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在Java中将元素从一个XML文件复制并重命名为另一个XML文件

xml java

dan2k3k4 · 技术社区 · 14 年前

我可以使用 “org.w3c.dom.Document.文档。 importNode(节点importedNode,布尔深度) "

我有点像:

<SomeCustomNode randomAttribute="aValue" another="10/10/2010">  
    <Information>  
        <Yellow name="banana"/>  
        <Orange name="orange"/>  
        <Red name="strawberry"/>  
    </Information>  
    <Some>  
        <IgnoredNode/>  
    </Some>  
</SomeCustomNode>

像这样的:
文件列表.xml

<ListOfNodes date="12/10/2010">  
    <aCopy name="fruit" version="10">  
        <Yellow name="banana"/>  
        <Orange name="orange"/>  
        <Red name="strawberry"/>  
    </aCopy>  
    <aCopy name="vegetables" version="3">  
        <Yellow name="sweetcorn"/>  
        <Orange name="carrot"/>  
        <Red name="tomato"/>  
    </aCopy>  
</ListOfNodes>

所以,我要做的是从 File1.xml 把它插入 FileList.xml Element 并向元素添加一些属性。
问询处 aCopy name="fruit" version="10"

我目前正在使用一个XPath表达式来获取 问询处 节点作为NodeList(仅1个结果),然后将其导入到File2中,如下所示:

Document docFile1 = XMLDocumentStore.getDoc("/path/to/File1.xml");  
Document docFileList = XMLDocumentStore.getDoc("/path/to/FileList.xml");  
NodeList result = XPathAPI.selectNodeList(docFile1.getFirstChild(), ".//Information");  
Node importNode = docFileList.importNode(result.item(0), true);  
// We want to replace aCopy fruit with the updated version found in File1  
NodeList fruitNode = XPathAPI.selectNodeList(docFileList.getFirstChild(), ".//aCopy[@name=\"fruit\"]");  
Node replaceNode = fruitNode.item(0).getParentNode().replaceChild(importNode, fruitNode.item(0)); // probably a better way to do this  
// Now we want to replace the Element name as it is still set to Information  
docFileList.renameNode(replaceNode, null, "aCopy"); // Error: oracle.xml.parser.v2.XMLDOMException: cannot add attribute belonging to another element

如果我稍微移动一下代码,就会出现其他错误,比如: 等。

干杯!

1 回复 | 直到 8 年前

dogbane 14 年前

为什么要使用Oracle XML解析器?

如果使用 javax.xml

import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import com.sun.org.apache.xpath.internal.XPathAPI;

public static void main(String[] args) throws Exception {

    Document docFile1 = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(new File("File1.xml"));
    Document docFileList = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(new File("FileList.xml"));
    NodeList result = XPathAPI.selectNodeList(docFile1.getFirstChild(), ".//Information");
    Node importNode = docFileList.importNode(result.item(0), true);  
    // We want to replace aCopy fruit with the updated version found in File1  
    NodeList fruitNode = XPathAPI.selectNodeList(docFileList.getFirstChild(), ".//aCopy[@name=\"fruit\"]");
    Node replaceNode = fruitNode.item(0).getParentNode().replaceChild(importNode, fruitNode.item(0)); // probably a better way to do this  
    // Now we want to replace the Element name as it is still set to Information  
    docFileList.renameNode(replaceNode, null, "aCopy"); 

    print(docFileList);

}

打印出来:

<ListOfNodes date="12/10/2010">  
    <Information>  
        <Yellow name="banana"/>  
        <Orange name="orange"/>  
        <Red name="strawberry"/>  
    </Information>  
    <aCopy name="vegetables" version="3">  
        <Yellow name="sweetcorn"/>  
        <Orange name="carrot"/>  
        <Red name="tomato"/>  
    </aCopy>  
</ListOfNodes>