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
使用jdom2: http://www.jdom.org/
import java.io.StringReader;
import org.jdom2.input.SAXBuilder;
import org.jdom2.output.Format;
import org.jdom2.output.XMLOutputter;
String prettyXml = new XMLOutputter(Format.getPrettyFormat()).
outputString(new SAXBuilder().build(new StringReader(uglyXml)));
其他回答
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……
使用jdom2: http://www.jdom.org/
import java.io.StringReader;
import org.jdom2.input.SAXBuilder;
import org.jdom2.output.Format;
import org.jdom2.output.XMLOutputter;
String prettyXml = new XMLOutputter(Format.getPrettyFormat()).
outputString(new SAXBuilder().build(new StringReader(uglyXml)));
以上所有的解决方案都不适合我，然后我找到了这个http://myshittycode.com/2014/02/10/java-properly-indenting-xml-string/
线索就是用XPath删除空格
String xml = "<root>" +
"\n " +
"\n<name>Coco Puff</name>" +
"\n <total>10</total> </root>";
try {
Document document = DocumentBuilderFactory.newInstance()
.newDocumentBuilder()
.parse(new InputSource(new ByteArrayInputStream(xml.getBytes("utf-8"))));
XPath xPath = XPathFactory.newInstance().newXPath();
NodeList nodeList = (NodeList) xPath.evaluate("//text()[normalize-space()='']",
document,
XPathConstants.NODESET);
for (int i = 0; i < nodeList.getLength(); ++i) {
Node node = nodeList.item(i);
node.getParentNode().removeChild(node);
}
Transformer transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");
StringWriter stringWriter = new StringWriter();
StreamResult streamResult = new StreamResult(stringWriter);
transformer.transform(new DOMSource(document), streamResult);
System.out.println(stringWriter.toString());
}
catch (Exception e) {
e.printStackTrace();
}
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
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();
}
