CODEWITHSUNDEEP
haskell
JSchwartz
JSchwartz

Reputation: 2734

Reading string data into haskell data type?

Short-and-sweet, I have a text file that looks like this:

ID1|ID2|DATE|SUM
0|0|20/03/2014|100.00
0|1|20/04/2014|99.00

I have a custom data type that looks like this:

data DBData = DBData { id1   :: Int
                     , id2   :: Int
                     , date  :: String
                     , sum   :: Int
                     } deriving (Eq, Read, Show)

How do I get this into that?

What I have been toying with so far is using something like this:

parseRow :: [String] -> DBData 
parseRow = let (id1:id2:date:sum) = (splitWhen (=='|')) s
                i = read id1
             in DBData {id1 = i}

But I can't seem to get the syntax right ...

related to my other post: https://stackoverflow.com/questions/25477554/using-splitwhen-to-split-string-by-delimiter-and-trying-to-figure-out-how-to-sto

Upvotes: 1

Views: 1172

Answers (1)

David Young
David Young

Reputation: 10793

In Haskell, all data constructors and concrete types must begin with a capital letter:

data DbData = DbData ...

Also, if you want to use the read method to read in something in a custom format like that, you must make a the data type an instance of Read:

instance Read DbData where
  read s = ...

Inside the instance definition you can define read as you would any other Haskell function.

Also, the proper let syntax (outside a do block) is

let binding = val
in
... body ...

When you create a data type, you (usually) make one or more constructors that you can use to make values of that type. Here is an example that is sort of similar to yours

data Example = Example { a :: Int
                       , b :: Char
                       , c :: String
                       }

We can make a value of type Example using the Example constructor (note that these don't need to have the same name):

exampleValue :: Example
example = Example 1 'z' "abcdef"

Upvotes: 4

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