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

下面的代码工作得很好

import javax.xml.transform.OutputKeys;
import javax.xml.transform.Source;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;

String formattedXml1 = prettyFormat("<root><child>aaa</child><child/></root>");

public static String prettyFormat(String input) {
    return prettyFormat(input, "2");
}

public static String prettyFormat(String input, String indent) {
    Source xmlInput = new StreamSource(new StringReader(input));
    StringWriter stringWriter = new StringWriter();
    try {
        TransformerFactory transformerFactory = TransformerFactory.newInstance();
        Transformer transformer = transformerFactory.newTransformer();
        transformer.setOutputProperty(OutputKeys.INDENT, "yes");
        transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", indent);
        transformer.transform(xmlInput, new StreamResult(stringWriter));

        String pretty = stringWriter.toString();
        pretty = pretty.replace("\r\n", "\n");
        return pretty;              
    } catch (Exception e) {
        throw new RuntimeException(e);
    }
}

其他回答

现在已经是2012年了,Java可以比以前的XML做更多的事情,我想在我公认的答案之外添加一个替代方案。这在Java 6之外没有依赖关系。

import org.w3c.dom.Node;
import org.w3c.dom.bootstrap.DOMImplementationRegistry;
import org.w3c.dom.ls.DOMImplementationLS;
import org.w3c.dom.ls.LSSerializer;
import org.xml.sax.InputSource;

import javax.xml.parsers.DocumentBuilderFactory;
import java.io.StringReader;

/**
 * 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 String format(String xml) {

        try {
            final InputSource src = new InputSource(new StringReader(xml));
            final Node document = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(src).getDocumentElement();
            final Boolean keepDeclaration = Boolean.valueOf(xml.startsWith("<?xml"));

        //May need this: System.setProperty(DOMImplementationRegistry.PROPERTY,"com.sun.org.apache.xerces.internal.dom.DOMImplementationSourceImpl");


            final DOMImplementationRegistry registry = DOMImplementationRegistry.newInstance();
            final DOMImplementationLS impl = (DOMImplementationLS) registry.getDOMImplementation("LS");
            final LSSerializer writer = impl.createLSSerializer();

            writer.getDomConfig().setParameter("format-pretty-print", Boolean.TRUE); // Set this to true if the output needs to be beautified.
            writer.getDomConfig().setParameter("xml-declaration", keepDeclaration); // Set this to true if the declaration is needed to be outputted.

            return writer.writeToString(document);
        } catch (Exception 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));
    }
}

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;
}

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();
} 

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。)