我有一个包含XML的Java字符串,没有换行或缩进。我想把它变成一个字符串与格式良好的XML。我怎么做呢?
String unformattedXml = "<tag><nested>hello</nested></tag>";
String formattedXml = new [UnknownClass]().format(unformattedXml);
注意:我的输入是一个字符串。输出是一个字符串。
(基本)模拟结果:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<tag>
<nested>hello</nested>
</tag>
</root>
我把它们混合在一起,写了一个小程序。它从xml文件中读取并打印出来。而不是xzy给出你的文件路径。
public static void main(String[] args) throws Exception {
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setValidating(false);
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(new FileInputStream(new File("C:/Users/xyz.xml")));
prettyPrint(doc);
}
private static String prettyPrint(Document document)
throws TransformerException {
TransformerFactory transformerFactory = TransformerFactory
.newInstance();
Transformer transformer = transformerFactory.newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "2");
transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "no");
DOMSource source = new DOMSource(document);
StringWriter strWriter = new StringWriter();
StreamResult result = new StreamResult(strWriter);transformer.transform(source, result);
System.out.println(strWriter.getBuffer().toString());
return strWriter.getBuffer().toString();
}
我用Scala看到了一个答案,所以这里有另一个用Groovy的答案,以防有人觉得有趣。默认缩进为2步,XmlNodePrinter构造函数也可以传递另一个值。
def xml = "<tag><nested>hello</nested></tag>"
def stringWriter = new StringWriter()
def node = new XmlParser().parseText(xml);
new XmlNodePrinter(new PrintWriter(stringWriter)).print(node)
println stringWriter.toString()
如果groovy jar在类路径中,则使用Java
String xml = "<tag><nested>hello</nested></tag>";
StringWriter stringWriter = new StringWriter();
Node node = new XmlParser().parseText(xml);
new XmlNodePrinter(new PrintWriter(stringWriter)).print(node);
System.out.println(stringWriter.toString());