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
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
其他回答
Transformer transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "2");
// initialize StreamResult with File object to save to file
StreamResult result = new StreamResult(new StringWriter());
DOMSource source = new DOMSource(doc);
transformer.transform(source, result);
String xmlString = result.getWriter().toString();
System.out.println(xmlString);
注意:根据Java版本的不同，结果可能有所不同。搜索特定于您的平台的解决方案。
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
基于这个答案的一个更简单的解决方案:
public static String prettyFormat(String input, int indent) {
try {
Source xmlInput = new StreamSource(new StringReader(input));
StringWriter stringWriter = new StringWriter();
StreamResult xmlOutput = new StreamResult(stringWriter);
TransformerFactory transformerFactory = TransformerFactory.newInstance();
transformerFactory.setAttribute("indent-number", indent);
transformerFactory.setAttribute(XMLConstants.ACCESS_EXTERNAL_DTD, "");
transformerFactory.setAttribute(XMLConstants.ACCESS_EXTERNAL_STYLESHEET, "");
Transformer transformer = transformerFactory.newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.transform(xmlInput, xmlOutput);
return xmlOutput.getWriter().toString();
} catch (Exception e) {
throw new RuntimeException(e); // simple exception handling, please review it
}
}
public static String prettyFormat(String input) {
return prettyFormat(input, 2);
}
testcase:
prettyFormat("<root><child>aaa</child><child/></root>");
返回:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<child>aaa</child>
<child/>
</root>
//忽略:原始编辑只需要在代码中的类名中缺少s。为了在SO上获得超过6个字符的验证，添加了多余的6个字符
如果您确信您有一个有效的XML，那么这个很简单，并且避免了XML DOM树。可能有一些错误，如果你看到任何错误，请评论
public String prettyPrint(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("<?")) {
// xml version tag
pretty.append(row + "\n");
} else if (row.startsWith("</")) {
// closing tag
String indent = repeatString(" ", --stack);
pretty.append(indent + row + "\n");
} else if (row.startsWith("<")) {
// starting tag
String indent = repeatString(" ", stack++);
pretty.append(indent + row + "\n");
} else {
// tag data
String indent = repeatString(" ", stack);
pretty.append(indent + row + "\n");
}
}
return pretty.toString().trim();
}
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>
