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

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。)

其他回答

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());
}

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……

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

下面是一种使用dom4j的方法:

进口:

import org.dom4j.Document;  
import org.dom4j.DocumentHelper;  
import org.dom4j.io.OutputFormat;  
import org.dom4j.io.XMLWriter;

代码:

String xml = "<your xml='here'/>";  
Document doc = DocumentHelper.parseText(xml);  
StringWriter sw = new StringWriter();  
OutputFormat format = OutputFormat.createPrettyPrint();  
XMLWriter xw = new XMLWriter(sw, format);  
xw.write(doc);  
String result = sw.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));
    }

}