您好我想从字符串中删除所有无效的 XML 字符。我想使用带有 string.replace 方法的正则表达式。
像
line.replace(regExp,"");
什么是正确的正则表达式？
无效的 XML 字符是不是这样的所有内容：
[#x1-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF]
谢谢。
问问题
82088 次
9 回答
86
Java 的正则表达式支持补充字符，因此您可以使用两个 UTF-16 编码的字符来指定这些高范围。
以下是删除XML 1.0中非法字符的模式：
// XML 1.0
// #x9 | #xA | #xD | [#x20-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF]
String xml10pattern = "[^"
+ "\u0009\r\n"
+ "\u0020-\uD7FF"
+ "\uE000-\uFFFD"
+ "\ud800\udc00-\udbff\udfff"
+ "]";
大多数人会想要 XML 1.0 版本。
以下是删除XML 1.1中非法字符的模式：
// XML 1.1
// [#x1-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF]
String xml11pattern = "[^"
+ "\u0001-\uD7FF"
+ "\uE000-\uFFFD"
+ "\ud800\udc00-\udbff\udfff"
+ "]+";
您将需要使用String.replaceAll(...)而不是String.replace(...).
String illegal = "Hello, World!\0";
String legal = illegal.replaceAll(pattern, "");
于 2010-11-21T12:58:40.870 回答
10
我们应该考虑代理字符吗？否则 '(current >= 0x10000) && (current <= 0x10FFFF)' 永远不会为真。
还测试了正则表达式方式似乎比以下循环慢。
if (null == text || text.isEmpty()) {
return text;
}
final int len = text.length();
char current = 0;
int codePoint = 0;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < len; i++) {
current = text.charAt(i);
boolean surrogate = false;
if (Character.isHighSurrogate(current)
&& i + 1 < len && Character.isLowSurrogate(text.charAt(i + 1))) {
surrogate = true;
codePoint = text.codePointAt(i++);
} else {
codePoint = current;
}
if ((codePoint == 0x9) || (codePoint == 0xA) || (codePoint == 0xD)
|| ((codePoint >= 0x20) && (codePoint <= 0xD7FF))
|| ((codePoint >= 0xE000) && (codePoint <= 0xFFFD))
|| ((codePoint >= 0x10000) && (codePoint <= 0x10FFFF))) {
sb.append(current);
if (surrogate) {
sb.append(text.charAt(i));
}
}
}
于 2012-07-26T15:31:56.187 回答
8
到目前为止，所有这些答案都只替换了字符本身。但有时 XML 文档会包含无效的 XML 实体序列，从而导致错误。例如，如果您的 xml 中有 java xml 解析器，则会抛出Illegal character entity: expansion character (code 0x2 at ....
这是一个简单的java程序，可以替换那些无效的实体序列。
public final Pattern XML_ENTITY_PATTERN = Pattern.compile("\\&\\#(?:x([0-9a-fA-F]+)|([0-9]+))\\;");
/**
* Remove problematic xml entities from the xml string so that you can parse it with java DOM / SAX libraries.
*/
String getCleanedXml(String xmlString) {
Matcher m = XML_ENTITY_PATTERN.matcher(xmlString);
Set<String> replaceSet = new HashSet<>();
while (m.find()) {
String group = m.group(1);
int val;
if (group != null) {
val = Integer.parseInt(group, 16);
if (isInvalidXmlChar(val)) {
replaceSet.add("&#x" + group + ";");
}
} else if ((group = m.group(2)) != null) {
val = Integer.parseInt(group);
if (isInvalidXmlChar(val)) {
replaceSet.add("&#" + group + ";");
}
}
}
String cleanedXmlString = xmlString;
for (String replacer : replaceSet) {
cleanedXmlString = cleanedXmlString.replaceAll(replacer, "");
}
return cleanedXmlString;
}
private boolean isInvalidXmlChar(int val) {
if (val == 0x9 || val == 0xA || val == 0xD ||
val >= 0x20 && val <= 0xD7FF ||
val >= 0x10000 && val <= 0x10FFFF) {
return false;
}
return true;
}
于 2017-07-20T18:55:37.767 回答
3
君的解决方案，简化。使用StringBuffer#appendCodePoint(int)，我不需要char currentor String#charAt(int)。codePoint我可以通过检查是否大于来判断代理对0xFFFF。
（没有必要做 i++，因为低代理不会通过过滤器。但是随后将代码重新用于不同的代码点并且它会失败。我更喜欢编程而不是黑客攻击。）
StringBuilder sb = new StringBuilder();
for (int i = 0; i < text.length(); i++) {
int codePoint = text.codePointAt(i);
if (codePoint > 0xFFFF) {
i++;
}
if ((codePoint == 0x9) || (codePoint == 0xA) || (codePoint == 0xD)
|| ((codePoint >= 0x20) && (codePoint <= 0xD7FF))
|| ((codePoint >= 0xE000) && (codePoint <= 0xFFFD))
|| ((codePoint >= 0x10000) && (codePoint <= 0x10FFFF))) {
sb.appendCodePoint(codePoint);
}
}
于 2015-02-02T17:33:18.227 回答
2
String xmlData = xmlData.codePoints().filter(c -> isValidXMLChar(c)).collect(StringBuilder::new,
StringBuilder::appendCodePoint, StringBuilder::append).toString();
private boolean isValidXMLChar(int c) {
if((c == 0x9) ||
(c == 0xA) ||
(c == 0xD) ||
((c >= 0x20) && (c <= 0xD7FF)) ||
((c >= 0xE000) && (c <= 0xFFFD)) ||
((c >= 0x10000) && (c <= 0x10FFFF)))
{
return true;
}
return false;
}
于 2018-01-23T09:03:37.273 回答
1
/**
* This method ensures that the output String has only
* valid XML unicode characters as specified by the
* XML 1.0 standard. For reference, please see
* <a href="http://www.w3.org/TR/2000/REC-xml-20001006#NT-Char">the
* standard</a>. This method will return an empty
* String if the input is null or empty.
*
* @param in The String whose non-valid characters we want to remove.
* @return The in String, stripped of non-valid characters.
*/
public static String stripNonValidXMLCharacters(String in) {
StringBuffer out = new StringBuffer(); // Used to hold the output.
char current; // Used to reference the current character.
if (in == null || ("".equals(in))) return ""; // vacancy test.
for (int i = 0; i < in.length(); i++) {
current = in.charAt(i); // NOTE: No IndexOutOfBoundsException caught here; it should not happen.
if ((current == 0x9) ||
(current == 0xA) ||
(current == 0xD) ||
((current >= 0x20) && (current <= 0xD7FF)) ||
((current >= 0xE000) && (current <= 0xFFFD)) ||
((current >= 0x10000) && (current <= 0x10FFFF)))
out.append(current);
}
return out.toString();
}
于 2012-06-05T09:20:15.960 回答
0
String xmlEscapeText(String t) {
StringBuilder sb = new StringBuilder();
for(int i = 0; i < t.length(); i++){
char c = t.charAt(i);
switch(c){
case '<': sb.append("<"); break;
case '>': sb.append(">"); break;
case '\"': sb.append("""); break;
case '&': sb.append("&"); break;
case '\'': sb.append("'"); break;
default:
if(c>0x7e) {
sb.append("&#"+((int)c)+";");
}else
sb.append(c);
}
}
return sb.toString();
}
于 2015-11-10T16:43:44.810 回答
0
如果您想以类似 XML 的形式存储带有禁止字符的文本元素，您可以使用 XPL 代替。开发工具包提供并发 XPL 到 XML 和 XML 处理 - 这意味着从 XPL 到 XML 的转换没有时间成本。或者，如果您不需要 XML（命名空间）的全部功能，您可以只使用 XPL。
于 2017-04-07T13:09:41.320 回答
-1
相信下面的文章可以帮到你。
http://commons.apache.org/lang/api-2.1/org/apache/commons/lang/StringEscapeUtils.html http://www.javapractices.com/topic/TopicAction.do?Id=96
不久，尝试使用 Jakarta 项目中的 StringEscapeUtils。
于 2010-11-21T12:26:00.210 回答