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

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());
}

其他回答

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

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));
    }
}

使用scala:

import xml._
val xml = XML.loadString("<tag><nested>hello</nested></tag>")
val formatted = new PrettyPrinter(150, 2).format(xml)
println(formatted)

如果你依赖scala-library.jar,你也可以在Java中这样做。它是这样的:

import scala.xml.*;

public class FormatXML {
    public static void main(String[] args) {
        String unformattedXml = "<tag><nested>hello</nested></tag>";
        PrettyPrinter pp = new PrettyPrinter(150, 3);
        String formatted = pp.format(XML.loadString(unformattedXml), TopScope$.MODULE$);
        System.out.println(formatted);
    }
}

PrettyPrinter对象是用两个整数构造的,第一个是最大行长,第二个是缩进步骤。

Since you are starting with a String, you can convert to a DOM object (e.g. Node) before you use the Transformer. However, if you know your XML string is valid, and you don't want to incur the memory overhead of parsing a string into a DOM, then running a transform over the DOM to get a string back - you could just do some old fashioned character by character parsing. Insert a newline and spaces after every </...> characters, keep and indent counter (to determine the number of spaces) that you increment for every <...> and decrement for every </...> you see.

免责声明-我对下面的函数做了剪切/粘贴/文本编辑,所以它们可能不能按原样编译。

public static final Element createDOM(String strXML) 
    throws ParserConfigurationException, SAXException, IOException {

    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    dbf.setValidating(true);
    DocumentBuilder db = dbf.newDocumentBuilder();
    InputSource sourceXML = new InputSource(new StringReader(strXML));
    Document xmlDoc = db.parse(sourceXML);
    Element e = xmlDoc.getDocumentElement();
    e.normalize();
    return e;
}

public static final void prettyPrint(Node xml, OutputStream out)
    throws TransformerConfigurationException, TransformerFactoryConfigurationError, TransformerException {
    Transformer tf = TransformerFactory.newInstance().newTransformer();
    tf.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
    tf.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
    tf.setOutputProperty(OutputKeys.INDENT, "yes");
    tf.transform(new DOMSource(xml), new StreamResult(out));
}

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。