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
试试这个:
try
{
TransformerFactory transFactory = TransformerFactory.newInstance();
Transformer transformer = null;
transformer = transFactory.newTransformer();
StringWriter buffer = new StringWriter();
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.transform(new DOMSource(element),
new StreamResult(buffer));
String str = buffer.toString();
System.out.println("XML INSIDE IS #########################################"+str);
return element;
}
catch (TransformerConfigurationException e)
{
e.printStackTrace();
}
catch (TransformerException e)
{
e.printStackTrace();
}
其他回答
如果您确信您有一个有效的XML，那么这个很简单，并且避免了XML DOM树。可能有一些错误，如果你看到任何错误，请评论
public String prettyPrint(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("<?")) {
// xml version tag
pretty.append(row + "\n");
} else if (row.startsWith("</")) {
// closing tag
String indent = repeatString(" ", --stack);
pretty.append(indent + row + "\n");
} else if (row.startsWith("<")) {
// starting tag
String indent = repeatString(" ", stack++);
pretty.append(indent + row + "\n");
} else {
// tag data
String indent = repeatString(" ", stack);
pretty.append(indent + row + "\n");
}
}
return pretty.toString().trim();
}
I have found that in Java 1.6.0_32 the normal method to pretty print an XML string (using a Transformer with a null or identity xslt) does not behave as I would like if tags are merely separated by whitespace, as opposed to having no separating text. I tried using <xsl:strip-space elements="*"/> in my template to no avail. The simplest solution I found was to strip the space the way I wanted using a SAXSource and XML filter. Since my solution was for logging I also extended this to work with incomplete XML fragments. Note the normal method seems to work fine if you use a DOMSource but I did not want to use this because of the incompleteness and memory overhead.
public static class WhitespaceIgnoreFilter extends XMLFilterImpl
{
@Override
public void ignorableWhitespace(char[] arg0,
int arg1,
int arg2) throws SAXException
{
//Ignore it then...
}
@Override
public void characters( char[] ch,
int start,
int length) throws SAXException
{
if (!new String(ch, start, length).trim().equals(""))
super.characters(ch, start, length);
}
}
public static String prettyXML(String logMsg, boolean allowBadlyFormedFragments) throws SAXException, IOException, TransformerException
{
TransformerFactory transFactory = TransformerFactory.newInstance();
transFactory.setAttribute("indent-number", new Integer(2));
Transformer transformer = transFactory.newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");
StringWriter out = new StringWriter();
XMLReader masterParser = SAXHelper.getSAXParser(true);
XMLFilter parser = new WhitespaceIgnoreFilter();
parser.setParent(masterParser);
if(allowBadlyFormedFragments)
{
transformer.setErrorListener(new ErrorListener()
{
@Override
public void warning(TransformerException exception) throws TransformerException
{
}
@Override
public void fatalError(TransformerException exception) throws TransformerException
{
}
@Override
public void error(TransformerException exception) throws TransformerException
{
}
});
}
try
{
transformer.transform(new SAXSource(parser, new InputSource(new StringReader(logMsg))), new StreamResult(out));
}
catch (TransformerException e)
{
if(e.getCause() != null && e.getCause() instanceof SAXParseException)
{
if(!allowBadlyFormedFragments || !"XML document structures must start and end within the same entity.".equals(e.getCause().getMessage()))
{
throw e;
}
}
else
{
throw e;
}
}
out.flush();
return out.toString();
}
我试图实现类似的东西，但没有任何外部依赖。应用程序已经在使用DOM来格式化xml了!
下面是我的示例片段
public void formatXML(final String unformattedXML) {
final int length = unformattedXML.length();
final int indentSpace = 3;
final StringBuilder newString = new StringBuilder(length + length / 10);
final char space = ' ';
int i = 0;
int indentCount = 0;
char currentChar = unformattedXML.charAt(i++);
char previousChar = currentChar;
boolean nodeStarted = true;
newString.append(currentChar);
for (; i < length - 1;) {
currentChar = unformattedXML.charAt(i++);
if(((int) currentChar < 33) && !nodeStarted) {
continue;
}
switch (currentChar) {
case '<':
if ('>' == previousChar && '/' != unformattedXML.charAt(i - 1) && '/' != unformattedXML.charAt(i) && '!' != unformattedXML.charAt(i)) {
indentCount++;
}
newString.append(System.lineSeparator());
for (int j = indentCount * indentSpace; j > 0; j--) {
newString.append(space);
}
newString.append(currentChar);
nodeStarted = true;
break;
case '>':
newString.append(currentChar);
nodeStarted = false;
break;
case '/':
if ('<' == previousChar || '>' == unformattedXML.charAt(i)) {
indentCount--;
}
newString.append(currentChar);
break;
default:
newString.append(currentChar);
}
previousChar = currentChar;
}
newString.append(unformattedXML.charAt(length - 1));
System.out.println(newString.toString());
}
我过去使用org.dom4j.io.OutputFormat.createPrettyPrint()方法打印过
public String prettyPrint(final String xml){
if (StringUtils.isBlank(xml)) {
throw new RuntimeException("xml was null or blank in prettyPrint()");
}
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();
}
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。)
