如何使用 Xpath 检索 XML 树的节点后的节点?

时间:2023-03-18
本文介绍了如何使用 Xpath 检索 XML 树的节点后的节点?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

首先,我必须说,我发现 Xpath 是一个非常好的解析器,并且在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 .

给定以下代码:

  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,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>

我会这样做的:

 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 1 轮的 second 节点 &那么门1:

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

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

但是我如何使用循环来做到这一点,因为我不知道我有多少 Value-nodes ,这意味着我如何使用循环来做到这一点,每次迭代我都会检索 3(我的意思是 xVal1xVal2pVal 值)更多值节点的值!?

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

我不想每次都声明一个新的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.只需要一次评估,但很难知道该值属于哪个圆形或门元素.

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 - 分别迭代每个 Round、Door 和 Value 元素.需要更多评估,但上下文很容易知道.如果需要索引,很容易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 - 根据您的要求将上述方法组合起来

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

请注意,我删除了表达式编译步骤以缩短示例.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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