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
试试这个:
try
{
TransformerFactory transFactory = TransformerFactory.newInstance();
Transformer transformer = null;
transformer = transFactory.newTransformer();
StringWriter buffer = new StringWriter();
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.transform(new DOMSource(element),
new StreamResult(buffer));
String str = buffer.toString();
System.out.println("XML INSIDE IS #########################################"+str);
return element;
}
catch (TransformerConfigurationException e)
{
e.printStackTrace();
}
catch (TransformerException e)
{
e.printStackTrace();
}
其他回答
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。)
试试这个:
try
{
TransformerFactory transFactory = TransformerFactory.newInstance();
Transformer transformer = null;
transformer = transFactory.newTransformer();
StringWriter buffer = new StringWriter();
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.transform(new DOMSource(element),
new StreamResult(buffer));
String str = buffer.toString();
System.out.println("XML INSIDE IS #########################################"+str);
return element;
}
catch (TransformerConfigurationException e)
{
e.printStackTrace();
}
catch (TransformerException e)
{
e.printStackTrace();
}
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>
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)));
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());
}
