正则表达式/ grep从字符串中拉出点路径

Question

Given the following string which may appear any number of times across various documents and in various different formats, I want to pull out the part between the quotes. I Only want strings which meet the following conditions. Starts and ends with " (between two quotes) and has 1 or more dots (.) within the string.

@CheckWith(value = PasswordCheck.class, message = "validation.password.blah.foo")

The following regex gives me the first three parts of the string validation.password.blah but misses the .foo

(\")([a-zA-Z]{1,}\.{1}){1,}

Answer 1

Try the following:

"([a-zA-Z]{1,}\.[a-zA-Z.]{1,})"

Note that instead of {1,} you can usually use + , but I'm not sure if grep supports that without an extended option.

Explanation:

"                  # match a literal '"' character
(                  # start capture group
  [a-zA-Z]{1,}       # one or more letters
  \.                 # match a literal '.' character
  [a-zA-Z.]{1,}      # one or more letters or '.' characters
)                  # end capture group
"                  # match a literal '"' character

Answer 2

I think the regex you are looking for is this: "[a-zA-Z]+(\\.[a-zA-Z]+)+

You should have in mind that it wont include last " and will only accept letters between the dots.

Answer 3

If you have egrep (or grep -E ), you can use

"([a-zA_Z]+\.)+[a-zA-Z]+"

(quotes are part of the regex, escape if needed).