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>
Since you are starting with a String, you can convert to a DOM object (e.g. Node) before you use the Transformer. However, if you know your XML string is valid, and you don't want to incur the memory overhead of parsing a string into a DOM, then running a transform over the DOM to get a string back - you could just do some old fashioned character by character parsing. Insert a newline and spaces after every </...> characters, keep and indent counter (to determine the number of spaces) that you increment for every <...> and decrement for every </...> you see.
免责声明-我对下面的函数做了剪切/粘贴/文本编辑,所以它们可能不能按原样编译。
public static final Element createDOM(String strXML)
throws ParserConfigurationException, SAXException, IOException {
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setValidating(true);
DocumentBuilder db = dbf.newDocumentBuilder();
InputSource sourceXML = new InputSource(new StringReader(strXML));
Document xmlDoc = db.parse(sourceXML);
Element e = xmlDoc.getDocumentElement();
e.normalize();
return e;
}
public static final void prettyPrint(Node xml, OutputStream out)
throws TransformerConfigurationException, TransformerFactoryConfigurationError, TransformerException {
Transformer tf = TransformerFactory.newInstance().newTransformer();
tf.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
tf.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
tf.setOutputProperty(OutputKeys.INDENT, "yes");
tf.transform(new DOMSource(xml), new StreamResult(out));
}
使用scala:
import xml._
val xml = XML.loadString("<tag><nested>hello</nested></tag>")
val formatted = new PrettyPrinter(150, 2).format(xml)
println(formatted)
如果你依赖scala-library.jar,你也可以在Java中这样做。它是这样的:
import scala.xml.*;
public class FormatXML {
public static void main(String[] args) {
String unformattedXml = "<tag><nested>hello</nested></tag>";
PrettyPrinter pp = new PrettyPrinter(150, 3);
String formatted = pp.format(XML.loadString(unformattedXml), TopScope$.MODULE$);
System.out.println(formatted);
}
}
PrettyPrinter对象是用两个整数构造的,第一个是最大行长,第二个是缩进步骤。
基于这个答案的一个更简单的解决方案:
public static String prettyFormat(String input, int indent) {
try {
Source xmlInput = new StreamSource(new StringReader(input));
StringWriter stringWriter = new StringWriter();
StreamResult xmlOutput = new StreamResult(stringWriter);
TransformerFactory transformerFactory = TransformerFactory.newInstance();
transformerFactory.setAttribute("indent-number", indent);
transformerFactory.setAttribute(XMLConstants.ACCESS_EXTERNAL_DTD, "");
transformerFactory.setAttribute(XMLConstants.ACCESS_EXTERNAL_STYLESHEET, "");
Transformer transformer = transformerFactory.newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.transform(xmlInput, xmlOutput);
return xmlOutput.getWriter().toString();
} catch (Exception e) {
throw new RuntimeException(e); // simple exception handling, please review it
}
}
public static String prettyFormat(String input) {
return prettyFormat(input, 2);
}
testcase:
prettyFormat("<root><child>aaa</child><child/></root>");
返回:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<child>aaa</child>
<child/>
</root>
//忽略:原始编辑只需要在代码中的类名中缺少s。为了在SO上获得超过6个字符的验证,添加了多余的6个字符
如果使用第三方XML库是可行的,那么您可以使用一些比目前票数最高的答案所建议的要简单得多的方法。
它声明输入和输出都应该是字符串,所以这里有一个实用程序方法,用XOM库实现:
import nu.xom.*;
import java.io.*;
[...]
public static String format(String xml) throws ParsingException, IOException {
ByteArrayOutputStream out = new ByteArrayOutputStream();
Serializer serializer = new Serializer(out);
serializer.setIndent(4); // or whatever you like
serializer.write(new Builder().build(xml, ""));
return out.toString("UTF-8");
}
我对它进行了测试,结果不依赖于JRE版本或类似的东西。要了解如何根据自己的喜好定制输出格式,请查看Serializer API。
这实际上比我想象的要长——需要一些额外的行,因为Serializer想要写入一个OutputStream。但是请注意,这里很少有用于实际XML处理的代码。
(这个答案是我对XOM的评估的一部分,在我关于替代dom4j的最佳Java XML库的问题中,XOM被建议作为一个选项。在dom4j中,您可以使用XMLWriter和OutputFormat轻松实现这一点。编辑:…正如mlo55的答案所示。)
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();
}