如何使用 Xpath 检索 XML 树的节点后的节点?
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 xVal1
、 xVal2
和 pVal
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:
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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