CODEWITHSUNDEEP
regexgrep
dstarh
dstarh

Reputation: 5086

Regex/grep to pull dot path out of string

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,}

Upvotes: 0

Views: 232

Answers (3)

Lev Levitsky
Lev Levitsky

Reputation: 65851

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).

Upvotes: 0

Diego
Diego

Reputation: 16714

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.

Upvotes: 0

Andrew Clark
Andrew Clark

Reputation: 208635

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

Upvotes: 1

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