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
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());
其他回答
下面的代码工作得很好
import javax.xml.transform.OutputKeys;
import javax.xml.transform.Source;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;
String formattedXml1 = prettyFormat("<root><child>aaa</child><child/></root>");
public static String prettyFormat(String input) {
return prettyFormat(input, "2");
}
public static String prettyFormat(String input, String indent) {
Source xmlInput = new StreamSource(new StringReader(input));
StringWriter stringWriter = new StringWriter();
try {
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", indent);
transformer.transform(xmlInput, new StreamResult(stringWriter));
String pretty = stringWriter.toString();
pretty = pretty.replace("\r\n", "\n");
return pretty;
} catch (Exception e) {
throw new RuntimeException(e);
}
}
凯文·哈肯森说: 但是，如果您知道您的XML字符串是有效的，并且您不想引起将字符串解析为DOM的内存开销，然后在DOM上运行转换以获得字符串—您可以通过字符解析进行一些老式的字符。在每个字符后插入换行符和空格，保持和缩进计数器(以确定空格的数量)，为每个<…>和递减你看到的每一个。”
同意了。这种方法要快得多，依赖关系也少得多。
示例解决方案:
/**
* XML utils, including formatting.
*/
public class XmlUtils
{
private static XmlFormatter formatter = new XmlFormatter(2, 80);
public static String formatXml(String s)
{
return formatter.format(s, 0);
}
public static String formatXml(String s, int initialIndent)
{
return formatter.format(s, initialIndent);
}
private static class XmlFormatter
{
private int indentNumChars;
private int lineLength;
private boolean singleLine;
public XmlFormatter(int indentNumChars, int lineLength)
{
this.indentNumChars = indentNumChars;
this.lineLength = lineLength;
}
public synchronized String format(String s, int initialIndent)
{
int indent = initialIndent;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++)
{
char currentChar = s.charAt(i);
if (currentChar == '<')
{
char nextChar = s.charAt(i + 1);
if (nextChar == '/')
indent -= indentNumChars;
if (!singleLine) // Don't indent before closing element if we're creating opening and closing elements on a single line.
sb.append(buildWhitespace(indent));
if (nextChar != '?' && nextChar != '!' && nextChar != '/')
indent += indentNumChars;
singleLine = false; // Reset flag.
}
sb.append(currentChar);
if (currentChar == '>')
{
if (s.charAt(i - 1) == '/')
{
indent -= indentNumChars;
sb.append("\n");
}
else
{
int nextStartElementPos = s.indexOf('<', i);
if (nextStartElementPos > i + 1)
{
String textBetweenElements = s.substring(i + 1, nextStartElementPos);
// If the space between elements is solely newlines, let them through to preserve additional newlines in source document.
if (textBetweenElements.replaceAll("\n", "").length() == 0)
{
sb.append(textBetweenElements + "\n");
}
// Put tags and text on a single line if the text is short.
else if (textBetweenElements.length() <= lineLength * 0.5)
{
sb.append(textBetweenElements);
singleLine = true;
}
// For larger amounts of text, wrap lines to a maximum line length.
else
{
sb.append("\n" + lineWrap(textBetweenElements, lineLength, indent, null) + "\n");
}
i = nextStartElementPos - 1;
}
else
{
sb.append("\n");
}
}
}
}
return sb.toString();
}
}
private static String buildWhitespace(int numChars)
{
StringBuilder sb = new StringBuilder();
for (int i = 0; i < numChars; i++)
sb.append(" ");
return sb.toString();
}
/**
* Wraps the supplied text to the specified line length.
* @lineLength the maximum length of each line in the returned string (not including indent if specified).
* @indent optional number of whitespace characters to prepend to each line before the text.
* @linePrefix optional string to append to the indent (before the text).
* @returns the supplied text wrapped so that no line exceeds the specified line length + indent, optionally with
* indent and prefix applied to each line.
*/
private static String lineWrap(String s, int lineLength, Integer indent, String linePrefix)
{
if (s == null)
return null;
StringBuilder sb = new StringBuilder();
int lineStartPos = 0;
int lineEndPos;
boolean firstLine = true;
while(lineStartPos < s.length())
{
if (!firstLine)
sb.append("\n");
else
firstLine = false;
if (lineStartPos + lineLength > s.length())
lineEndPos = s.length() - 1;
else
{
lineEndPos = lineStartPos + lineLength - 1;
while (lineEndPos > lineStartPos && (s.charAt(lineEndPos) != ' ' && s.charAt(lineEndPos) != '\t'))
lineEndPos--;
}
sb.append(buildWhitespace(indent));
if (linePrefix != null)
sb.append(linePrefix);
sb.append(s.substring(lineStartPos, lineEndPos + 1));
lineStartPos = lineEndPos + 1;
}
return sb.toString();
}
// other utils removed for brevity
}
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
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。
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>
