[英]regular expression to replace two (or more) consecutive characters by only one?
In java, which regular expression can be used to replace these, for example: 在java中,可以使用哪个正则表达式来替换它们,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!
In perl 在perl
s/(.)\1+/$1/g;
Does the trick, I assume if java has perl compatible regexps it should work too. 诀窍,我假设如果java有perl兼容的regexp它也应该工作。
Edit: Here is what it means 编辑:这就是它的含义
s {
(.) # match any charater ( and capture it )
\1 # if it is followed by itself
+ # One or more times
}{$1}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:正如其他人所指出的,Java中的语法会变成
original.replaceAll("(.)\\1+", "$1");
remember to escape the \\1 记得逃避\\ 1
String a = "aaabbb";
String b = a.replaceAll("(.)\\1+", "$1");
System.out.println("'" + a + "' -> '" + b + "'");
"14442345".replaceAll("(.)\\1+", "$1");
originalString.replaceAll( "(.)\\1+", "$1" );
match pattern (in Java/languages where \\ must be escaped): 匹配模式 (在Java /必须转义\\的语言中):
(.)\\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\\作为转义字符的字符串的语言中)
(.)\1+
replacement : 替换 :
$1
在TextEdit中(假设posix表达式)find:[a] + [b] +替换为:ab
In Perl: 在Perl中:
tr/a-z0-9//s;
Example: 例:
$ perl -E'@a = (aaabbb, 14442345); for(@a) { tr/a-z0-9//s; say }'
ab
142345
If Java has no tr
analog then: 如果Java没有
tr
模拟那么:
s/(.)\1+/$1/sg;
#NOTE: `s` modifier. It takes into account consecutive newlines.
Example: 例:
$ perl -E'@a = (aaabbb, 14442345); for(@a) { s/(.)\1+/$1/sg; say }'
ab
142345
Sugared with a Java 7 : Named Groups 添加了Java 7:命名组
static String cleanDuplicates(@NonNull final String val) {
assert val != null;
return val.replaceAll("(?<dup>.)\\k<dup>+","${dup}");
}
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