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>
当前回答
只是另一种适合我们的解决方法
import java.io.StringWriter;
import org.dom4j.DocumentHelper;
import org.dom4j.io.OutputFormat;
import org.dom4j.io.XMLWriter;
**
* Pretty Print XML String
*
* @param inputXmlString
* @return
*/
public static String prettyPrintXml(String xml) {
final StringWriter sw;
try {
final OutputFormat format = OutputFormat.createPrettyPrint();
final org.dom4j.Document document = DocumentHelper.parseText(xml);
sw = new StringWriter();
final XMLWriter writer = new XMLWriter(sw, format);
writer.write(document);
}
catch (Exception e) {
throw new RuntimeException("Error pretty printing xml:\n" + xml, e);
}
return sw.toString();
}
其他回答
只是另一种适合我们的解决方法
import java.io.StringWriter;
import org.dom4j.DocumentHelper;
import org.dom4j.io.OutputFormat;
import org.dom4j.io.XMLWriter;
**
* Pretty Print XML String
*
* @param inputXmlString
* @return
*/
public static String prettyPrintXml(String xml) {
final StringWriter sw;
try {
final OutputFormat format = OutputFormat.createPrettyPrint();
final org.dom4j.Document document = DocumentHelper.parseText(xml);
sw = new StringWriter();
final XMLWriter writer = new XMLWriter(sw, format);
writer.write(document);
}
catch (Exception e) {
throw new RuntimeException("Error pretty printing xml:\n" + xml, e);
}
return sw.toString();
}
稍微改进了milosmns的版本…
public static String getPrettyXml(String xml) {
if (xml == null || xml.trim().length() == 0) return "";
int stack = 0;
StringBuilder pretty = new StringBuilder();
String[] rows = xml.trim().replaceAll(">", ">\n").replaceAll("<", "\n<").split("\n");
for (int i = 0; i < rows.length; i++) {
if (rows[i] == null || rows[i].trim().length() == 0) continue;
String row = rows[i].trim();
if (row.startsWith("<?")) {
pretty.append(row + "\n");
} else if (row.startsWith("</")) {
String indent = repeatString(--stack);
pretty.append(indent + row + "\n");
} else if (row.startsWith("<") && row.endsWith("/>") == false) {
String indent = repeatString(stack++);
pretty.append(indent + row + "\n");
if (row.endsWith("]]>")) stack--;
} else {
String indent = repeatString(stack);
pretty.append(indent + row + "\n");
}
}
return pretty.toString().trim();
}
private static String repeatString(int stack) {
StringBuilder indent = new StringBuilder();
for (int i = 0; i < stack; i++) {
indent.append(" ");
}
return indent.toString();
}
我也有同样的问题，我用JTidy (http://jtidy.sourceforge.net/index.html)取得了巨大的成功。
例子:
Tidy t = new Tidy();
t.setIndentContent(true);
Document d = t.parseDOM(
new ByteArrayInputStream("HTML goes here", null);
OutputStream out = new ByteArrayOutputStream();
t.pprint(d, out);
String html = out.toString();
以上所有的解决方案都不适合我，然后我找到了这个http://myshittycode.com/2014/02/10/java-properly-indenting-xml-string/
线索就是用XPath删除空格
String xml = "<root>" +
"\n " +
"\n<name>Coco Puff</name>" +
"\n <total>10</total> </root>";
try {
Document document = DocumentBuilderFactory.newInstance()
.newDocumentBuilder()
.parse(new InputSource(new ByteArrayInputStream(xml.getBytes("utf-8"))));
XPath xPath = XPathFactory.newInstance().newXPath();
NodeList nodeList = (NodeList) xPath.evaluate("//text()[normalize-space()='']",
document,
XPathConstants.NODESET);
for (int i = 0; i < nodeList.getLength(); ++i) {
Node node = nodeList.item(i);
node.getParentNode().removeChild(node);
}
Transformer transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");
StringWriter stringWriter = new StringWriter();
StreamResult streamResult = new StreamResult(stringWriter);
transformer.transform(new DOMSource(document), streamResult);
System.out.println(stringWriter.toString());
}
catch (Exception e) {
e.printStackTrace();
}
请注意，排名靠前的答案需要使用xerces。
如果您不想添加这个外部依赖，那么您可以简单地使用标准jdk库(实际上是在内部使用xerces构建的)。
注意:jdk 1.5版本有一个bug，请参阅http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6296446，但现在已经解决了。
(注意，如果发生错误，将返回原始文本)
package com.test;
import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import javax.xml.transform.OutputKeys;
import javax.xml.transform.Source;
import javax.xml.transform.Transformer;
import javax.xml.transform.sax.SAXSource;
import javax.xml.transform.sax.SAXTransformerFactory;
import javax.xml.transform.stream.StreamResult;
import org.xml.sax.InputSource;
public class XmlTest {
public static void main(String[] args) {
XmlTest t = new XmlTest();
System.out.println(t.formatXml("<a><b><c/><d>text D</d><e value='0'/></b></a>"));
}
public String formatXml(String xml){
try{
Transformer serializer= SAXTransformerFactory.newInstance().newTransformer();
serializer.setOutputProperty(OutputKeys.INDENT, "yes");
//serializer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
serializer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "2");
//serializer.setOutputProperty("{http://xml.customer.org/xslt}indent-amount", "2");
Source xmlSource=new SAXSource(new InputSource(new ByteArrayInputStream(xml.getBytes())));
StreamResult res = new StreamResult(new ByteArrayOutputStream());
serializer.transform(xmlSource, res);
return new String(((ByteArrayOutputStream)res.getOutputStream()).toByteArray());
}catch(Exception e){
//TODO log error
return xml;
}
}
}
