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Re: [Haskell-beginners] remove XML tags using Text.Regex.Posix
by Jan Jakubuv 30 Sep '09

30 Sep '09

Hi Robert, On Tue, Sep 29, 2009 at 12:25:07PM -0700, Robert Ziemba wrote: > I have been working with the regular expression package (Text.Regex.Posix). > My hope was to find a simple way to remove a pair of XML tags from a short > string. > > I have something like this "<tag>Data</tag>" and would like to extract > 'Data'. There is only one tag pair, no nesting, and I know exactly what the > tag is. > This is so simple that I would not recommend anything other than regular expressions. Use the following pattern: pat = "<tag>(.*)</tag>" It creates a group withing the matched string containing the data (it is done using parenthesis). Use `[[String]]` as a result type and you receive a list of matches where each match is described by a list of strings whose first member is the whole matched string (including <tag> and </tag>) and it is followed by values of groups (in our case we have just one group). Thus: *Main> "text<tag>data</tag>text" =~ pat :: [[String]] [["<tag>data</tag>","data"]] It is easy extract the data using `(!!)` and `head`: *Main> (!! 1) . head $ ("text<tag>7</tag>text" =~ pat :: [[String]]) "7" > My first attempt was this: > > "<tag>123</tag>" =~ "[^<tag>].+[^</tag>]"::String > > result: "123" > The problem with your pattern is that `[^<tag>]` doesn't mean what you think it does. Its meaning is “one character which is not `<`, `t`, `a`, or `>`” as Patrick already described in his mail. > Upon further experimenting I realized that it only works with more than 2 > digits in 'Data'. I occured to me that my thinking on how this regular > expression works was not correct - but I don't understand why it works at > all for 3 or more digits. > It doesn't work for all 3 or more digits: *Main> "<tag>tag</tag>" =~ "[^<tag>].+[^</tag>]" :: String "" Briefly, it doesn't work when the data contains one of characters `<`, `t`, `a`, `g`, `>`. Finally, consider using pat = "<tag>([^<]*)</tag>" which works with more tags in the same line as well. Sincerely, jan. -- Heriot-Watt University is a Scottish charity registered under charity number SC000278.

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remove XML tags using Text.Regex.Posix
by Robert Ziemba 30 Sep '09

30 Sep '09

I have been working with the regular expression package (Text.Regex.Posix). My hope was to find a simple way to remove a pair of XML tags from a short string. I have something like this "<tag>Data</tag>" and would like to extract 'Data'. There is only one tag pair, no nesting, and I know exactly what the tag is. My first attempt was this: "<tag>123</tag>" =~ "[^<tag>].+[^</tag>]"::String result: "123" Upon further experimenting I realized that it only works with more than 2 digits in 'Data'. I occured to me that my thinking on how this regular expression works was not correct - but I don't understand why it works at all for 3 or more digits. Can anyone help me understand this result and perhaps suggest another strategy? Thank you.

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Can you identify this recursion pattern please?
by Colin Paul Adams 29 Sep '09

29 Sep '09

This looks like some kind of unfold to me. Can you help? I have the following code that works for the first two levels: -- get the list of all (recursive) child gallery names -- first, just the immediate children childNames <- childGalleries db (name gallery) -- now let's try the next level grandchildren <- mapM (childGalleries db) childNames mapM putStrLn (childNames ++ concat grandchildren) The database query childGalleries gives me a list of all the immediate children. So what I want is recursively, the names of all the descendants, as well as the original name, concatenated to a flat list. I'm struggling to identify the higher-order recursion function I need. -- Colin Adams Preston Lancashire

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Distributing executables
by Joe Van Dyk 29 Sep '09

29 Sep '09

How hard is it to distribute a haskell binary that contains a bunch of libraries embedded in the executable? Is it pretty easy to make a large statically-linked executable? -- Joe Van Dyk http://fixieconsulting.com

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Finance.Quote.Yahoo
by Hong Yang 28 Sep '09

28 Sep '09

Hi All, I was using a modified version of Finance.Quote.Yahoo module (I only added "setproxy"), and got unwanted print on the screen. My questions are as follows: 1) How can I get rid of this screen print? (I guess HTTP package called by the Finance.Quote.Yahoo module internally prints those info. Should I temporarily turn off STDOUT then turn it on? How to do this?) 2) "ghc-pkg list" shows HTTP-4000.0.8. How come the output shows "HTTP-4000.0.5"? Thanks, Hong P.S. The following is the screen output: Sending: GET http://download.finance.yahoo.com/d/quotes.csv?s=BAX&f=l1 HTTP/1.1 Content-Length: 0 User-Agent: hs-HTTP-4000.0.5 Host: download.finance.yahoo.com proxy uri host: xxxproxy, port: :74 Creating new connection to xxxproxy:74 Received: HTTP/1.1 200 OK Proxy-Connection: Keep-Alive Connection: Keep-Alive Transfer-Encoding: chunked Via: 1.1 SAUSISA02 Date: Mon, 28 Sep 2009 18:49:43 GMT Content-Type: application/octet-stream P3P: policyref="http://p3p.yahoo.com/w3c/p3p.xml", CP="CAO DSP COR CUR ADM DEV TAI PSA PSD IVAi IVDi CONi TELo OTPi OUR DELi SAMi OTRi UNRi PUBi IND PHY ONL UNI PUR FIN COM NAV INT DEM CNT STA POL HEA PRE GOV" Cache-Control: private Content-Length: 7

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Anyone use haskell + cassandra?
by Joe Van Dyk 28 Sep '09

28 Sep '09

I'm thinking of using Haskell + Cassandra + Rails to build a horizontally-scalable web app. Haskell + Cassandra would power the webservice that Rails communicates with. Has anyone done anything like that? -- Joe Van Dyk http://fixieconsulting.com

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Better way to retrieve data strucures from sql-db?
by Roman Naumann 28 Sep '09

28 Sep '09

The following function converts the data read from an sql database to a player data structure. toPlayer :: [SqlValue] -> Player toPlayer sx = Player { plID = fromSql (sx!!0), plAccount = fromSql (sx!!1), plForename = fromSql (sx!!2), plSurename = fromSql (sx!!3), plPos_x = fromSql (sx!!4), plPos_y = fromSql (sx!!5) } It appears to me, that I have to write this boilerplate code every time I introduce a new data structure that has to be read from the sql database. Can you think of a way to automate this? Thanks for your replies, Roman Naumann

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Haskell type system
by informationen 27 Sep '09

27 Sep '09

Hi, i am trying to understand the Haskell type system. I thought i understood it quite well until i encountered the three following exercises. As you can see, i have the answers already. But i don't understand, why they are correct. Could anybody tell me a good place where i could learn how to answers these kind of questions correctly or could give me some explanations, why these answers are correct? Any help is highly appreciated. Kind regards Chris Two functions f and g with the following signatures are given: f :: a -> a g :: b -> c -> b A) Give the type of the following expressions: 1) g [False] True :: 2) g [] True :: 3) g (f True) :: 4) g f :: Answers: 1 [Boolean] 2) [a] 3) c -> Bool 4) c -> (a -> a)1) g f 1 B) Which of the following statements is correct? 2) g (f 1) is type correct 3) g . (f 1) is type correct 4) g . f 1 is type correct 5) (g . f) 1 is type correct 6) none of the expressions is correct Answers: 1,2 and 5 are correct. C) A function h is given as: h p x = p (f x). Which of the following statements is correct. 1) h :: a -> b -> a -> b 2) h :: (a -> a) -> a -> a 3) h :: (a -> b) -> a -> b 4) h is equivalent to h' with h' p = p . f 5) h is equivalent to h' with h' p = p f 5) h is equivalent to h' with h' p x = p f x Answers: (I am not sure, if i remember correctly, but 3) and 4) should be correct.)

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proxy
by Hong Yang 24 Sep '09

24 Sep '09

Hi, I was using Finance.Quote.Yahoo, but got an error: user error (openTCPConnection: host lookup failure for " download.finance.yahoo.com"). I think it is due to I being behind a proxy. I then imported Network.Browser, ran setProxy, and tried to fetch data using Finance.Quote.Yahoo.getQuote. I still got the error. So I suggest the module author incorporate proxy setting inside the module. Thanks, Hong

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Question regarding infering the type of a function
by MoC 21 Sep '09

21 Sep '09

Hi everybody, today somebody on IRC (#haskell on Freenode) jokingly submitted the following to lambdabot: > (.) (.) It turned out that (.) (.) :: (a1 -> b -> c) -> a1 -> (a -> b) -> a -> c. I got interested in why that is so, but unfortunately I have not been able to figure this out by myself yet. I know that partial application is used, but don't know to what it would expand to and why. Could someone please explain? (I asked on the channel, but nobody did answer so I thought I'd try my luck here.) Regards, MoC

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