Java 正则表达式仅用一个替换两个（或多个）连续字符？
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regular expression to replace two (or more) consecutive characters by only one?
提问by Alex. S.
In java, which regular expression can be used to replace these, for example:
在java中，可以用哪些正则表达式来代替这些，例如：
before: aaabbb after: ab
之前：aaabbb 之后：ab
before: 14442345 after: 142345
之前：14442345 之后：142345
thanks!
谢谢！
采纳答案by Pat
In perl
在 perl 中
s/(.)+//g;
Does the trick, I assume if java has perl compatible regexps it should work too.
有什么技巧，我假设如果 java 有 perl 兼容的正则表达式它也应该工作。
Edit: Here is what it means
编辑：这就是它的意思
s {
(.) # match any charater ( and capture it )
# if it is followed by itself
+ # One or more times
}{}gx; # And replace the whole things by the first captured character (with g modifier to replace all occurences)
Edit: As others have pointed out, the syntax in Java would become
编辑：正如其他人指出的那样，Java 中的语法将变成
original.replaceAll("(.)\1+", "");
remember to escape the \1
记得要转义 \1
回答by Jorge Ferreira
String a = "aaabbb";
String b = a.replaceAll("(.)\1+", "");
System.out.println("'" + a + "' -> '" + b + "'");
回答by Matthew Crumley
"14442345".replaceAll("(.)\1+", "");
回答by Jeff Hillman
originalString.replaceAll( "(.)\1+", "" );
回答by Jeff Hillman
in TextEdit (assuming posix expressions) find: [a]+[b]+ replace with: ab
在 TextEdit（假设 posix 表达式）中找到：[a]+[b]+ 替换为：ab
回答by Imran
match pattern(in Java/languages where \ must be escaped):
匹配模式（在 Java/语言中 \ 必须被转义）：
(.)\1+
or (in languages where you can use strings which don't treat \ as escape character)
或（在您可以使用不将 \ 视为转义字符的字符串的语言中）
(.)+
replacement:
更换：
回答by jfs
In Perl:
在 Perl 中：
tr/a-z0-9//s;
Example:
例子：
$ perl -E'@a = (aaabbb, 14442345); for(@a) { tr/a-z0-9//s; say }'
ab
142345
If Java has no tranalog then:
如果 Java 没有tr模拟，那么：
s/(.)+//sg;
#NOTE: `s` modifier. It takes into account consecutive newlines.
Example:
例子：
$ perl -E'@a = (aaabbb, 14442345); for(@a) { s/(.)+//sg; say }'
ab
142345
回答by kunwar.sangram
Sugared with a Java 7 : Named Groups
用 Java 7 加糖：命名组
static String cleanDuplicates(@NonNull final String val) {
assert val != null;
return val.replaceAll("(?<dup>.)\k<dup>+","${dup}");
}
