简单的正则表达式匹配包含<n>字符的字符串

Question

我正在编写这个正则表达式,因为我需要一种方法来查找没有n点的字符串,我虽然负向前看将是最好的选择,到目前为止我的正则表达式是:

"^(?!\\.{3})$"

我读这个的方式是,在字符串的开始和结束之间,可以有多于或少于3个点而不是3.令人惊讶的是,这不匹配,hello.here.im.greetings 而是我期望匹配.我正在用Java写作,所以它的Perl就像味道一样,我没有逃避花括号,因为它不需要Java任何建议？

Answer 1

你走在正确的轨道上:

"^(?!(?:[^.]*\\.){3}[^.]*$)"

将按预期工作.

你的正则表达式意味着

^          # Match the start of the string
(?!\\.{3}) # Make sure that there aren't three dots at the current position
$          # Match the end of the string

所以它只能匹配空字符串.

我的正则表达式意味着:

^       # Match the start of the string
(?!     # Make sure it's impossible to match...
 (?:    # the following:
  [^.]* # any number of characters except dots
  \\.   # followed by a dot
 ){3}   # exactly three times.
 [^.]*  # Now match only non-dot characters
 $      # until the end of the string.
)       # End of lookahead

使用方法如下:

Pattern regex = Pattern.compile("^(?!(?:[^.]*\\.){3}[^.]*$)");
Matcher regexMatcher = regex.matcher(subjectString);
foundMatch = regexMatcher.find();