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。
喜欢
line.replace(regExp,"");
使用什么样的regExp？
无效的XML字符是不是这个的所有内容：
[#x1-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF]
感谢。
答案 0 :(得分：73)
Java's regex supports supplementary characters，因此您可以使用两个UTF-16编码的字符指定这些高范围。
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"
+ "]";
大多数人都想要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"
+ "]+";
您需要使用String.replaceAll(...)而不是String.replace(...)。
String illegal = "Hello, World!\0";
String legal = illegal.replaceAll(pattern, "");
答案 1 :(得分：6)
我们应该考虑替代字符吗？否则'（当前＆gt; = 0x10000）＆amp;＆amp; （当前＆lt; = 0x10FFFF）'永远不会成真。
还测试了正则表达式看起来比以下循环慢。
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));
}
}
}
答案 2 :(得分：2)
Jun的解决方案，简化了。使用StringBuffer#appendCodePoint(int)，我不需要char current或String#charAt(int)。我可以通过检查codePoint是否大于0xFFFF来判断代理对。
（没有必要使用i ++，因为低代理不会通过过滤器。但是然后人们会重新使用不同代码点的代码，它会失败。我更喜欢编程到黑客。 ）
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);
}
}
答案 3 :(得分：2)
到目前为止，所有这些答案只能取代角色本身。但有时XML文档会有无效的XML实体序列导致错误。例如，如果xml中有，则java xml解析器将抛出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;
}
答案 4 :(得分：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();
}
答案 5 :(得分：0)
来自Best way to encode text data for XML in Java?
<Component onChange={this.handleOnchange}/>
handleOnChnage(value){
//instead of this.setState({zoom:value});
this.setZoomState(value);//where I check the value and make the constaints
}
答案 6 :(得分：0)
如果要以类似XML的形式存储带有禁止字符的文本元素，则可以使用XPL。 dev-kit为XML和XML处理提供了并发XPL - 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（名称空间）的全部功能，则可以使用XPL。
答案 7 :(得分：0)
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;
}
答案 8 :(得分：-2)
我相信以下文章可能会对您有所帮助。
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。