大家都知道split方法是将一个字符串作为this的分隔符来传入方法中,而返回值是一个字符串数组。
而总有一些不安分的一些人,拿一些不常用符号当这个分隔符为“ . $ | ( ) [ { ^ ? * + \” 这些符合时,这个方法就会出现一个小问题。在split方法中的正则表达式中有可能会搜索不到这些特殊符号,导致拆分失败。
第一种情况:所有字符分开,每个字符为一个元素
第二种情况:字符串没有被分割
解决方案,给分隔符添加转义字符
当然了,最好的解决方法是不适用这些特殊的符号,并且split方法的源码的注释中已经声明了这些特殊情况
public String[] split(String regex, int limit) {
/* fastpath if the regex is a
(1)one-char String and this character is not one of the
RegEx's meta characters ".$|()[{^?*+\\", or
(2)two-char String and the first char is the backslash and
the second is not the ascii digit or ascii letter.
*/
char ch = 0;
if (((regex.value.length == 1 &&
".$|()[{^?*+\\".indexOf(ch = regex.charAt(0)) == -1) ||
(regex.length() == 2 &&
regex.charAt(0) == '\\' &&
(((ch = regex.charAt(1))-'0')|('9'-ch)) < 0 &&
((ch-'a')|('z'-ch)) < 0 &&
((ch-'A')|('Z'-ch)) < 0)) &&
(ch < Character.MIN_HIGH_SURROGATE ||
ch > Character.MAX_LOW_SURROGATE))
{
int off = 0;
int next = 0;
boolean limited = limit > 0;
ArrayList<String> list = new ArrayList<>();
while ((next = indexOf(ch, off)) != -1) {
if (!limited || list.size() < limit - 1) {
list.add(substring(off, next));
off = next + 1;
} else { // last one
//assert (list.size() == limit - 1);
list.add(substring(off, value.length));
off = value.length;
break;
}
}
// If no match was found, return this
if (off == 0)
return new String[]{this};
// Add remaining segment
if (!limited || list.size() < limit)
list.add(substring(off, value.length));
// Construct result
int resultSize = list.size();
if (limit == 0) {
while (resultSize > 0 && list.get(resultSize - 1).length() == 0) {
resultSize--;
}
}
String[] result = new String[resultSize];
return list.subList(0, resultSize).toArray(result);
}
return Pattern.compile(regex).split(this, limit);
}