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
java有一个静态方法U.formatXml(string)。生活的例子
import com.github.underscore.U;
public class MyClass {
public static void main(String args[]) {
String xml = "<tag><nested>hello</nested></tag>";
System.out.println(U.formatXml("<?xml version=\"1.0\" encoding=\"UTF-8\"?><root>" + xml + "</root>"));
}
}
输出:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<tag>
<nested>hello</nested>
</tag>
</root>
其他回答
java有一个静态方法U.formatXml(string)。生活的例子
import com.github.underscore.U;
public class MyClass {
public static void main(String args[]) {
String xml = "<tag><nested>hello</nested></tag>";
System.out.println(U.formatXml("<?xml version=\"1.0\" encoding=\"UTF-8\"?><root>" + xml + "</root>"));
}
}
输出:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<tag>
<nested>hello</nested>
</tag>
</root>
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();
}
我过去使用org.dom4j.io.OutputFormat.createPrettyPrint()方法打印过
public String prettyPrint(final String xml){
if (StringUtils.isBlank(xml)) {
throw new RuntimeException("xml was null or blank in prettyPrint()");
}
final StringWriter sw;
try {
final OutputFormat format = OutputFormat.createPrettyPrint();
final org.dom4j.Document document = DocumentHelper.parseText(xml);
sw = new StringWriter();
final XMLWriter writer = new XMLWriter(sw, format);
writer.write(document);
}
catch (Exception e) {
throw new RuntimeException("Error pretty printing xml:\n" + xml, e);
}
return sw.toString();
}
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());
}
