how to create JSON string in Erlang manually - json

I am new to Erlang and noticed that there is no native function to create json string from lists (Or is there?). I use this method to create json string in Erlang but do not know if it will not malfunction.
Here is an example of my method:
-module(index).
-export([test/0]).
test() ->
Ma = "Hello World", Mb = "Hello Erlang",
A = "{\"Messages\" : [\"" ++ Ma ++ "\", \""++Mb++"\"], \"Usernames\" : [\"Username1\", \"Username2\"]}",
A.
The result is:
388> test().
"{\"Messages\" : [\"Hello World\", \"Hello Erlang\"], \"Usernames\" : [\"Username1\", \"Username2\"]}"
389>
I think this is the expected result but is there any chance that this method may malfunction when included special characters, such as: <, >, & / \ " ??
What precautions should I take to make this method stronger?

If Ma or Mb contains double quotes or whatever control characters, the parsing from string to JSON will fail. This parsing may never occur in Erlang, as Erlang does not have string to JSON conversion built-in.
It's a good idea to use binaries (<<"I am a binary string">>), as lists consume a lot more resources.
We're using jiffy, which is implemented as a NIF and hence is reasonably fast and it allows for document construction like so:
jiffy:decode(<<"{\"foo\": \"bar\"}">>).
{[{<<"foo">>,<<"bar">>}]}
Doc = {[{foo, [<<"bing">>, 2.3, true]}]}.
{[{foo,[<<"bing">>,2.3,true]}]}
jiffy:encode(Doc).
<<"{\"foo\":[\"bing\",2.3,true]}">>

I had this very same problem, searched high and low and in the end came up with my own method. This is purely just pointing people in the right directions to finding a solution for themselves. Note: I tried jiffy but as I'm using rebar3 it's not currently compatible.
Im using MS sql server so i use the Erlang odbc module: http://erlang.org/doc/man/odbc.html
The odbc:sql_query/2 gives me back {selected, Columns, Results}
From here i can take the Columns which is a list of strings & the Results, a list of rows represented each as a tuple, then create a few functions to output valid Erlang code to be able to serialize correctly to Json based on a number of factors. Here's the full code:
make the initial query:
Sql = "SELECT * FROM alloys;",
Ref = connect(),
case odbc:sql_query(Ref, Sql) of
{selected, Columns, Results} ->
set_json_from_sql(Columns, Results, []);
{error, Reason} ->
{error, Reason}
end.
Then the input function is set_json_from_sql/3 that calls the below functions:
format_by_type(Item) ->
if
is_list(Item) -> list_to_binary(io_lib:format("~s", [Item]));
is_integer(Item) -> Item;
is_boolean(Item) -> io_lib:format("~a", [Item]);
is_atom(Item) -> Item
end.
json_by_type([H], [Hc], Data) ->
NewH = format_by_type(H),
set_json_flatten(Data, Hc, NewH);
json_by_type([H|T], [Hc|Tc], Data) ->
NewH = format_by_type(H),
NewData = set_json_flatten(Data, Hc, NewH),
json_by_type(T, Tc, NewData).
set_json_flatten(Data, Column, Row) ->
ResTuple = {list_to_binary(Column), Row},
lists:flatten(Data, [ResTuple]).
set_json_from_sql([], [], Data) -> jsone:encode([{<<"data">>, lists:reverse(Data)}]);
set_json_from_sql(Columns, [H], Data) ->
NewData = set_json_merge(H, Columns, Data),
set_json_from_sql([], [], NewData);
set_json_from_sql(Columns, [H|T], Data) ->
NewData = set_json_merge(H, Columns, Data),
set_json_from_sql(Columns, T, NewData).
set_json_merge(Row, Columns, Data) ->
TupleRow = json_by_type(tuple_to_list(Row), Columns, []),
lists:append([TupleRow], Data).
So set_json_from_sql/3 gives you your Json output after matching set_json_from_sql([], [], Data).
The key points here are that you need to call list_to_binary/1 for strings & atoms. Use jsone to encode Erlang objects to Json: https://github.com/sile/jsone
And, notice format_by_type/1 is used to match against Erlang object types, yes not ideal but works as long as you are aware of your DB's types or you can increase the extra guards to accommodate this.

This works for me
test()->
Ma = "Hello World", Mb = "Hello Erlang",
A = "{\"Messages\" : {{\"Ma\":\"" ++ Ma ++ "\"}, {\"Mb\":\""++Mb++"\"}}, {\"Usernames\" : {\"Username1\":\"usrname1\"}, {\"Username2\":\"usrname2\"}}",
io:format("~s~n",[A]).
Output
10> io:format("~s~n",[A]).
{"Messages" : {{"Ma":Hello World"}, {"Mb":Hello Erlang"}}, {"Usernames" : {"Username1":"usrname1"}, {"Username2":"usrname2"}}
ok
or use one of many libraries on github to convert erlang terms to json. My Tuple to JSON module is simple but effective.

Do it like a pro
-define(JSON_WRAPPER(Proplist), {Proplist}).
-spec from_list(json_proplist()) -> object().
from_list([]) -> new();
from_list(L) when is_list(L) -> ?JSON_WRAPPER(L).
-spec to_binary(atom() | string() | binary() | integer() | float() | pid() | iolist()) -> binary().
to_binary(X) when is_float(X) -> to_binary(mochinum:digits(X));
to_binary(X) when is_integer(X) -> list_to_binary(integer_to_list(X));
to_binary(X) when is_atom(X) -> list_to_binary(atom_to_list(X));
to_binary(X) when is_list(X) -> iolist_to_binary(X);
to_binary(X) when is_pid(X) -> to_binary(pid_to_list(X));
to_binary(X) when is_binary(X) -> X.
-spec recursive_from_proplist(any()) -> object().
recursive_from_proplist([]) -> new();
recursive_from_proplist(List) when is_list(List) ->
case lists:all(fun is_integer/1, List) of
'true' -> List;
'false' ->
from_list([{to_binary(K) ,recursive_from_proplist(V)}
|| {K,V} <- List
])
end;
recursive_from_proplist(Other) -> Other.

Related

Unable to define a parser in Haskell: Not in scope: type variable ‘a’

I am trying to define a parser in Haskell. I am a total beginner and somehow didn't manage to find any solution to my problem at all.
For the first steps I tried to follow the instructions on the slides of a powerpoint presentation. But I constantly get the error "Not in scope: type variable ‘a’":
type Parser b = a -> [(b,a)]
item :: Parser Char
item = \inp -> case inp of
[] -> []
(x:xs) -> [(x:xs)]
error: Not in scope: type variable ‘a’
|
11 | type Parser b = a -> [(b,a)]
| ^
I don't understand the error but moreover I don't understand the first line of the code as well:
type Parser b = a -> [(b,a)]
What is this supposed to do? On the slide it just tells me that in Haskell, Parsers can be defined as functions. But that doesn't look like a function definition to me. What is "type" doing here? If it s used to specify the type, why not use "::" like in second line above? And "Parser" seems to be a data type (because we can use it in the type definition of "item"). But that doesn't make sense either.
The line derives from:
type Parser = String -> (String, Tree)
The line I used in my code snippet above is supposed to be a generalization of that.
Your help would be much appreciated. And please bear in mind that I hardly know anything about Haskell, when you write an answer :D
There is a significant difference between the type alias type T = SomeType and the type annotation t :: SomeType.
type T = Int simply states that T is just another name for the type Int. From now on, every time we use T, it will be replaced with Int by the compiler.
By contrast, t :: Int indicates that t is some value of type Int. The exact value is to be specified by an equation like t = 42.
These two concepts are very different. On one hand we have equations like T = Int and t = 42, and we can replace either side with the other side, replacing type with types and values with values. On the other hand, the annotation t :: Int states that a value has a given type, not that the value and the type are the same thing (which is nonsensical: 42 and Int have a completely different nature, a value and a type).
type Parser = String -> (String, Tree)
This correctly defines a type alias. We can make it parametric by adding a parameter:
type Parser a = String -> (String, a)
In doing so, we can not use variables in the right hand side that are not parameters, for the same reason we can not allow code like
f x = x + y -- error: y is not in scope
Hence you need to use the above Parser type, or some variation like
type Parser a = String -> [(String, a)]
By contrast, writing
type Parser a = b -> [(b, a)] -- error
would use an undeclared type b, and is an error. At best, we could have
type Parser a b = b -> [(b, a)]
which compiles. I wonder, though, is you really need to make the String type even more general than it is.
So, going back to the previous case, a possible way to make your code run is:
type Parser a = String -> [(a, String)]
item :: Parser Char
item = \inp -> case inp of
[] -> []
(x:xs) -> [(x, xs)]
Note how [(x, xs)] is indeed of type [(Char, String)], as needed.
If you really want to generalize String as well, you need to write:
type Parser a b = b -> [(b, a)]
item :: Parser Char String
item = \inp -> case inp of
[] -> []
(x:xs) -> [(xs, x)]

How to store a JSON with null values with cmsgpack and Lua in Redis?

I've been reading the documentation about cjson and know about the conversion of null values to cjson.null values (because of no possible way to store nil in Lua tables. This works well if I use cjson.decode and cjson.encode.
However, when I try to pack and unpack the contents of the table with cmsgpack, the keys with null values are not present. I use code similar to this one.
How do you solve this problem? Do you replace null values with a especial value ({isnull: true} before calling cmspack.pack and later, replace back {isnull: true} to null) or use any other technique?
I did a quick research and didn't find any msgpack library that handles null values in hash tables. I'm not saying that such a library doesn't exist, though.
I've chosen another approach. Since lua-cmsgpack doesn't do anything special for handling null values when decoding MessagePack binary data and nil values are problem in tables only, we can add some special case code for hash → table decoder.
Actually, the patch is pretty trivial:
--- a/lua_cmsgpack.c
+++ b/lua_cmsgpack.c
## -565,6 +565,10 ## void mp_decode_to_lua_hash(lua_State *L, mp_cur *c, size_t len) {
mp_decode_to_lua_type(L,c); /* key */
if (c->err) return;
mp_decode_to_lua_type(L,c); /* value */
+ if (lua_isnil(L, -1)) {
+ lua_pop(L, 1);
+ lua_pushlightuserdata(L, NULL);
+ }
if (c->err) return;
lua_settable(L,-3);
}
Here the second call of mp_decode_to_lua_type converts a hash entry value to a corresponding Lua type value, and when the value is nil (that is, null in MessagePack), we replace it with a light userdata (NULL pointer). The same userdata (NULL pointer) is used as the cjson.null value.
We don't need to patch the encoder because it encodes any unsupported value including any userdata to null, that is, you can use cjson.null userdata to represent null values (and cjson.decode does just that). When the hash is decoded, you'll get the same userdata.
The same trick can be applied to array-like tables but it is not necessary if you know a size of a table — you can iterate the table from the first to the last index.
local cjson = require('cjson')
local cmsgpack = require('cmsgpack')
local encoded_hash = cmsgpack.pack({k1 = nil, k2 = cjson.null, k3 = 'baz'})
local decoded_hash = cmsgpack.unpack(encoded_hash)
print('decoded_hash.k2 == cjson.null?', decoded_hash.k2 == cjson.null)
for k, v in pairs(decoded_hash) do
print(k, '->', v)
end
local encoded_array = cmsgpack.pack({'foo', nil, cjson.null, 'bar'})
local decoded_array = cmsgpack.unpack(encoded_array)
print('#decoded_array =', #decoded_array)
for i = 1, 4 do
print(i, '->', decoded_array[i])
end
Output:
decoded_hash.k2 == cjson.null? true
k3 -> baz
k2 -> userdata: (nil)
#decoded_array = 4
1 -> foo
2 -> nil
3 -> userdata: (nil)
4 -> bar
Thanks to LuaRocks we don't even need to support the fork of the library for such a trivial patch. Patches can be included directly in rockspec file. I've created a gist with modified rockspec: https://gist.github.com/un-def/d6b97f5ef6b47edda7f65e0c6144663c
You can install a patched version of the library with the following command:
luarocks install https://gist.github.com/un-def/d6b97f5ef6b47edda7f65e0c6144663c/raw/8fe2f1ad70c75bfe2532a1f4cf9213f4e28423d3/lua-cmsgpack-0.4.0-0.rockspec

How to serialize function type to json in haskell?

data Task = Task
{ id :: String
, description :: String
, dependsOn :: [String]
, dependentTasks :: [String]
} deriving (Eq, Show, Generic, ToJSON, FromJSON)
type Storage = Map String Task
s :: Storage
s = empty
addTask :: Task -> Storage -> Storage
addTask (Task id desc dep dept) = insert id (Task id desc dep dept)
removeTask :: String -> Storage -> Storage
removeTask tid = delete tid
changes = [addTask (Task "1" "Description" [] []), removeTask "1"]
main = putStrLn . show $ foldl (\s c -> c s) s changes
Suppose I have the following code. I want to store changes list in a json file. But I don't know how to do that with Aeson, aside probably from writing a custom parser and there must be a better way to do that obviously. Like maybe using language extension to derive (Generic, ToJSON, FromJSON) for addTask and removeTask etc...
EDIT. For all people that say "You can't serialize function".
Read the comments to an answer to this question.
Instance Show for function
That said, it's not possible to define Show to actually give you more
? detail about the function. – Louis Wasserman May 12 '12 at 14:51
Sure it is. It can show the type (given via Typeable); or it can show some of the inputs and outputs (as is done in QuickCheck).
EDIT2. Okay, I got that I can't have function name in serialization. But can this be done via template Haskell? I see that aeson supports serialization via template Haskell, but as newcomer to Haskell can't figure out how to do that.
Reading between the lines a bit, a recurring question here is, "Why can't I serialize a function (easily)?" The answer -- which several people have mentioned, but not explained clearly -- is that Haskell is dedicated to referential transparency. Referential transparency says that you can replace a definition with its defined value (and vice versa) without changing the meaning of the program.
So now, let's suppose we had a hypothetical serializeFunction, which in the presence of this code:
foo x y = x + y + 3
Would have this behavior:
> serializeFunction (foo 5)
"foo 5"
I guess you wouldn't object too strenuously if I also claimed that in the presence of
bar x y = x + y + 3
we would "want" this behavior:
> serializeFunction (bar 5)
"bar 5"
And now we have a problem, because by referential transparency
serializeFunction (foo 5)
= { definition of foo }
serializeFunction (\y -> 5 + y + 3)
= { definition of bar }
serializeFunction (bar 5)
but "foo 5" does not equal "bar 5".
The obvious followup question is: why do we demand referential transparency? There are at least two good reasons: first, it allows equational reasoning like above, hence eases the burden of refactoring; and second, it reduces the amount of runtime information that's needed, hence improving performance.
Of course, if you can come up with a representation of functions that respects referential transparency, that poses no problems. Here are some ideas in that direction:
printing the type of the function
instance (Typeable a, Typeable b) => Show (a -> b) where
show = show . typeOf
-- can only write a Read instance for trivial functions
printing the input-output behavior of the function (which can also be read back in)
creating a data type that combines a function with its name, and then printing that name
data Named a = Named String a
instance Show (Named a) where
show (Named n _) = n
-- perhaps you could write an instance Read (Map String a -> Named a)
(and see also cloud haskell for a more complete working of this idea)
constructing an algebraic data type that can represent all the expressions you care about but contains only basic types that already have a Show instance and serializing that (e.g. as described in the other answer)
But printing a bare function's name is in conflict with referential transparency.
Make a data type for your functions and an evaluation function:
data TaskFunction = AddTask Task | RemoveTask String
deriving (Eq, Show, Generic, ToJSON, FromJSON)
eval :: TaskFunction -> Storage -> Storage
eval (AddTask t) = addTask t
eval (RemoveTask t) = removeTask t
changes = [AddTask (Task "1" "Description" [] []), RemoveTask "1"]
main = putStrLn . show $ foldl (\s c -> c s) s (eval <$> changes)

How to read value of property depending on an argument

How can I get the value of a property given a string argument.
I have a Object CsvProvider.Row which has attributes a,b,c.
I want to get the attribute value depending on property given as a string argument.
I tried something like this:
let getValue (tuple, name: string) =
snd tuple |> Seq.averageBy (fun (y: CsvProvider<"s.csv">.Row) -> y.```name```)
but it gives me the following error:
Unexpected reserved keyword in lambda expression. Expected incomplete
structured construct at or before this point or other token.
Simple invocation of function should look like this:
getValue(tuple, "a")
and it should be equivalent to the following function:
let getValue (tuple) =
snd tuple |> Seq.averageBy (fun (y: CsvProvider<"s.csv">.Row) -> y.a)
Is something like this is even possible?
Thanks for any help!
The CSV type provider is great if you are accessing data by column names statically, because you get nice auto-completion with type inference and checking.
However, for a dynamic access, it might be easier to use the underlying CsvFile (also a part of F# Data) directly, rather than using the type provider:
// Read the given file
let file = CsvFile.Load("c:/test.csv")
// Look at the parsed headers and find the index of column "A"
let aIdx = file.Headers.Value |> Seq.findIndex (fun k -> k = "A")
// Iterate over rows and print A values
for r in file.Rows do
printfn "%A" (r.Item(aIdx))
The only unfortunate thing is that the items are accessed by index, so you need to build some lookup table if you want to easily access them by their name.

F# Inline Function Specialization

My current project involves lexing and parsing script code, and as such I'm using fslex and fsyacc. Fslex LexBuffers can come in either LexBuffer<char> and LexBuffer<byte> varieties, and I'd like to have the option to use both.
In order to user both, I need a lexeme function of type ^buf -> string. Thus far, my attempts at specialization have looked like:
let inline lexeme (lexbuf: ^buf) : ^buf -> string where ^buf : (member Lexeme: char array) =
new System.String(lexbuf.Lexeme)
let inline lexeme (lexbuf: ^buf) : ^buf -> string where ^buf : (member Lexeme: byte array) =
System.Text.Encoding.UTF8.GetString(lexbuf.Lexeme)
I'm getting a type error stating that the function body should be of type ^buf -> string, but the inferred type is just string. Clearly, I'm doing something (majorly?) wrong.
Is what I'm attempting even possible in F#? If so, can someone point me to the proper path?
Thanks!
Functions and members marked as inline cannot be overloaded, so your original strategy won't work. You need to write different code for both of the declarations, so you need to use overloading (if you want to write this without boxing and dynamic type tests).
If you're using standard F# tools, then the type you'll get as a buffer will always be LexBuffer<'T> and you want to have two overloads based on the type argument. In this case, you don't need the static member constraints at all and can write just:
type Utils =
static member lexeme(buf:LexBuffer<char>) =
new System.String(buf.Lexeme)
static member lexeme(buf:LexBuffer<byte>) =
System.Text.Encoding.UTF8.GetString(buf.Lexeme)
Are you sure this strategy of redefining inline functions with different argument types can work? Looks like you're trying to overload to me...
type LexBuffer<'a>(data : 'a []) =
member this.Lexeme = data
let lexeme (buf : LexBuffer<'a>) =
match box buf.Lexeme with
| :? (char array) as chArr ->
new System.String(chArr)
| :? (byte array) as byArr ->
System.Text.Encoding.UTF8.GetString(byArr)
| _ -> invalidArg "buf" "must be either char or byte LexBuffer"
new LexBuffer<byte>([| 97uy; 98uy; 99uy |])
|> lexeme
|> printfn "%A"
new LexBuffer<char>([| 'a'; 'b'; 'c' |])
|> lexeme
|> printfn "%A"