Parsec how to find "matches" within a string

Question

How can I use parsec to parse all matched input in a string and discard the rest?

Example: I have a simple number parser, and I can find all the numbers if I know what separates them:

num :: Parser Int
num = read <$> many digit

parse (num `sepBy` space) "" "111 4 22"

But what if I don't know what is between the numbers?

"I will live to be 111 years <b>old</b> if I work out 4 days a week starting at 22."

many anyChar doesn't work as a separator, because it consumes everything.

So how can I get things that match an arbitrary parser surrounded by things I want to ignore?

EDIT: Note that in the real problem, my parser is more complicated:

optionTag :: Parser Fragment
optionTag = do
    string "<option"
    manyTill anyChar (string "value=")
    n <- many1 digit
    manyTill anyChar (char '>')
    chapterPrefix
    text <- many1 (noneOf "<>")
    return $ Option (read n) text
  where
    chapterPrefix = many digit >> char '.' >> many space

By the way, do you really need to choose Parsec for this task? Could you use a simple regular expression like `^$`? — Yuuri, Apr 10 '15 at 08:22

AJF · Accepted Answer · 2016-12-24T23:00:49.607

8

For an arbitrary parser myParser, it's quite easy:

solution = many (let one = myParser <|> (anyChar >> one) in one)

It might be clearer to write it this way:

solution = many loop
    where 
        loop = myParser <|> (anyChar >> loop)

Essentially, this defines a recursive parser (called loop) that will continue searching for the first thing that can be parsed by myParser. many will simply search exhaustively until failure, ie: EOF.

edited Dec 24 '16 at 23:00

answered Apr 10 '15 at 17:38

AJF

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This answer is close, but you have to backtrack `myParser` if it consumes some input and fails, as @neil-smith notes in his answer. Also you have to consider if `myParser` succeeds without consuming any input, because then this will loop forever. [`sepCap`](https://hackage.haskell.org/package/replace-megaparsec/docs/Replace-Megaparsec.html#v:sepCap) will handle these and other subtleties. – James Brock Aug 28 '19 at 01:01

score 2 · Answer 2 · answered Apr 09 '15 at 21:39

2

You can use

 many ( noneOf "0123456789")

i'm not sure about "noneOf" and "digit" types but you can give e try also to

many $ noneOf digit

answered Apr 09 '15 at 21:39

Gabriel Ciubotaru

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I guess you need to do it after you extract the – Gabriel Ciubotaru Apr 10 '15 at 09:28
That's very useful information. I wasn't sure if parsed was the right tool for this. Thanks for your help! – Sean Clark Hess Apr 10 '15 at 13:43
This is fine and dandy, but working for arbitrary parsers is more important in my eyes. – AJF Apr 10 '15 at 17:39

score 2 · Answer 3 · answered Dec 08 '16 at 14:50

To find the item in the string, the item is either at the start of the string, or consume one character and look for the item in the now-shorter string. If the item isn't right at the start of the string, you'll need to un-consume the characters used while looking for it, so you'll need a try block.

hasItem = prefixItem <* (many anyChar)
preafixItem = (try item) <|> (anyChar >> prefixItem)
item = <parser for your item here>

This code looks for just one occurrence of item in the string.

(AJFarmar almost has it.)

James Brock · Answer 4 · 2019-08-27T14:36:01.883

The replace-megaparsec package allows you to split up a string into sections which match your pattern and sections which don't match by using the sepCap parser combinator.

import Replace.Megaparsec
import Text.Megaparsec
import Text.Megaparsec.Char

let num :: Parsec Void String Int
    num = read <$> many digitChar

>>> parseTest (sepCap num) "I will live to be 111 years <b>old</b> if I work out 4 days a week starting at 22."
[Left "I will live to be "
,Right 111
,Left " years <b>old</b> if I work out "
,Right 4
,Left " days a week starting at "
,Right 22
,Left "."
]

Parsec how to find "matches" within a string

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