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
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
其他回答
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();
}
我把它们混合在一起，写了一个小程序。它从xml文件中读取并打印出来。而不是xzy给出你的文件路径。
public static void main(String[] args) throws Exception {
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setValidating(false);
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(new FileInputStream(new File("C:/Users/xyz.xml")));
prettyPrint(doc);
}
private static String prettyPrint(Document document)
throws TransformerException {
TransformerFactory transformerFactory = TransformerFactory
.newInstance();
Transformer transformer = transformerFactory.newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "2");
transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "no");
DOMSource source = new DOMSource(document);
StringWriter strWriter = new StringWriter();
StreamResult result = new StreamResult(strWriter);transformer.transform(source, result);
System.out.println(strWriter.getBuffer().toString());
return strWriter.getBuffer().toString();
}
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();
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……
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。)
