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
使用jdom2: http://www.jdom.org/
import java.io.StringReader;
import org.jdom2.input.SAXBuilder;
import org.jdom2.output.Format;
import org.jdom2.output.XMLOutputter;
String prettyXml = new XMLOutputter(Format.getPrettyFormat()).
outputString(new SAXBuilder().build(new StringReader(uglyXml)));
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();
}
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());
}
有一个非常好的命令行XML实用程序叫做xmlstarlet(http://xmlstar.sourceforge.net/)，它可以做很多事情，很多人都在使用它。
您可以使用Runtime以编程方式执行此程序。然后读入格式化的输出文件。它具有比几行Java代码所能提供的更多选项和更好的错误报告。
下载xmlstarlet: http://sourceforge.net/project/showfiles.php?group_id=66612&package_id=64589
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();
