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scala:
import xml._
val xml = XML.loadString("<tag><nested>hello</nested></tag>")
val formatted = new PrettyPrinter(150, 2).format(xml)
println(formatted)
如果你依赖scala-library.jar，你也可以在Java中这样做。它是这样的:
import scala.xml.*;
public class FormatXML {
public static void main(String[] args) {
String unformattedXml = "<tag><nested>hello</nested></tag>";
PrettyPrinter pp = new PrettyPrinter(150, 3);
String formatted = pp.format(XML.loadString(unformattedXml), TopScope$.MODULE$);
System.out.println(formatted);
}
}
PrettyPrinter对象是用两个整数构造的，第一个是最大行长，第二个是缩进步骤。
其他回答
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());
}
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();
}
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();
}
对于那些寻找快速和肮脏的解决方案的人——它不需要XML是100%有效的。例如，在REST / SOAP日志的情况下(你永远不知道其他人发送了什么;-))
我发现并改进了一个我在网上找到的代码剪辑，我认为这仍然是一个有效的可能的方法:
public static String prettyPrintXMLAsString(String xmlString) {
/* Remove new lines */
final String LINE_BREAK = "\n";
xmlString = xmlString.replaceAll(LINE_BREAK, "");
StringBuffer prettyPrintXml = new StringBuffer();
/* Group the xml tags */
Pattern pattern = Pattern.compile("(<[^/][^>]+>)?([^<]*)(</[^>]+>)?(<[^/][^>]+/>)?");
Matcher matcher = pattern.matcher(xmlString);
int tabCount = 0;
while (matcher.find()) {
String str1 = (null == matcher.group(1) || "null".equals(matcher.group())) ? "" : matcher.group(1);
String str2 = (null == matcher.group(2) || "null".equals(matcher.group())) ? "" : matcher.group(2);
String str3 = (null == matcher.group(3) || "null".equals(matcher.group())) ? "" : matcher.group(3);
String str4 = (null == matcher.group(4) || "null".equals(matcher.group())) ? "" : matcher.group(4);
if (matcher.group() != null && !matcher.group().trim().equals("")) {
printTabs(tabCount, prettyPrintXml);
if (!str1.equals("") && str3.equals("")) {
++tabCount;
}
if (str1.equals("") && !str3.equals("")) {
--tabCount;
prettyPrintXml.deleteCharAt(prettyPrintXml.length() - 1);
}
prettyPrintXml.append(str1);
prettyPrintXml.append(str2);
prettyPrintXml.append(str3);
if (!str4.equals("")) {
prettyPrintXml.append(LINE_BREAK);
printTabs(tabCount, prettyPrintXml);
prettyPrintXml.append(str4);
}
prettyPrintXml.append(LINE_BREAK);
}
}
return prettyPrintXml.toString();
}
private static void printTabs(int count, StringBuffer stringBuffer) {
for (int i = 0; i < count; i++) {
stringBuffer.append("\t");
}
}
public static void main(String[] args) {
String x = new String(
"<soap:Envelope xmlns:soap=\"http://schemas.xmlsoap.org/soap/envelope/\"><soap:Body><soap:Fault><faultcode>soap:Client</faultcode><faultstring>INVALID_MESSAGE</faultstring><detail><ns3:XcbSoapFault xmlns=\"\" xmlns:ns3=\"http://www.someapp.eu/xcb/types/xcb/v1\"><CauseCode>20007</CauseCode><CauseText>INVALID_MESSAGE</CauseText><DebugInfo>Problems creating SAAJ object model</DebugInfo></ns3:XcbSoapFault></detail></soap:Fault></soap:Body></soap:Envelope>");
System.out.println(prettyPrintXMLAsString(x));
}
输出如下:
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
<soap:Body>
<soap:Fault>
<faultcode>soap:Client</faultcode>
<faultstring>INVALID_MESSAGE</faultstring>
<detail>
<ns3:XcbSoapFault xmlns="" xmlns:ns3="http://www.someapp.eu/xcb/types/xcb/v1">
<CauseCode>20007</CauseCode>
<CauseText>INVALID_MESSAGE</CauseText>
<DebugInfo>Problems creating SAAJ object model</DebugInfo>
</ns3:XcbSoapFault>
</detail>
</soap:Fault>
</soap:Body>
</soap:Envelope>
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());
