匹配跳过字符的出现次数

Question

我试图匹配一个模式，它是三个 y，两边都有一个数字：

1yyy5

上面的例子是可行的：

\d{1}y{3}\d{1}

现在，如果我在 y 之一之间添加一个额外的字符，它会失败：

1yyay5

如何使用 {}（或其他东西？）来匹配单个数字之间的出现，即使它们不是连续的顺序？两个数字之间可以有无限多的字符，只要在

之间恰好存在三个y

期望的结果：

 1yyy5 //should match because three y between 2 numbers  

 1yyaaay5 // should match because there are three y between two numbers 

 3..!y3777 // would fail, only one y

 ..@#9naymnymmmyptjr8 // pass, there are exactly 3 y between 9 and 8

 1yyyy2  /fail, 1 to many y. must be exactly 3

Answer 1

这个可以胜任：

\d(?:[^y\d]*y){3}[^y\d]*\d

解释：

\d          # a digit.
(?:         # start non capture group.
  [^y\d]*   # 0 or more non y or digit.
  y         # 1 y.
){3}        # end group, must appear 3 times.
[^y\d]*     # 0 or more non y or digit.
\d          # a digit.

匹配跳过字符的出现次数

Matching number of occurrences skipping chars

regex

pcre