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  • How to structure my GUI agnostic project?

    - by Nezreli
    I have a project which loads from database a XML file which defines a form for some user. XML is transformed into a collection of objects whose classes derive from single parent. Something like Control - EditControl - TextBox Control - ContainterControl - Panel Those classes are responsible for creation of GUI controls for three different enviroments: WinForms, DevExpress XtraReports and WebForms. All three frameworks share mostly the same control tree and have a common single parent (Windows.Forms.Control, XrControl and WebControl). So, how to do it? Solution a) Control class has abstract methods Control CreateWinControl(); XrControl CreateXtraControl(); WebControl CreateWebControl(); This could work but the project has to reference all three frameworks and the classes are going to be fat with methods which would support all three implementations. Solution b) Each framework implementation is done in separate projects and have the exact class tree like the Core project. All three implementations are connected using a interface to the Core class. This seems clean but I'm having a hard time wrapping my head around it. Does anyone have a simpler solution or a suggestion how should I approach this task?

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  • How do I build a matrix to translate one set of points to another?

    - by dotminic
    I've got 3 points in space that define a triangle. I've also got a vertex buffer made up of three vertices, that also represent a triangle that I will refer to as a "model". How can I can I find the matrix M that will transform vertex in my buffer to those 3 points in space ? For example, let's say my three points A, B, C are at locations: A.x = 10, A.y = 16, A.z = 8 B.x = 12, B.y = 11, B.z = 1 C.x = 19, C.y = 12, C.z = 3 given these coordinates how can I build a matrix that will translate and rotate my model such that both triangles have the exact same world space ? That is, I want the first vertex in my triangle model to have the same coordinates as A, the second to have the same coordinates as B, and same goes for C. nb: I'm using instanced rendering so I can't just give each vertex the same position as my 3 points. I have a set of three points defining a triangle, and only three vertices in my vertex buffer.

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  • Why don't my Google Analytics custom segmentation visit numbers match up?

    - by Hates_
    I have three main areas of my site and want to track total usage as well as breakdowns of the three parts. I am trying to segment the "type" of use on each page using a custom variable as such: ['_setCustomVar',1,'Visitor Type','Unknown',1] Visitor type can be one of three values: "Unknown", "Reader" or "publisher". Every page has this value set. Now when I look at my analytics chart and chose all three segments, the individual values do not match the sum. I've double checked the pages to make sure the custom var is there.

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  • Refer to te current directory in a shell script

    - by One Two Three
    How do I refer to the current directory in a shell script So I have this script which calls another script in the same directory #! /bin/sh #Call the other script ./foo.sh # do something ... For this I got ./foo.sh: No such file or directory So I changed it to: #! /bin/sh #Call the other script foo.sh # do something ... But this would call the foo script which is, by default, in the PATH. This is not what I want. So the question is, what's the syntax of doing './` in a shell script?

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  • Android:Multi touch doesn't work as expected?

    - by user187532
    Hi folks, Help me in resolving the below issue. I have three image buttons on screen. All these three buttons controlled under ontouchlistner as below. buttonOne.setOnTouchListener(this); buttonTwo.setOnTouchListener(this); buttonThree.setOnTouchListener(this); I override "public boolean onTouch(View v, MotionEvent event)". Under this i check for these three image buttons touch events like below. ImageButton imageBtn = (ImageButton) v; if ( imageBtn == buttonOne ) // first button touch ..Log.. else if ( imageBtn == buttonTwo ) ..Log.. else if ( imageBtn == buttonThree ) // first button touch ..Log.. My problem is, as it is under multi touch event handler like above, it does not detect when touch all three button at a time to try to produce multi touch effect, instead it detects only one imagebutton touch at a time even though i touch all three image buttons. As i am developing this project on Android 1.6 SDK, is there any problem accessing my requirement(multi touch) (or) it is a known issue? I am hoping that, when it works for single button touch, why shouldn't it work when clicking three imagebuttons at a time to produce three logs printed as per my above code? How do i resolve it for my case? Please don't question me why i am still developing on 1.6 for such a requirement. Thank you. Appreciate your suggestions !

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  • Possible to rank partial matches in Postgres full text search?

    - by Joe
    I'm trying to calculate a ts_rank for a full-text match where some of the terms in the query may not be in the ts_vector against which it is being matched. I would like the rank to be higher in a match where more words match. Seems pretty simple? Because not all of the terms have to match, I have to | the operands, to give a query such as to_tsquery('one|two|three') (if it was &, all would have to match). The problem is, the rank value seems to be the same no matter how many words match. In other words, it's maxing rather than multiplying the clauses. select ts_rank('one two three'::tsvector, to_tsquery('one')); gives 0.0607927. select ts_rank('one two three'::tsvector, to_tsquery('one|two|three|four')); gives the expected lower value of 0.0455945 because 'four' is not the vector. But select ts_rank('one two three'::tsvector, to_tsquery('one|two')); gives 0.0607927 and likewise select ts_rank('one two three'::tsvector, to_tsquery('one|two|three')); gives 0.0607927 I would like the result of ts_rank to be higher if more terms match. Possible? To counter one possible response: I cannot calculate all possible subsequences of the search query as intersections and then union them all in a query because I am going to be working with large queries. I'm sure there are plenty of arguments against this anyway! Edit: I'm aware of ts_rank_cd but it does not solve the above problem.

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  • Sticky/static variable references in for loops

    - by pthulin
    In this example I create three buttons 'one' 'two' 'three'. When clicked I want them to alert their number: <html> <head> <script type="application/javascript" src="jquery.js"></script> <script type="application/javascript"> $(document).ready(function() { var numbers = ['one', 'two', 'three']; for (i in numbers) { var nr = numbers[i]; var li = $('<li>' + nr + '</li>'); li.click(function() { var newVariable = String(nr); alert(i); // 2 alert(nr); // three alert(newVariable); // three alert(li.html()); // three }); $('ul').append(li); } }); </script> </head> <body> <ul> </ul> </body> </html> The problem is, when any of these are clicked, the last value of the loop's variables is used, i.e. alert box always says 'three'. In JavaScript, variables inside for-loops seem to be 'static' in the C language sense. Is there some way to create separate variables for each click function, i.e. not using the same reference? Thanks!

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  • preg_match to find the current directory in a URL

    - by Ian
    I'm trying to detect the current section of a site that a user is viewing by checking for the final directory in the URL. I'm using a PHP and regex to do it and I think I'm close but unfortunately not quite there yet. Here's what I currently have: <?php $url = $_SERVER['REQUEST_URI_PATH'] = preg_replace('/\\?.*/', '', $_SERVER['REQUEST_URI']); $one = '/one/'; $two = '/three/'; $three = '/three/'; $four = '/four/'; $five = '/five/'; echo $url; if (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($one)) == $one) { // URI path starts with "/one/" echo "The section is one."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($two)) == $two) { // URI path starts with "/two/" echo "The section is two."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($three)) == $three) { // URI path starts with "/three/" echo "The section is three."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($four)) == $four) { // URI path starts with "/four/" echo "The section is four."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($five)) == $five) { // URI path starts with "/five/" echo "The section is five."; } ?> I've placed in the echo before the if statements just to get confirmation of the value of $url. This outputs /currentdirectory/file.php However the conditions themselves don't match anything and my individual echo for each section never displays. Also if there's a simpler way of doing it then I'm open to suggestions. Thanks

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  • Problem installing CanonMF5880dn

    - by Paul
    Just got a CanonMF5880dn and cannot print to it from Suse 11.1 MacBook prints w/o issue ping 192.168.1.103 no problem cups sees it as Canon MF5880/MF5840 PCL at URI socket://192.168.1.103:9100 cups test print appears to submit and complete job but no action from printer Yast also seems to install printer correctly CQue2 also seems to install printer correctly all attempts to print yield same results: Suse indicates job processed correctly and completely but no printing happens. firewall is off http://192.168.1.103 in FF gives me the printer config menus correctly What have I failed to do?

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  • Enabling hardware acceleration and Xinerama for multi-monitor/multi-GPU in Linux

    - by mynameiscoffey
    My current setup is three monitors connected as follows (monitors listed from left to right): GPU0 (nVidia GTX 280): - Dell 2405FPW (1920x1200) - Dell U2410 (1920x1200) GPU1 (nVidia 210): - Dell 2405FPW (1920x1200) Works like a charm in Windows 7, not so much in Linux. I seem to only have three real options: Run all three monitors as a seperate X screen, I get hardware acceleration but as they are all independent X sessions I cannot move windows between them and can only have firefox open on one at any given time. Run the two on GPU0 in TwinView mode and have GPU1 as a seperate X screen. Same limitation as 1 but at least two monitors work together ok. I did have an issue where occasionally Linux saw both monitors on GPU0 as a single large monitor however. Enable Xinerama and have everything work as I want it to but hardware acceleration is gone and the display is Windows 95 style choppy. My ideal solution would be to have all screens working as they do under Xinerama without the limitation of having hardware acceleration disabled. I don't even care if that means rendering all three on GPU0 and somehow farming out the display of the third monitor to GPU1, whatever works. My question is this: is there any way to accomplish this? I don't feel like my use case is so out there that there shouldn't be at least some form of support (beyond the three limited options presented above), or is my best option going to be to just suck it up and pick up a better card to replace both that can handle three outputs by itself?

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  • nSticky/static variable references in for loops

    - by pthulin
    In this example I create three buttons 'one' 'two' 'three'. When clicked I want them to alert their number: <html> <head> <script type="application/javascript" src="jquery.js"></script> <script type="application/javascript"> $(document).ready(function() { var numbers = ['one', 'two', 'three']; for (i in numbers) { var nr = numbers[i]; var li = $('<li>' + nr + '</li>'); li.click(function() { var newVariable = String(nr); alert(i); // 2 alert(nr); // three alert(newVariable); // three alert(li.html()); // three }); $('ul').append(li); } }); </script> </head> <body> <ul> </ul> </body> </html> The problem is, when any of these are clicked, the last value of the loop's variables is used, i.e. alert box always says 'three'. In JavaScript, variables inside for-loops seem to be 'static' in the C language sense. Is there some way to create separate variables for each click function, i.e. not using the same reference? Thanks! Edit: The solution is to use jQuery.data to associate arbitrary data with each element: <html> <head> <script type="application/javascript" src="jquery.js"></script> <script type="application/javascript"> $(document).ready(function() { var numbers = ['one', 'two', 'three']; for (i in numbers) { var nr = numbers[i]; var li = $('<li>' + nr + '</li>'); li.data('nr', nr); li.click(function() { alert($(this).data('nr')); }); $('ul').append(li); } }); </script> </head> <body> <ul> </ul> </body> </html>

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  • How do I get an HP 1018 laser printer to work on 12.04?

    - by MInner
    edit : already printed on windows-machine Ubuntu 12.04, LJ 1018. I've tried making foo2zjs from sourse. I've downloaded firmware and placed it where necessarily manually from here I've also tried this - same as manual dowload but with some patches like stoping cups, etc.. I've fixed usbfs: process 11200 (usb) did not claim interface 0 before use problems using sudo rmmod usblp as described here. And now I have Aug 10 11:51:59 pav /usr/sbin/hplj1018: foo2zjs: usb://HP/LaserJet%201018?serial=KP07RLV... download failed. Aug 10 11:51:59 pav /usr/sbin/hplj1018: foo2zjs: loading HP LaserJet 1018 firmware /lib/firmware/hp/sihp1018.dl to CUPS USB device ... error. Before that prinet was stucking on "Sending data to printer". Now it's code is "Not connected". Unless ben@pav:~$ lsusb ... Bus 001 Device 015: ID 03f0:4117 Hewlett-Packard LaserJet 1018 Thank you.

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  • Help with Kodak esp 3250 printer driver on Lubuntu 12.1 SOLVED!

    - by user108608
    First my system: pentium 4 -don't remember the speed-, 1g ram, dual boot to separate physical drives, Fdos and Lubuntu 12.1 second my lan: I have four computers operating for the same printer. 1. Intel quad core i5, 4g ram, running Windoze 7 64 bit, printer connected and shared from here. Kodak ESP 3250 2. Gateway 17" laptop running Windoze 7 32bit 3. Asus tablet (small laptop) running Lumbutu 12.1 4. My dual boot system running Fdos and Lubuntu 12.1 The problem: I downloaded c2esp_25c-1_i386.deb, tried to install it using DEBI Package Installer, it loads the files, looks for cups driver and ends with an error: "Dependancy is not satisfiable: libcupsdriver1 (=1.4.0)" What do I do now? Is there some place that I can get the correct cups driver? further information: The Asus tablet was running Ubuntu 12.1 (very slowly and with a few crashes) and could print from the lan printer with no problems. Is there something in Ubuntu that can be loaded into Lubuntu? noobee user hoping for answers, Paul

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  • How can I set 'Print to File' as my default printing option?

    - by edm
    At the moment when I print, my Deskjet-3050 is selected as the default printer. I would like 'Print to File' to be the default 'printer' without using cups-pdf I specifically do not want to use cups-pdf because of the way it renders text (see below). I am not entirely sure what it is doing but it seems as though it renders the text as bitmaps and embeds them in pdf (as I am not able to highlight/copy/search embedded text as I am using a standard Print to File pdf). N.B. this is not a dupe of: Can I make PDF the default for 'print to file'

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  • How to resolve "Network host is busy; will retry in 30 seconds"?

    - by k0pernikus
    I have added a network printer, a Konica Minolta C253, via cups by its IP (http://10.0.0.42) under the http://localhost:631/ cups webfrontend on an Ubuntu 11.10 client. I used the documented KOC353U.ppd in the process. I was able to print exactly one page. Afterwards, all I get in the printer's status is the message: "Network host '10.0.0.42' is busy; will retry in 30 seconds". What is the problem here and how do I resolve it?

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  • How do I install a driver for a Kodak esp 3250 printer?

    - by user108608
    First my system: pentium 4 -don't remember the speed-, 1g ram, dual boot to separate physical drives, Fdos and Lubuntu 12.1 second my lan: I have four computers operating for the same printer. Intel quad core i5, 4g ram, running Windoze 7 64 bit, printer connected and shared from here. Kodak ESP 3250 Gateway 17" laptop running Windoze 7 32bit Asus tablet (small laptop) running Lumbutu 12.1 My dual boot system running Fdos and Lubuntu 12.1 The problem: I downloaded c2esp_25c-1_i386.deb, tried to install it using DEBI Package Installer, it loads the files, looks for cups driver and ends with an error: "Dependancy is not satisfiable: libcupsdriver1 (=1.4.0)" What do I do now? Is there some place that I can get the correct cups driver? further information: The Asus tablet was running Ubuntu 12.1 (very slowly and with a few crashes) and could print from the lan printer with no problems. Is there something in Ubuntu that can be loaded into Lubuntu?

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  • Trouble with printer canon lbp2900-Ubuntu 14.04

    - by user288922
    I use Ubuntu 14.04. I installed the printer Canon LBP2900 following the guide and did all steps and now the printer works. I added CCPD in Startup Application (with command: sudo /etc/init.d/ccpd start). But after I log out or restart my pc, the printer can't print until I have to open terminal and run sudo /etc/init.d/ccpd restart && sudo /etc/init.d/cups restart. Once more, after restarting my pc I’ve just only printed 1 job (of course it can't print), and I run sudo /etc/init.d/ccpd restart && sudo /etc/init.d/cups restart, the printer will work but it prints the same job three times.

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  • I can not print on Windows 7 shared printer

    - by lrichard
    For a while I can not print from my Ubuntu 12.04 64 bit pc on a Windows 7 shared printer. Before that everything was ok. The printer is HP LaserJet 1100. The error message is: "File "/usr/lib/cups/backend/smb" not available: No such file or directory" I have tried to reinstall the samba, I have configured the workgroup properly. I have tried to reinstall the printer on the Ubuntu machine with the proper address, or with lpd (I have read somewhere). If I restart the printer on the cups web interface ("Resume Printer") the problem seems to be solved, but not. (The printer seems to be idle, and if I want to print, I still get the error message.) I have checked the share rights on the Windows 7 pc, and I think it is ok.

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  • testing Clojure in Maven

    - by Ralph
    I am new at Maven and even newer at Clojure. As an exercise to learn the language, I am writing a spider solitaire player program. I also plan on writing a similar program in Scala to compare the implementations (see my post http://stackoverflow.com/questions/2571267/modern-java-alternatives-closed). I have configured a Maven directory structure containing the usual src/main/clojure and src/test/clojure directories. My pom.xml file includes the clojure-maven-plugin. When I run "mvn test", it displays "No tests to run", despite my having test code in the src/test/clojure directory. As I misnaming something? Here is my pom.xml file: <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>SpiderPlayer</groupId> <artifactId>SpiderPlayer</artifactId> <version>1.0.0-SNAPSHOT</version> <inceptionYear>2010</inceptionYear> <packaging>jar</packaging> <properties> <maven.build.timestamp.format>yyMMdd.HHmm</maven.build.timestamp.format> <main.dir>org/dogdaze/spider_player</main.dir> <main.package>org.dogdaze.spider_player</main.package> <main.class>${main.package}.Main</main.class> </properties> <build> <sourceDirectory>src/main/clojure</sourceDirectory> <testSourceDirectory>src/main/clojure</testSourceDirectory> <plugins> <plugin> <groupId>com.theoryinpractise</groupId> <artifactId>clojure-maven-plugin</artifactId> <version>1.3.1</version> </plugin> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-antrun-plugin</artifactId> <version>1.3</version> <executions> <execution> <goals> <goal>run</goal> </goals> <phase>generate-sources</phase> <configuration> <tasks> <echo file="${project.build.sourceDirectory}/${main.dir}/Version.clj" message="(ns ${main.package})${line.separator}"/> <echo file="${project.build.sourceDirectory}/${main.dir}/Version.clj" append="true" message="(def version &quot;${maven.build.timestamp}&quot;)${line.separator}"/> </tasks> </configuration> </execution> </executions> </plugin> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-assembly-plugin</artifactId> <version>2.1</version> <executions> <execution> <goals> <goal>single</goal> </goals> <phase>package</phase> <configuration> <descriptorRefs> <descriptorRef>jar-with-dependencies</descriptorRef> </descriptorRefs> <archive> <manifest> <mainClass>${main.class}</mainClass> </manifest> </archive> </configuration> </execution> </executions> </plugin> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-surefire-plugin</artifactId> <configuration> <redirectTestOutputToFile>true</redirectTestOutputToFile> <skipTests>false</skipTests> <skip>false</skip> </configuration> <executions> <execution> <id>surefire-it</id> <phase>integration-test</phase> <goals> <goal>test</goal> </goals> <configuration> <skip>false</skip> </configuration> </execution> </executions> </plugin> </plugins> </build> <dependencies> <dependency> <groupId>commons-cli</groupId> <artifactId>commons-cli</artifactId> <version>1.2</version> <scope>compile</scope> </dependency> </dependencies> </project> Here is my Clojure source file (src/main/clojure/org/dogdaze/spider_player/Deck.clj): ; Copyright 2010 Dogdaze (ns org.dogdaze.spider_player.Deck (:use [clojure.contrib.seq-utils :only (shuffle)])) (def suits [:clubs :diamonds :hearts :spades]) (def ranks [:ace :two :three :four :five :six :seven :eight :nine :ten :jack :queen :king]) (defn suit-seq "Return 4 suits: if number-of-suits == 1: :clubs :clubs :clubs :clubs if number-of-suits == 2: :clubs :diamonds :clubs :diamonds if number-of-suits == 4: :clubs :diamonds :hearts :spades." [number-of-suits] (take 4 (cycle (take number-of-suits suits)))) (defstruct card :rank :suit) (defn unshuffled-deck "Create an unshuffled deck containing all cards from the number of suits specified." [number-of-suits] (for [rank ranks suit (suit-seq number-of-suits)] (struct card rank suit))) (defn deck "Create a shuffled deck containing all cards from the number of suits specified." [number-of-suits] (shuffle (unshuffled-deck number-of-suits))) Here is my test case (src/test/clojure/org/dogdaze/spider_player/TestDeck.clj): ; Copyright 2010 Dogdaze (ns org.dogdaze.spider_player (:use clojure.set clojure.test org.dogdaze.spider_player.Deck)) (deftest test-suit-seq (is (= (suit-seq 1) [:clubs :clubs :clubs :clubs])) (is (= (suit-seq 2) [:clubs :diamonds :clubs :diamonds])) (is (= (suit-seq 4) [:clubs :diamonds :hearts :spades]))) (def one-suit-deck [{:rank :ace, :suit :clubs} {:rank :ace, :suit :clubs} {:rank :ace, :suit :clubs} {:rank :ace, :suit :clubs} {:rank :two, :suit :clubs} {:rank :two, :suit :clubs} {:rank :two, :suit :clubs} {:rank :two, :suit :clubs} {:rank :three, :suit :clubs} {:rank :three, :suit :clubs} {:rank :three, :suit :clubs} {:rank :three, :suit :clubs} {:rank :four, :suit :clubs} {:rank :four, :suit :clubs} {:rank :four, :suit :clubs} {:rank :four, :suit :clubs} {:rank :five, :suit :clubs} {:rank :five, :suit :clubs} {:rank :five, :suit :clubs} {:rank :five, :suit :clubs} {:rank :six, :suit :clubs} {:rank :six, :suit :clubs} {:rank :six, :suit :clubs} {:rank :six, :suit :clubs} {:rank :seven, :suit :clubs} {:rank :seven, :suit :clubs} {:rank :seven, :suit :clubs} {:rank :seven, :suit :clubs} {:rank :eight, :suit :clubs} {:rank :eight, :suit :clubs} {:rank :eight, :suit :clubs} {:rank :eight, :suit :clubs} {:rank :nine, :suit :clubs} {:rank :nine, :suit :clubs} {:rank :nine, :suit :clubs} {:rank :nine, :suit :clubs} {:rank :ten, :suit :clubs} {:rank :ten, :suit :clubs} {:rank :ten, :suit :clubs} {:rank :ten, :suit :clubs} {:rank :jack, :suit :clubs} {:rank :jack, :suit :clubs} {:rank :jack, :suit :clubs} {:rank :jack, :suit :clubs} {:rank :queen, :suit :clubs} {:rank :queen, :suit :clubs} {:rank :queen, :suit :clubs} {:rank :queen, :suit :clubs} {:rank :king, :suit :clubs} {:rank :king, :suit :clubs} {:rank :king, :suit :clubs} {:rank :king, :suit :clubs}]) (def two-suits-deck [{:rank :ace, :suit :clubs} {:rank :ace, :suit :diamonds} {:rank :ace, :suit :clubs} {:rank :ace, :suit :diamonds} {:rank :two, :suit :clubs} {:rank :two, :suit :diamonds} {:rank :two, :suit :clubs} {:rank :two, :suit :diamonds} {:rank :three, :suit :clubs} {:rank :three, :suit :diamonds} {:rank :three, :suit :clubs} {:rank :three, :suit :diamonds} {:rank :four, :suit :clubs} {:rank :four, :suit :diamonds} {:rank :four, :suit :clubs} {:rank :four, :suit :diamonds} {:rank :five, :suit :clubs} {:rank :five, :suit :diamonds} {:rank :five, :suit :clubs} {:rank :five, :suit :diamonds} {:rank :six, :suit :clubs} {:rank :six, :suit :diamonds} {:rank :six, :suit :clubs} {:rank :six, :suit :diamonds} {:rank :seven, :suit :clubs} {:rank :seven, :suit :diamonds} {:rank :seven, :suit :clubs} {:rank :seven, :suit :diamonds} {:rank :eight, :suit :clubs} {:rank :eight, :suit :diamonds} {:rank :eight, :suit :clubs} {:rank :eight, :suit :diamonds} {:rank :nine, :suit :clubs} {:rank :nine, :suit :diamonds} {:rank :nine, :suit :clubs} {:rank :nine, :suit :diamonds} {:rank :ten, :suit :clubs} {:rank :ten, :suit :diamonds} {:rank :ten, :suit :clubs} {:rank :ten, :suit :diamonds} {:rank :jack, :suit :clubs} {:rank :jack, :suit :diamonds} {:rank :jack, :suit :clubs} {:rank :jack, :suit :diamonds} {:rank :queen, :suit :clubs} {:rank :queen, :suit :diamonds} {:rank :queen, :suit :clubs} {:rank :queen, :suit :diamonds} {:rank :king, :suit :clubs} {:rank :king, :suit :diamonds} {:rank :king, :suit :clubs} {:rank :king, :suit :diamonds}]) (def four-suits-deck [{:rank :ace, :suit :clubs} {:rank :ace, :suit :diamonds} {:rank :ace, :suit :hearts} {:rank :ace, :suit :spades} {:rank :two, :suit :clubs} {:rank :two, :suit :diamonds} {:rank :two, :suit :hearts} {:rank :two, :suit :spades} {:rank :three, :suit :clubs} {:rank :three, :suit :diamonds} {:rank :three, :suit :hearts} {:rank :three, :suit :spades} {:rank :four, :suit :clubs} {:rank :four, :suit :diamonds} {:rank :four, :suit :hearts} {:rank :four, :suit :spades} {:rank :five, :suit :clubs} {:rank :five, :suit :diamonds} {:rank :five, :suit :hearts} {:rank :five, :suit :spades} {:rank :six, :suit :clubs} {:rank :six, :suit :diamonds} {:rank :six, :suit :hearts} {:rank :six, :suit :spades} {:rank :seven, :suit :clubs} {:rank :seven, :suit :diamonds} {:rank :seven, :suit :hearts} {:rank :seven, :suit :spades} {:rank :eight, :suit :clubs} {:rank :eight, :suit :diamonds} {:rank :eight, :suit :hearts} {:rank :eight, :suit :spades} {:rank :nine, :suit :clubs} {:rank :nine, :suit :diamonds} {:rank :nine, :suit :hearts} {:rank :nine, :suit :spades} {:rank :ten, :suit :clubs} {:rank :ten, :suit :diamonds} {:rank :ten, :suit :hearts} {:rank :ten, :suit :spades} {:rank :jack, :suit :clubs} {:rank :jack, :suit :diamonds} {:rank :jack, :suit :hearts} {:rank :jack, :suit :spades} {:rank :queen, :suit :clubs} {:rank :queen, :suit :diamonds} {:rank :queen, :suit :hearts} {:rank :queen, :suit :spades} {:rank :king, :suit :clubs} {:rank :king, :suit :diamonds} {:rank :king, :suit :hearts} {:rank :king, :suit :spades}]) (deftest test-unshuffled-deck (is (= (unshuffled-deck 1) one-suit-deck)) (is (= (unshuffled-deck 2) two-suits-deck)) (is (= (unshuffled-deck 4) four-suits-deck))) (deftest test-shuffled-deck (is (= (set (deck 1)) (set one-suit-deck))) (is (= (set (deck 2)) (set two-suits-deck))) (is (= (set (deck 4)) (set four-suits-deck)))) (run-tests) Any idea why the test is not running? BTW, feel free to suggest improvements to the Clojure code. Thanks, Ralph

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  • Investigation: Can different combinations of components effect Dataflow performance?

    - by jamiet
    Introduction The Dataflow task is one of the core components (if not the core component) of SQL Server Integration Services (SSIS) and often the most misunderstood. This is not surprising, its an incredibly complicated beast and we’re abstracted away from that complexity via some boxes that go yellow red or green and that have some lines drawn between them. Example dataflow In this blog post I intend to look under that facade and get into some of the nuts and bolts of the Dataflow Task by investigating how the decisions we make when building our packages can affect performance. I will do this by comparing the performance of three dataflows that all have the same input, all produce the same output, but which all operate slightly differently by way of having different transformation components. I also want to use this blog post to challenge a common held opinion that I see perpetuated over and over again on the SSIS forum. That is, that people assume adding components to a dataflow will be detrimental to overall performance. Its not surprising that people think this –it is intuitive to think that more components means more work- however this is not a view that I share. I have always been of the opinion that there are many factors affecting dataflow duration and the number of components is actually one of the less important ones; having said that I have never proven that assertion and that is one reason for this investigation. I have actually seen evidence that some people think dataflow duration is simply a function of number of rows and number of components. I’ll happily call that one out as a myth even without any investigation!  The Setup I have a 2GB datafile which is a list of 4731904 (~4.7million) customer records with various attributes against them and it contains 2 columns that I am going to use for categorisation: [YearlyIncome] [BirthDate] The data file is a SSIS raw format file which I chose to use because it is the quickest way of getting data into a dataflow and given that I am testing the transformations, not the source or destination adapters, I want to minimise external influences as much as possible. In the test I will split the customers according to month of birth (12 of those) and whether or not their yearly income is above or below 50000 (2 of those); in other words I will be splitting them into 24 discrete categories and in order to do it I shall be using different combinations of SSIS’ Conditional Split and Derived Column transformation components. The 24 datapaths that occur will each input to a rowcount component, again because this is the least resource intensive means of terminating a datapath. The test is being carried out on a Dell XPS Studio laptop with a quad core (8 logical Procs) Intel Core i7 at 1.73GHz and Samsung SSD hard drive. Its running SQL Server 2008 R2 on Windows 7. The Variables Here are the three combinations of components that I am going to test:     One Conditional Split - A single Conditional Split component CSPL Split by Month of Birth and income category that will use expressions on [YearlyIncome] & [BirthDate] to send each row to one of 24 outputs. This next screenshot displays the expression logic in use: Derived Column & Conditional Split - A Derived Column component DER Income Category that adds a new column [IncomeCategory] which will contain one of two possible text values {“LessThan50000”,”GreaterThan50000”} and uses [YearlyIncome] to determine which value each row should get. A Conditional Split component CSPL Split by Month of Birth and Income Category then uses that new column in conjunction with [BirthDate] to determine which of the same 24 outputs to send each row to. Put more simply, I am separating the Conditional Split of #1 into a Derived Column and a Conditional Split. The next screenshots display the expression logic in use: DER Income Category         CSPL Split by Month of Birth and Income Category       Three Conditional Splits - A Conditional Split component that produces two outputs based on [YearlyIncome], one for each Income Category. Each of those outputs will go to a further Conditional Split that splits the input into 12 outputs, one for each month of birth (identical logic in each). In this case then I am separating the single Conditional Split of #1 into three Conditional Split components. The next screenshots display the expression logic in use: CSPL Split by Income Category         CSPL Split by Month of Birth 1& 2       Each of these combinations will provide an input to one of the 24 rowcount components, just the same as before. For illustration here is a screenshot of the dataflow containing three Conditional Split components: As you can these dataflows have a fair bit of work to do and remember that they’re doing that work for 4.7million rows. I will execute each dataflow 10 times and use the average for comparison. I foresee three possible outcomes: The dataflow containing just one Conditional Split (i.e. #1) will be quicker There is no significant difference between any of them One of the two dataflows containing multiple transformation components will be quicker Regardless of which of those outcomes come to pass we will have learnt something and that makes this an interesting test to carry out. Note that I will be executing the dataflows using dtexec.exe rather than hitting F5 within BIDS. The Results and Analysis The table below shows all of the executions, 10 for each dataflow. It also shows the average for each along with a standard deviation. All durations are in seconds. I’m pasting a screenshot because I frankly can’t be bothered with the faffing about needed to make a presentable HTML table. It is plain to see from the average that the dataflow containing three conditional splits is significantly faster, the other two taking 43% and 52% longer respectively. This seems strange though, right? Why does the dataflow containing the most components outperform the other two by such a big margin? The answer is actually quite logical when you put some thought into it and I’ll explain that below. Before progressing, a side note. The standard deviation for the “Three Conditional Splits” dataflow is orders of magnitude smaller – indicating that performance for this dataflow can be predicted with much greater confidence too. The Explanation I refer you to the screenshot above that shows how CSPL Split by Month of Birth and salary category in the first dataflow is setup. Observe that there is a case for each combination of Month Of Date and Income Category – 24 in total. These expressions get evaluated in the order that they appear and hence if we assume that Month of Date and Income Category are uniformly distributed in the dataset we can deduce that the expected number of expression evaluations for each row is 12.5 i.e. 1 (the minimum) + 24 (the maximum) divided by 2 = 12.5. Now take a look at the screenshots for the second dataflow. We are doing one expression evaluation in DER Income Category and we have the same 24 cases in CSPL Split by Month of Birth and Income Category as we had before, only the expression differs slightly. In this case then we have 1 + 12.5 = 13.5 expected evaluations for each row – that would account for the slightly longer average execution time for this dataflow. Now onto the third dataflow, the quick one. CSPL Split by Income Category does a maximum of 2 expression evaluations thus the expected number of evaluations per row is 1.5. CSPL Split by Month of Birth 1 & CSPL Split by Month of Birth 2 both have less work to do than the previous Conditional Split components because they only have 12 cases to test for thus the expected number of expression evaluations is 6.5 There are two of them so total expected number of expression evaluations for this dataflow is 6.5 + 6.5 + 1.5 = 14.5. 14.5 is still more than 12.5 & 13.5 though so why is the third dataflow so much quicker? Simple, the conditional expressions in the first two dataflows have two boolean predicates to evaluate – one for Income Category and one for Month of Birth; the expressions in the Conditional Split in the third dataflow however only have one predicate thus they are doing a lot less work. To sum up, the difference in execution times can be attributed to the difference between: MONTH(BirthDate) == 1 && YearlyIncome <= 50000 and MONTH(BirthDate) == 1 In the first two dataflows YearlyIncome <= 50000 gets evaluated an average of 12.5 times for every row whereas in the third dataflow it is evaluated once and once only. Multiply those 11.5 extra operations by 4.7million rows and you get a significant amount of extra CPU cycles – that’s where our duration difference comes from. The Wrap-up The obvious point here is that adding new components to a dataflow isn’t necessarily going to make it go any slower, moreover you may be able to achieve significant improvements by splitting logic over multiple components rather than one. Performance tuning is all about reducing the amount of work that needs to be done and that doesn’t necessarily mean use less components, indeed sometimes you may be able to reduce workload in ways that aren’t immediately obvious as I think I have proven here. Of course there are many variables in play here and your mileage will most definitely vary. I encourage you to download the package and see if you get similar results – let me know in the comments. The package contains all three dataflows plus a fourth dataflow that will create the 2GB raw file for you (you will also need the [AdventureWorksDW2008] sample database from which to source the data); simply disable all dataflows except the one you want to test before executing the package and remember, execute using dtexec, not within BIDS. If you want to explore dataflow performance tuning in more detail then here are some links you might want to check out: Inequality joins, Asynchronous transformations and Lookups Destination Adapter Comparison Don’t turn the dataflow into a cursor SSIS Dataflow – Designing for performance (webinar) Any comments? Let me know! @Jamiet

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  • Parallelism in .NET – Part 18, Task Continuations with Multiple Tasks

    - by Reed
    In my introduction to Task continuations I demonstrated how the Task class provides a more expressive alternative to traditional callbacks.  Task continuations provide a much cleaner syntax to traditional callbacks, but there are other reasons to switch to using continuations… Task continuations provide a clean syntax, and a very simple, elegant means of synchronizing asynchronous method results with the user interface.  In addition, continuations provide a very simple, elegant means of working with collections of tasks. Prior to .NET 4, working with multiple related asynchronous method calls was very tricky.  If, for example, we wanted to run two asynchronous operations, followed by a single method call which we wanted to run when the first two methods completed, we’d have to program all of the handling ourselves.  We would likely need to take some approach such as using a shared callback which synchronized against a common variable, or using a WaitHandle shared within the callbacks to allow one to wait for the second.  Although this could be accomplished easily enough, it requires manually placing this handling into every algorithm which requires this form of blocking.  This is error prone, difficult, and can easily lead to subtle bugs. Similar to how the Task class static methods providing a way to block until multiple tasks have completed, TaskFactory contains static methods which allow a continuation to be scheduled upon the completion of multiple tasks: TaskFactory.ContinueWhenAll. This allows you to easily specify a single delegate to run when a collection of tasks has completed.  For example, suppose we have a class which fetches data from the network.  This can be a long running operation, and potentially fail in certain situations, such as a server being down.  As a result, we have three separate servers which we will “query” for our information.  Now, suppose we want to grab data from all three servers, and verify that the results are the same from all three. With traditional asynchronous programming in .NET, this would require using three separate callbacks, and managing the synchronization between the various operations ourselves.  The Task and TaskFactory classes simplify this for us, allowing us to write: var server1 = Task.Factory.StartNew( () => networkClass.GetResults(firstServer) ); var server2 = Task.Factory.StartNew( () => networkClass.GetResults(secondServer) ); var server3 = Task.Factory.StartNew( () => networkClass.GetResults(thirdServer) ); var result = Task.Factory.ContinueWhenAll( new[] {server1, server2, server3 }, (tasks) => { // Propogate exceptions (see below) Task.WaitAll(tasks); return this.CompareTaskResults( tasks[0].Result, tasks[1].Result, tasks[2].Result); }); .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; } This is clean, simple, and elegant.  The one complication is the Task.WaitAll(tasks); statement. Although the continuation will not complete until all three tasks (server1, server2, and server3) have completed, there is a potential snag.  If the networkClass.GetResults method fails, and raises an exception, we want to make sure to handle it cleanly.  By using Task.WaitAll, any exceptions raised within any of our original tasks will get wrapped into a single AggregateException by the WaitAll method, providing us a simplified means of handling the exceptions.  If we wait on the continuation, we can trap this AggregateException, and handle it cleanly.  Without this line, it’s possible that an exception could remain uncaught and unhandled by a task, which later might trigger a nasty UnobservedTaskException.  This would happen any time two of our original tasks failed. Just as we can schedule a continuation to occur when an entire collection of tasks has completed, we can just as easily setup a continuation to run when any single task within a collection completes.  If, for example, we didn’t need to compare the results of all three network locations, but only use one, we could still schedule three tasks.  We could then have our completion logic work on the first task which completed, and ignore the others.  This is done via TaskFactory.ContinueWhenAny: var server1 = Task.Factory.StartNew( () => networkClass.GetResults(firstServer) ); var server2 = Task.Factory.StartNew( () => networkClass.GetResults(secondServer) ); var server3 = Task.Factory.StartNew( () => networkClass.GetResults(thirdServer) ); var result = Task.Factory.ContinueWhenAny( new[] {server1, server2, server3 }, (firstTask) => { return this.ProcessTaskResult(firstTask.Result); }); .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; } Here, instead of working with all three tasks, we’re just using the first task which finishes.  This is very useful, as it allows us to easily work with results of multiple operations, and “throw away” the others.  However, you must take care when using ContinueWhenAny to properly handle exceptions.  At some point, you should always wait on each task (or use the Task.Result property) in order to propogate any exceptions raised from within the task.  Failing to do so can lead to an UnobservedTaskException.

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  • AJAX, PHP, XML, and cascading drop-down lists

    - by Dave Jarvis
    What PHP libraries would you recommend to implement the following: Three dependent drop-down lists Three XML data sources AJAX-based Essentially, I'd like to create an XML database and wire up a form that allows the user to select three different dependent parameters: User clicks Region User clicks District (filtered by Region) User clicks Station (filtered by District) Even though I would like to use PHP and XML, the general problem is: One XHTML form Three dependent, cascading drop-down lists Three flat files (no relational database) for the list data The solution must be efficient, simple, reliable, and cross-browser. What technologies would you recommend to solve the problem? Thank you!

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