Re: [Haskell-cafe] Where to start about writing web/HTTP apps ?

11 Sep 2005

      Thanks for the info. Looking forward to see it. What I
would like to see is also some form of templating
system like HTMLTemplate in Python, that is reading in
HTML then merge in data using the DOM model, on the
server(may be the XML package in haskell already does
it). As for AJAX, I would assume that once there is
built-in support for HTTP, it is just a matter of
handling XMLRPC/JSONRPC calls from the client, instead
of HTTP calls, that should be solved automatically by
then, given the extremely strong parser support in
haskell.

--- Brian McQueen  wrote:
...
I emailed the WASH guy the other day and he is
currently working on
integrating it with a web server.  There is talk
about Haskell server
pages.  It sounds quite interesting.  I'm impressed
with WASH as it
is, though.  I too am hoping to see it when it moves
from a plain CGI
model to the something that's built right into the
server.  There are
some very useful and intersting features such as the
way you program
as if it were a resident server, rather than a
remote, swtateless exec
and restart CGI.  I like the way handlers for the
input fields are
integrated with the definition of the field, as well
as with the error
handling for the field.  I saw something about
caching of the static
portion of the pages too.
I'd like to see some means of integrating the new,
and quite awesome
AJAX techniques too.
Brian McQueen
...
...
From: gary ng 
To: haskell-cafe@haskell.org
Subject: [Haskell-cafe] Where to start about
writing web/HTTP apps ?
Date: Sat, 10 Sep 2005 04:15:45 -0700 (PDT)
Hi,
Hi,
I just start learning haskell and have to say
...
...
is stunning in how precise it can be(coming from
a
background of C then python/perl/js).
I want to write apps for WEB and have briefly
read
WASH. However, that seems to be a CGI based
solution.
What I want is a native HTTP server(written in
haskell), like Twisted/Cherrypy in Python. Are
On 9/10/05, Thomas Spriggs
 wrote:
that it
there
...
...
any boilerplate(I know I should scrap the
boilerplate
but I need to have something to get start) for
reference ?
In addition, during my learning process, I keep
on
using my old experience as reference such as
writing
simple programs that needs functions like
ltrim/rtrim/substr etc. that is in almost any
language
I have used. But it seems that haskell doesn't
have
it. I know that a haskell expert can write them
in no
time with things like dropWhile and reverse, it
is a
bit frustrating for new comers from a imperative
background.
Oh, while I am still here, I am reading "The
Evolution
of a Haskell Programmer"
...
...
and learning the various way to tackle the same
problem. Obviously, there are lots of things I
don't
...
know what it is about and I would tackle them as
time
go by. But I have problem even with the seems to
be
simple one like this :
I don't think its that simple.
fac n = foldr (\x g n -> g (x*n)) id [1..n] 1
I can understand a foldl or foldr version but
have
problem with this, especially the "id" and the
The id is the identity function defined
id x = x
trailing "1" and the function being folded takes
3
parameters instead of 2(as in standard
foldr/foldl
solution).
For a start the previous standard fold solutions
take 3 parameters not 2.
Its just the "fac n = foldr (\x g n -> g (x*n)) id
[1..n] 1" example appears
to take 4. However it doesn't, foldr always takes
3 arguments.
foldr :: (a -> b -> b) -> b -> [a] -> b (from zvon
reference)
In fact the foldr function actually returns a
function in this case. It
could be rewritten as
fac n = (foldr (\x g n -> g (x*n)) id [1..n]) 1
with the additional brackets added for clarity.
...
Would appreciate if someone can knock on my head
and
tell me what is going on in it.
Well, here goes. The way the lambda function/foldr
...
resulting foldl like function needs a new identity
...
additional "1". To demonstrate something like how
...
low number eg 3: (Please would someone correct me
if I have made a mistake
in this)
fac 3 = (foldr (\x g n -> g (x*n)) id [1..3]) 1
fac 3 = (foldr (\x g n -> g (x*n)) id [1,2,3]) 1
fac 3 = (foldr (\x g n -> g (x*n)) (\n -> id
(3*n)) [1,2])) 1
fac 3 = (foldr (\x g n -> g (x*n)) (\n -> (\n ->
id (3*n)) (2*n)) [1]) 1
fac 3 = (foldr (\x g n -> g (x*n)) (\n -> (\n ->
(\n -> id (3*n)) (2*n))
(1*n)) []) 1
fac 3 = (\n -> (\n -> (\n -> id (3*n)) (2*n))
(1*n)) 1
fac 3 = (\n -> (\n -> id (3*n)) (2*n)) (1*1)
fac 3 = (\n -> (\n -> id (3*n)) (2*n)) 1
fac 3 = (\n -> id (3*n)) (2*1)
fac 3 = (\n -> id (3*n)) 2
fac 3 = id (3*2)
fac 3 = id 6
fac 3 = 6
I would suggest that you use something other than
...
haskell programmer" to learn haskell as the
versions of factorial get
complicated very quickly and its largely use less
as you should probably
just use:
fac n = product [1..n]
anyway. A better introduction would be something
http://www.willamette.edu/~fruehr/haskell/evolution.html
thing evaluates, the
parameter hence the
this is evaluated for a
the "evolution of a
like
...
http://www.cse.unsw.edu.au/~cs1011/05s2/ and use
http://zvon.org/other/haskell/Outputglobal/index.html
...
and
...
http://www.haskell.org/tutorial/ if you want to
learn something in specific
or are strugling. All links from
http://www.haskell.org/learning.html of
course.
...
thanks for help in advance.
You're welcome.
...
regards,
gary
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...
...
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...
Well good luck furthering your knowledge of
haskell,
...
Thomas
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