将新属性添加到 Java 中的现有 XML 节点？
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Add new attributes to an existing XML node in java?
提问by user2966458
I want to add an attribute to an existing xml node.I don't want to add new elements (new nodes) to my xml file, I just want to add a new attribute. How can I do this?
我想向现有的 xml 节点添加一个属性。我不想向我的 xml 文件添加新元素（新节点），我只想添加一个新属性。我怎样才能做到这一点？
In particular I've tried this lines of code:
特别是我试过这行代码：
Element process = doc.getElementsById("id");
process.setAttribute("modelgroup", "");
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
DOMSource source = new DOMSource(doc);
StreamResult result = new StreamResult(new File("C:\Users\Blerta\workspaceKEPLER\XML_to_JSON\SampleExample.xml"));
transformer.transform(source, result);
But I get the following exception:
但我得到以下异常：
Exception in thread "main" java.lang.NullPointerException
at Main.appendAttributes(Main.java:172)
at Main.displayNodes(Main.java:65)
at Main.displayNodes(Main.java:138)
at Main.main(Main.java:42)**
采纳答案by subash
in DOM parser it is very easy. get your node and simply use this function.
在 DOM 解析器中，这很容易。获取您的节点并简单地使用此功能。
((Element)node).setAttribute("attr_name","attr_value");
then finally update your document. like this..
然后最后更新您的文档。像这样..
Transformer transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty(OutputKeys.METHOD, "xml");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "5");
DOMSource source = new DOMSource(document);
StreamResult result = new StreamResult(new File(tablePath));
transformer.transform(source, result);
回答by stripybadger
You could do it in a few lines using xslt. Oracle have a half decent tutorial with all the code snippets http://docs.oracle.com/javase/tutorial/jaxp/xslt/transformingXML.html
您可以使用 xslt 在几行中完成。Oracle 有一个半体面的教程，其中包含所有代码片段http://docs.oracle.com/javase/tutorial/jaxp/xslt/transformingXML.html
The key bit for your xslt would be something like the following:
xslt 的关键位如下所示：
<xsl:template match="elementToAddNewAttrTo">
<xsl:attribute name="newAttrName">NewAttrValue</xsl:attribute>
</xsl:template>
回答by Sergey Tarasov
The easiest and the shortest is to cast the node to org.w3c.dom.Elementand then invoke setAttributeon it:
最简单和最短的是将节点转换为org.w3c.dom.Element，然后在其上调用setAttribute：
((Element)aNode).setAttribute("name", "value");
回答by Yeti
Recommended approach:
推荐方法：
Node node = ...;
if(node.getNodeType() == Node.ELEMENT_NODE)
{
((Element) node).setAttribute("name", "value");
}
Situational approach:
情境方法：
try
{
// ...
Node node = ...;
((Element) node).setAttribute("name", "value");
// ...
}
catch(ClassCastException e)
{
// Handle exception
}
Only use the try-catch approach if you already know that all the nodes that you process should be of type 'Element' (and thus any other case is an "exception" and should break from the code).
只有在您已经知道您处理的所有节点都应该是“元素”类型的情况下才使用 try-catch 方法（因此任何其他情况都是“例外”并且应该与代码中断）。
