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();
    }

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。)

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();
} 

这是我自己问题的答案。我将各种结果的答案结合起来，编写了一个输出XML的类。

不保证它如何响应无效的XML或大型文档。

package ecb.sdw.pretty;

import org.apache.xml.serialize.OutputFormat;
import org.apache.xml.serialize.XMLSerializer;
import org.w3c.dom.Document;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import java.io.IOException;
import java.io.StringReader;
import java.io.StringWriter;
import java.io.Writer;

/**
 * Pretty-prints xml, supplied as a string.
 * <p/>
 * eg.
 * <code>
 * String formattedXml = new XmlFormatter().format("<tag><nested>hello</nested></tag>");
 * </code>
 */
public class XmlFormatter {

    public XmlFormatter() {
    }

    public String format(String unformattedXml) {
        try {
            final Document document = parseXmlFile(unformattedXml);

            OutputFormat format = new OutputFormat(document);
            format.setLineWidth(65);
            format.setIndenting(true);
            format.setIndent(2);
            Writer out = new StringWriter();
            XMLSerializer serializer = new XMLSerializer(out, format);
            serializer.serialize(document);

            return out.toString();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

    private Document parseXmlFile(String in) {
        try {
            DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
            DocumentBuilder db = dbf.newDocumentBuilder();
            InputSource is = new InputSource(new StringReader(in));
            return db.parse(is);
        } catch (ParserConfigurationException e) {
            throw new RuntimeException(e);
        } catch (SAXException e) {
            throw new RuntimeException(e);
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

    public static void main(String[] args) {
        String unformattedXml =
                "<?xml version=\"1.0\" encoding=\"UTF-8\"?><QueryMessage\n" +
                        "        xmlns=\"http://www.SDMX.org/resources/SDMXML/schemas/v2_0/message\"\n" +
                        "        xmlns:query=\"http://www.SDMX.org/resources/SDMXML/schemas/v2_0/query\">\n" +
                        "    <Query>\n" +
                        "        <query:CategorySchemeWhere>\n" +
                        "   \t\t\t\t\t         <query:AgencyID>ECB\n\n\n\n</query:AgencyID>\n" +
                        "        </query:CategorySchemeWhere>\n" +
                        "    </Query>\n\n\n\n\n" +
                        "</QueryMessage>";

        System.out.println(new XmlFormatter().format(unformattedXml));
    }

}

嗯…面对这样的事情，这是一个已知的bug… 只需添加这个OutputProperty ..

transformer.setOutputProperty(OutputPropertiesFactory.S_KEY_INDENT_AMOUNT, "8");

希望这对你有所帮助……

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。