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>
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
}
如果使用第三方XML库是可行的,那么您可以使用一些比目前票数最高的答案所建议的要简单得多的方法。
它声明输入和输出都应该是字符串,所以这里有一个实用程序方法,用XOM库实现:
import nu.xom.*;
import java.io.*;
[...]
public static String format(String xml) throws ParsingException, IOException {
ByteArrayOutputStream out = new ByteArrayOutputStream();
Serializer serializer = new Serializer(out);
serializer.setIndent(4); // or whatever you like
serializer.write(new Builder().build(xml, ""));
return out.toString("UTF-8");
}
我对它进行了测试,结果不依赖于JRE版本或类似的东西。要了解如何根据自己的喜好定制输出格式,请查看Serializer API。
这实际上比我想象的要长——需要一些额外的行,因为Serializer想要写入一个OutputStream。但是请注意,这里很少有用于实际XML处理的代码。
(这个答案是我对XOM的评估的一部分,在我关于替代dom4j的最佳Java XML库的问题中,XOM被建议作为一个选项。在dom4j中,您可以使用XMLWriter和OutputFormat轻松实现这一点。编辑:…正如mlo55的答案所示。)