我总是使用下面的函数:

public static String prettyPrintXml(String xmlStringToBeFormatted) {
    String formattedXmlString = null;
    try {
        DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory.newInstance();
        documentBuilderFactory.setValidating(true);
        DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder();
        InputSource inputSource = new InputSource(new StringReader(xmlStringToBeFormatted));
        Document document = documentBuilder.parse(inputSource);

        Transformer transformer = TransformerFactory.newInstance().newTransformer();
        transformer.setOutputProperty(OutputKeys.INDENT, "yes");
        transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "2");

        StreamResult streamResult = new StreamResult(new StringWriter());
        DOMSource dOMSource = new DOMSource(document);
        transformer.transform(dOMSource, streamResult);
        formattedXmlString = streamResult.getWriter().toString().trim();
    } catch (Exception ex) {
        StringWriter sw = new StringWriter();
        ex.printStackTrace(new PrintWriter(sw));
        System.err.println(sw.toString());
    }
    return formattedXmlString;
}

关于“您必须首先构建DOM树”的评论:不，您不需要也不应该这样做。

相反，创建一个StreamSource(new StreamSource(new StringReader(str))，并将其提供给前面提到的标识转换器。这将使用SAX解析器，结果将快得多。 在这种情况下，构建中间树纯粹是开销。 否则，排名第一的答案是好的。

凯文·哈肯森说: 但是，如果您知道您的XML字符串是有效的，并且您不想引起将字符串解析为DOM的内存开销，然后在DOM上运行转换以获得字符串—您可以通过字符解析进行一些老式的字符。在每个字符后插入换行符和空格，保持和缩进计数器(以确定空格的数量)，为每个<…>和递减你看到的每一个。”

同意了。这种方法要快得多，依赖关系也少得多。

示例解决方案:

/**
 * XML utils, including formatting.
 */
public class XmlUtils
{
  private static XmlFormatter formatter = new XmlFormatter(2, 80);

  public static String formatXml(String s)
  {
    return formatter.format(s, 0);
  }

  public static String formatXml(String s, int initialIndent)
  {
    return formatter.format(s, initialIndent);
  }

  private static class XmlFormatter
  {
    private int indentNumChars;
    private int lineLength;
    private boolean singleLine;

    public XmlFormatter(int indentNumChars, int lineLength)
    {
      this.indentNumChars = indentNumChars;
      this.lineLength = lineLength;
    }

    public synchronized String format(String s, int initialIndent)
    {
      int indent = initialIndent;
      StringBuilder sb = new StringBuilder();
      for (int i = 0; i < s.length(); i++)
      {
        char currentChar = s.charAt(i);
        if (currentChar == '<')
        {
          char nextChar = s.charAt(i + 1);
          if (nextChar == '/')
            indent -= indentNumChars;
          if (!singleLine)   // Don't indent before closing element if we're creating opening and closing elements on a single line.
            sb.append(buildWhitespace(indent));
          if (nextChar != '?' && nextChar != '!' && nextChar != '/')
            indent += indentNumChars;
          singleLine = false;  // Reset flag.
        }
        sb.append(currentChar);
        if (currentChar == '>')
        {
          if (s.charAt(i - 1) == '/')
          {
            indent -= indentNumChars;
            sb.append("\n");
          }
          else
          {
            int nextStartElementPos = s.indexOf('<', i);
            if (nextStartElementPos > i + 1)
            {
              String textBetweenElements = s.substring(i + 1, nextStartElementPos);

              // If the space between elements is solely newlines, let them through to preserve additional newlines in source document.
              if (textBetweenElements.replaceAll("\n", "").length() == 0)
              {
                sb.append(textBetweenElements + "\n");
              }
              // Put tags and text on a single line if the text is short.
              else if (textBetweenElements.length() <= lineLength * 0.5)
              {
                sb.append(textBetweenElements);
                singleLine = true;
              }
              // For larger amounts of text, wrap lines to a maximum line length.
              else
              {
                sb.append("\n" + lineWrap(textBetweenElements, lineLength, indent, null) + "\n");
              }
              i = nextStartElementPos - 1;
            }
            else
            {
              sb.append("\n");
            }
          }
        }
      }
      return sb.toString();
    }
  }

  private static String buildWhitespace(int numChars)
  {
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < numChars; i++)
      sb.append(" ");
    return sb.toString();
  }

  /**
   * Wraps the supplied text to the specified line length.
   * @lineLength the maximum length of each line in the returned string (not including indent if specified).
   * @indent optional number of whitespace characters to prepend to each line before the text.
   * @linePrefix optional string to append to the indent (before the text).
   * @returns the supplied text wrapped so that no line exceeds the specified line length + indent, optionally with
   * indent and prefix applied to each line.
   */
  private static String lineWrap(String s, int lineLength, Integer indent, String linePrefix)
  {
    if (s == null)
      return null;

    StringBuilder sb = new StringBuilder();
    int lineStartPos = 0;
    int lineEndPos;
    boolean firstLine = true;
    while(lineStartPos < s.length())
    {
      if (!firstLine)
        sb.append("\n");
      else
        firstLine = false;

      if (lineStartPos + lineLength > s.length())
        lineEndPos = s.length() - 1;
      else
      {
        lineEndPos = lineStartPos + lineLength - 1;
        while (lineEndPos > lineStartPos && (s.charAt(lineEndPos) != ' ' && s.charAt(lineEndPos) != '\t'))
          lineEndPos--;
      }
      sb.append(buildWhitespace(indent));
      if (linePrefix != null)
        sb.append(linePrefix);

      sb.append(s.substring(lineStartPos, lineEndPos + 1));
      lineStartPos = lineEndPos + 1;
    }
    return sb.toString();
  }

  // other utils removed for brevity
}

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();

如果你不需要缩进那么多，但一些换行，这可能是足够的简单regex…

String leastPrettifiedXml = uglyXml.replaceAll("><", ">\n<");

代码很好，而不是因为缺少缩进而导致的结果。

(对于有缩进的解，请参见其他答案。)

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());