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
关于“您必须首先构建DOM树”的评论:不，您不需要也不应该这样做。
相反，创建一个StreamSource(new StreamSource(new StringReader(str))，并将其提供给前面提到的标识转换器。这将使用SAX解析器，结果将快得多。 在这种情况下，构建中间树纯粹是开销。 否则，排名第一的答案是好的。
其他回答
我用Scala看到了一个答案，所以这里有另一个用Groovy的答案，以防有人觉得有趣。默认缩进为2步，XmlNodePrinter构造函数也可以传递另一个值。
def xml = "<tag><nested>hello</nested></tag>"
def stringWriter = new StringWriter()
def node = new XmlParser().parseText(xml);
new XmlNodePrinter(new PrintWriter(stringWriter)).print(node)
println stringWriter.toString()
如果groovy jar在类路径中，则使用Java
String xml = "<tag><nested>hello</nested></tag>";
StringWriter stringWriter = new StringWriter();
Node node = new XmlParser().parseText(xml);
new XmlNodePrinter(new PrintWriter(stringWriter)).print(node);
System.out.println(stringWriter.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));
}
}
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();
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());
}
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();
}
