
                <xVal2>-50</xVal2>
                <pVal>0.04</pVal>
            </Value>
            <Value>
                <xVal1>-50</xVal1>
                <xVal2>-75</xVal2>
                <pVal>0.19</pVal>
            </Value>
            <Value>
                <xVal1>-75</xVal1>
                <xVal2>-100</xVal2>
                <pVal>0.46</pVal>
            </Value>
        </Door>
        <Door>
            <doorName>cc</doorName>
            <Value>
                <xVal1>-60</xVal1>
                <xVal2>-62</xVal2>
                <pVal>0.3</pVal>
            </Value>
            <Value>
                <xVal1>-70</xVal1>
                <xVal2>-78</xVal2>
                <pVal>0.7</pVal>
            </Value>
        </Door>
    </Round>
</Game>

我将这样做:

 XPathExpression expr = xpath.compile("//Round[1]/Door[1]/Value[1]/*/text()");      
  Object result = expr.evaluate(doc, XPathConstants.NODESET);
  NodeList nodes = (NodeList) result;

，如果我想要第一回合&的 second 节点，然后是1号门:

and if I wanted the second node of Round 1 & Door 1 then :

XPathExpression expr = xpath.compile("//Round[1]/Door[1]/Value[2]/*/text()");  

但是如何使用循环执行此操作，因为我不知道我有多少 Value-nodes ，这意味着我如何使用循环执行此操作，其中每次迭代检索3(我的意思是 xVal1 ， xVal2 和 pVal 值)值节点的更多值！

but how do I do this using a loop , since I don't know how much Value-nodes I have , meaning how can I do this using a loop , where each iteration I retrieve 3 (I mean the xVal1 , xVal2 and pVal values ) more values of a Value node !?

要求这样做的原因是:

1. 我不知道我有多少 Round -s

我不知道我有多少 Value -s

I don't know how much Value-s I have

我不想每次都声明新的 XPathExpression

I don't want to declare every time a new XPathExpression

谢谢.

推荐答案

选项1 -遍历文档中的所有Value元素.仅需要进行一次评估，但很难知道该值属于哪个Round或Door元素.

Option 1 - Iterate over all Value elements in the document. Only one evaluation required, but difficult to know which Round or Door element the Value belongs to.

NodeList result = (NodeList) xpath.evaluate("//Round/Door/Value/*/text()", doc, XPathConstants.NODESET);

选项2 -分别遍历每个Round，Door和Value元素.需要更多评估，但是上下文很容易理解.如果需要索引，可以很容易地在循环中添加一个计数器.

Option 2 - Iterate over each Round, Door and Value elements separately. Requires more evaluations but the context is easily known. If index is required, it is easy to add a counter to the loops.

// Get all rounds and iterate over them
NodeList rounds = (NodeList) xpath.evaluate("//Round", doc, XPathConstants.NODESET);
for (Node round : rounds) {
  // Get all doors and iterate over them
  NodeList doors = (NodeList) xpath.evaluate("Door", round, XPathConstants.NODESET);
  for (Node door : doors) {
    // Get all values and iterate over them
    NodeList values = (NodeList) xpath.evaluate("Value/*/text()", door, XPathConstants.NODESET);
    for (Node value : values) {
      // Do something
    }
  }
}

选项3 -根据您的要求进行上述操作的组合

Option 3 - Do some combination of the above depending on your requirements

请注意，为了简化示例，我删除了表达式编译步骤.应该重新添加它以提高性能.

Note that I've removed the expression compilation step to shorten the example. It should be re-added to improve performance.

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