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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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  • 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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  • Negative number representation across multiple architechture

    - by Donotalo
    I'm working with OKI 431 micro controller. It can communicate with PC with appropriate software installed. An EEPROM is connected in the I2C bus of the micro which works as permanent memory. The PC software can read from and write to this EEPROM. Consider two numbers, B and C, each is two byte integer. B is known to both the PC software and the micro and is a constant. C will be a number so close to B such that B-C will fit in a signed 8 bit integer. After some testing, appropriate value for C will be determined by PC and will be stored into the EEPROM of the micro for later use. Now the micro can store C in two ways: The micro can store whole two byte representing C The micro can store B-C as one byte signed integer, and can later derive C from B and B-C I think that two's complement representation of negative number is now universally accepted by hardware manufacturers. Still I personally don't like negative numbers to be stored in a storage medium which will be accessed by two different architectures because negative number can be represented in different ways. For you information, 431 also uses two's complement. Should I get rid of the headache that negative number can be represented in different ways and accept the one byte solution as my other team member suggested? Or should I stick to the decision of the two byte solution because I don't need to deal with negative numbers? Which one would you prefer and why?

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  • Apache subdomain not working

    - by tandu
    I'm running apache on my local machine and I'm trying to create a subdomain, but it's not working. Here is what I have (stripped down): <VirtualHost *:80> DocumentRoot /var/www/one ServerName one.localhost </VirtualHost> <VirtualHost *:80> DocumentRoot /var/www/two ServerName two.localhost </VirtualHost> I recently added one. The two entry has been around for a while, and it still works fine (displays the webpage when I go to two.localhost). In fact, I copied the entire two.localhost entry and simply changed two to one, but it's not working. I have tried each of the following: * `apachectl -k graceful` * `apachectl -k restart` * `/etc/init.d/apache2 restart` * `/etc/init.d/apache2 stop && !#:0 start` Apache will complain if /var/www/one does not exist, so I know it's doing something, but when I visit one.localhost in my browser, the browser complains that nothing is there. I put an index.html file there and also tried going to one.localhost/index.html directly, and the browser still won't fine it. This is very perplexing since the entry I copied from two.localhost is exactly the same .. not only that, but if something were wrong I would expect to get a 500 rather than the browser not being able to find anything. The error_log also has nothing extra.

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  • Why does just splitting an Ethernet cable not work?

    - by Sin Jeong-hun
    I thought the Ethernet is logically a one-line communication bus (for argument's sake, I am excluding hubs). All machines attached on the bus hears the same signals and the machines themselves try to avoid collisions by randomly backing off. http://computer.howstuffworks.com/ethernet6.htm If so, why would splitting one Ethernet line from my home router into two and connecting two computers not work? Why do I have to add a switch to it? *What the Internet said would not work. [4 port home router] ------[one Ethernet cable]-----[simple splitter]======[two computers] *What the Internet said I should do [4 port home router] ------[one Ethernet cable]-----[switch]======[two computers] Is this because of the signal degradation (reduced electric current)? Thank you for all the answers! The reason why I did not just use the two ports of my home router is... The 4-port gigabit router is in my room, and I had put a computer in another room (also my room, though). Since a wired network is far more reliable and secure, I had bought a long Ethernet cable and and connected the computer to the router. Now I was thinking about adding another computer to that room. I could buy another long Ethernet cable, but then there will be two cables between the rooms. The one line already is a minor annoyance, so I thought if I could share the one line between the two computers in that room. A switch would work, but it requires power and is a little bit pricey. That is why I wondered why it would not work to simply split the physical Ethernet cable. Apparently I do not completely understand how Ethernet and a switch work. I just have some bit of knowledge I heard in my college class.

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  • Windows: How to make programs think they're not running in a terminal server session?

    - by sinni800
    I am using the program "SoftXPand 2011 Duo" by Miniframe on my Windows 7 PC. It makes two workstations out of one computer. It uses the terminal services built into Windows to create the additional session. I use two screens, two keyboards and two mice to create this "illusion" of two computers. It works quite well and I can even play two different 3D games on the two screens attached to this single machine (using a Radeon HD5770 and a Core i5 2500k with 8 Gbytes RAM). There are a few downsides to this. I just found about one that is hidden on the first look. The sessions you are in (even on the first workstation) will identify as a terminal server session! Now some programs will run with limited effects (graphical), and some won't run at all. This also resulted in some games not running at all. They just say "Cannot be run in a terminal server session" and exit. I have already proven that top modern games (DirectX 10, 11) run just as good as on the same machine without SoftXPand, so this is a pretty artificial limitation! So, can I somehow hack my current session so it doesn't look like a terminal server session anymore? I. E. #include <windows.h> #pragma comment(lib, "user32.lib") BOOL IsRemoteSession(void) { return GetSystemMetrics( SM_REMOTESESSION ); } Will return FALSE? (Not a programming question! Just an example how programs detect if they're in a terminal server session!)

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  • IBM BladeCenter S: Disk Configuration

    - by gravyface
    Have just the one storage bay right now (SAS 15K 600GB x 6) and have configured one storage pool in RAID 10 with 4 disks (and two global spares). For each blade, I've created a volume and mapped accordingly: Blade #1 400 GB Blade #2 200 GB Blade #3 100 GB Blade #4 100 GB When I boot up Blade 1 and enter into the UEFI Setup (F1) followed by the Adapters and UEFI Drivers LSI Logic Fusion MPT SAS Driver Utility, I see 4 disks: two are the on-board 73GB drives, the other two are 200GB each and assume I'm being presented with two logical disks from the volume I created and mapped to this blade. I was a bit surprised by this: I figured I would've been presented with one logical drive per volume, not two. I'm assuming I can just configure whatever RAID level I wish that supports two disks, but not really sure what the benefits/trade-offs here. Should I go with RAID 10 on top of RAID 10? RAID 0? Software RAID 0/1/10? Does it even matter? If this is "normal" to see two disks, then I'm going to likely just do some benchmarking and see if it makes a difference changing the RAID levels (my guess is no); if this is not normal, well, please let me know. :)

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  • Why just splitting an Ethernet cable does not work?

    - by Sin Jeong-hun
    I thought the Ethernet is logically one-line communication bus (for argument's sake, I am excluding hubs). All machines attached in the bus hears the same signals and the machines themselves try to avoid collisions by randomly backing off. http://computer.howstuffworks.com/ethernet6.htm If so, why splitting one Ethernet line from my home router into two and connecting two computers would not work? Why do I have to add a switch to it? *What the Internet said would not work. [4 port home router] ------[one Ethernet cable]-----[simple splitter]======[two computers] *What the Internet said I should do [4 port home router] ------[one Ethernet cable]-----[switch]======[two computers] Is this because of the signal degradation (reduced electric current)? Thank you for all the answers! The reason why I did not just use the two ports of my home router is... The 4-port gigabit router is in my room and I had put a computer in another room (also my room, though). Since wired network is far more reliable and secure, I had bought a long Ethernet cable and and connected the computer to the router. Now I was thinking about adding another computer to that room. I could buy another long Ethernet cable, but then there will be two cables between the rooms. The one line already is a minor annoyance, so I thought if I could share the one line between the two computers in that room. A switch would work, but it requires power and is a little bit pricey. That is why I wondered why it would not work to simply split the physical Ethernet cable. Apparently I do not completely understand how Ethernet and a switch work. I just have some bit of knowledge I heard in my college class.

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  • C# Alternating threads

    - by Mutoh
    Imagine a situation in which there are one king and n number of minions submissed to him. When the king says "One!", one of the minions says "Two!", but only one of them. That is, only the fastest minion speaks while the others must wait for another call of the king. This is my try: using System; using System.Threading; class Program { static bool leaderGO = false; void Leader() { do { lock(this) { //Console.WriteLine("? {0}", leaderGO); if (leaderGO) Monitor.Wait(this); Console.WriteLine("> One!"); Thread.Sleep(200); leaderGO = true; Monitor.Pulse(this); } } while(true); } void Follower (char chant) { do { lock(this) { //Console.WriteLine("! {0}", leaderGO); if (!leaderGO) Monitor.Wait(this); Console.WriteLine("{0} Two!", chant); leaderGO = false; Monitor.Pulse(this); } } while(true); } static void Main() { Console.WriteLine("Go!\n"); Program m = new Program(); Thread king = new Thread(() => m.Leader()); Thread minion1 = new Thread(() => m.Follower('#')); Thread minion2 = new Thread(() => m.Follower('$')); king.Start(); minion1.Start(); minion2.Start(); Console.ReadKey(); king.Abort(); minion1.Abort(); minion2.Abort(); } } The expected output would be this (# and $ representing the two different minions): > One! # Two! > One! $ Two! > One! $ Two! ... The order in which they'd appear doesn't matter, it'd be random. The problem, however, is that this code, when compiled, produces this instead: > One! # Two! $ Two! > One! # Two! > One! $ Two! # Two! ... That is, more than one minion speaks at the same time. This would cause quite the tumult with even more minions, and a king shoudln't allow a meddling of this kind. What would be a possible solution?

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  • How do I disable MEDIUM and WEAK/LOW strength ciphers in Apache + mod_ssl?

    - by superwormy
    A PCI Compliance scan has suggested that we disable Apache's MEDIUM and LOW/WEAK strength ciphers for security. Can someone tell me how to disable these ciphers? Apache v2.2.14 mod_ssl v2.2.14 This is what they've told us: Synopsis : The remote service supports the use of medium strength SSL ciphers. Description : The remote host supports the use of SSL ciphers that offer medium strength encryption, which we currently regard as those with key lengths at least 56 bits and less than 112 bits. Solution: Reconfigure the affected application if possible to avoid use of medium strength ciphers. Risk Factor: Medium / CVSS Base Score : 5.0 (CVSS2#AV:N/AC:L/Au:N/C:P/I:N/A:N) [More] Synopsis : The remote service supports the use of weak SSL ciphers. Description : The remote host supports the use of SSL ciphers that offer either weak encryption or no encryption at all. See also : http://www.openssl.org/docs/apps/ciphers .html Solution: Reconfigure the affected application if possible to avoid use of weak ciphers. Risk Factor: Medium / CVSS Base Score : 5.0 (CVSS2#AV:N/AC:L/Au:N/C:P/I:N/A:N) [More]

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  • Windows Vista claims wireless key is the wrong length

    - by humble coffee
    A family member of mine is house sitting and has been given the details of their wifi. The access point is an Airport Express, it has WEP encryption (I think) and they've been given a passphrase to use. I know it's a passphrase and not the encrypted key as it's an English word. The passphrase is 10 characters long. The problem is that Vista complains that it's not a valid key as it must be a 5 or 13 character non-hex key or a 10 or 26 character hex key. (From what I've read this suggests the encryption is WEP?) I've found a couple of suggested solutions, but I'm not actually at the house at the moment so I wanted to make sure I have a good chance of getting it to work when I'm there but have no internets to ask. Solution 1: Vista needs to be told explicitly what kind of encryption and key is being used. Specify in the connection settings that you are using WEP and that it is a "shared key". Solution2: Try converting the passphrase to hexadecimal using an ASCII-hex converter and entering that.

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  • How do I get a Wireless N PCi card to connect to a wireless G router?

    - by Andy
    I'm having some problems setting up a new wireless PCI card on a WinXP SP3 PC. I know that the router is configured correctly. It is a Linksys WRT54GL, using 802.11b/g. Security mode is WPA2 Personal with TKIP+AES encryption. I am able to connect to this fine using my laptop (first gen MacBook with a 802.11b built in card). The new PCI card is also Linksys, but it supports 802.11n. Card seems to be installed ok (Windows sees it fine, doesn't list any errors in Device Manager), however when it scans for available wireless networks it can't find my wireless network (the router is set to broadcast the SSID). I tried to enter the network SSID manually, but that didn't seem to help. I chose WPA2-PSK for network authentication. The only options for encryption are TKIP or AES - I've tried both, neither worked. I am sure that I typed in my wireless key correctly. At this point, I don't think the problem is with encryption, but something else. It almost seems like I need to switch the wireless card into g mode, but I haven't found a way to do that (if that is even possible/necessary - I thought n was fully backwards compatible with g). Also, the PC is in the same room as the router, and my laptop, so I don't think that it is an interference issue. Any ideas what I'm doing wrong? I'm running out of things to try at this point. :(

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  • New, separate window in PowerPoint

    - by bobobobo
    I'm trying to open two PowerPoint 2007 documents, and they are open, but they're STUCK in the same window. I can't look at both presentations simultaneously, which is what I want to do. I want to open each presentation in ITS OWN, SEPARATE WINDOW, like in MS-Word how you can have two documents open and they'd be in two separate, draggable windows. I want OUT of the MDI and just have two completely separate windows! How?

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  • Exchange 2007 and migrating only some users under a shared domain name

    - by DomoDomo
    I'm in the process of moving two law firms to hosted Exchange 2007, a service that the consulting company I work for offers. Let's call these two firms Crane Law and Poole Law. These two firms were ONE firm just six months ago, but split. So they have three email domains: Old Firm: craneandpoole.com New Firm 1: cranelaw.com New Firm 2: poolelaw.com Both Firm 1 & Firm 2 use craneandpoole.com email addresses, as for the other two domains, only people who work at the respective firm use that firm's domain name, natch. Currently these two firms are still using the same pre-split internal Exchange 2007 server, where MX records for all three domains point. Here's the problem. I'm not moving both companies at the same time. I'm moving Crane Law two weeks before Poole Law. During this two weeks, both companies need to be able to: Continue to receive emails addressed to craneandpoole.com Send emails between firms, using cranelaw.com and poolelaw.com accounts I also have a third problem: I'd like to setup all three domains in my hosting infrastructure way ahead of time, to make my own life easier What would solve all my problems would be, if there is some way I can tell Exchange 2007, even though this domain exists locally forward on the message to the outside world using public MX record as a basis for where to send it (or if I could somehow create a route for it statically that would work too). If this doesn't work, to address points #1 when I migrate Crane Law, I will delete all references locally to cranelaw.com on their current Exchange server, and setup individual forwards for each of their craneandpool.com mailboxes to forward to our hosted exchange server. This will also take care of point #2, since the cranelaw.com won't be there locally, when poolelaw.com tries to send to cranelaw.com, public MX records will be used for mail routing decisions and go to my hosted exchange. The bummer of that though is, I won't be able to setup poolelaw.com ahead of time in hosted Exchange, will have to wait to do it day of :( Sorry for the long and confusing post. Just wondering if there is a better or simpler way to do what I want? Three tier forests and that kind of thing are out, this is just a two week window where they won't be in the same place.

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  • Duplication of Architecture State means physically extra?

    - by Doopy Doo
    Hyper-Threading Technology makes a single physical processor appear as two logical processors; the physical execution resources are shared and the architecture state is duplicated for the two logical processors. So, this means that there are two sets of basic registers such as Next Instruction Pointer, processor registers like AX, BX, CX etc physically embedded in the micro-processor chip, OR they(arch. state) are made to look two sets by some low level duplication by software/OS.

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  • How to link specific ports to specific domains with Apache virtual hosts?

    - by theJoe
    We have a forward-facing linux box running Apache HTTP server that is acting as a reverse proxy for several back-end servers. The servers are accessed through specific domain names and ports and are set up as virtual hosts within Apache as such: Listen 8001 Listen 8002 <Virtualhost *:8001> ServerName service.one.mycompany.com ProxyPass / http://internal.one.mycompany.com:8001/ ProxyPassReverse / http://internal.one.mycompany.com:8001/ RewriteEngine On RewriteCond %{REQUEST_METHOD} ^(TRACE|TRACK) RewriteRule .* - [F] </Virtualhost> <Virtualhost *:8002> ServerName service.two.mycompany.com ProxyPass / http://internal.two.mycompany.com:8002/ ProxyPassReverse / http://internal.two.mycompany.com:8002/ RewriteEngine On RewriteCond %{REQUEST_METHOD} ^(TRACE|TRACK) RewriteRule .* - [F] </Virtualhost> The proxy server has only one IP address, and both domains are pointing to it. Accessing internal.one via service.one works fine, as does accessing internal.two via service.two. Now the problem is that Apache does not take the requesting domain into account when accessing the virtual hosts. What I mean is that both domains work for both ports: requests for service.one:8002 proxies to internal.two:8002, and requests for service.two:8001 proxies to internal.one:8001, where ideally both these requests should be denied. I can get around this by creating more virtual hosts that explicitly deny these requests: NameVirtualHost *:8001 NameVirtualHost *:8002 <Virtualhost *:8001> ServerName service.two.mycompany.com Redirect permanent / http://errorpage.mycompany.com/ </Virtualhost> <Virtualhost *:8002> ServerName service.one.mycompany.com Redirect permanent / http://errorpage.mycompany.com/ </Virtualhost> But this is not an ideal solution, since we plan to add more services to the proxy, and each new port would need to be explicitly denied on all the other domains, and each new domain would need to be explicitly denied on all ports it is not utilizing. As we add more services, the number of virtual hosts can get out of hand quickly. My question, then, is whether there is a better way? Can we explicitly tie specific ports to specific domains in a virtual host so that only that domain-port combination is processed, and all other combinations are not? Things I’ve tried: Adding NameVirtualHost *:8001, etc. without the additional virtual hosts. Setting ProxyRequests On and Off, as well as ProxyPreserveHost On and Off Adding the server name or IP address to the virtual host header, e.g. <VirtualHost service.one.mycompany.com:8001> Using the <proxy> directive inside the virtual host directive. Lots and lots of googling. The proxy server is running CentOS 6.2 64-bit, Apache HTTPD server 2.2.15. As mentioned, the proxy server has only one IP address, and all the domains we are using are pointing to it.

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  • C# how to get current encoding type used by C# to write/read configuration for config file?

    - by 5YrsLaterDBA
    I am doing connection string encryption. we use our own encryption key with AES algorithm to do this. during the process, we need to convert string to byte array and then convert byte array back to string. I found the encoding play an important role on those conversions. So I need to know the encoding C# is using to get above conversion right. Any idea how to get current encoding programmably? thanks,

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  • AesManaged and RijndaelManaged

    - by xscape
    Im currently developing a Silverlight application that connects to an old webservice. Our old webservice uses an encryption tool which silverlight does not support. Finally, we decided to used AesManaged for encryption, however, our webservice does not support AesManaged. Is their a way to decrypt an AesManaged to RijndaelManaged? If yes, can you please post a sample snippet? Your feedback is highly needed. Thank you.

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  • UITableView: Juxtaposing row, header, and footer insertions/deletions

    - by jdandrea
    Consider a very simple UITableView with one of two states. First state: One (overall) table footer One section containing two rows, a section header, and a section footer Second state: No table footer One section containing four rows and no section header/footer In both cases, each row is essentially one of four possible UITableViewCell objects, each containing its own UITextField. We don't even bother with reuse or caching, since we're only dealing with four known cells in this case. They've been created in an accompanying XIB, so we already have them all wired up and ready to go. Now consider we want to toggle between the two states. Sounds easy enough. Let's suppose our view controller's right bar button item provides the toggling support. We'll also track the current state with an ivar and enumeration. To be explicit for a sec, here's how one might go from state 1 to 2. (Presume we handle the bar button item's title as well.) In short, we want to clear out our table's footer view, then insert the third and fourth rows. We batch this inside an update block like so: // Brute forced references to the third and fourth rows in section 0 NSUInteger row02[] = {0, 2}; NSUInteger row03[] = {0, 3}; [self.tableView beginUpdates]; state = tableStateTwo; // 'internal' iVar, not a property self.tableView.tableFooterView = nil; [self.tableView insertRowsAtIndexPaths:[NSArray arrayWithObjects: [NSIndexPath indexPathWithIndexes:row02 length:2], [NSIndexPath indexPathWithIndexes:row03 length:2], nil] withRowAnimation:UITableViewRowAnimationFade]; [self.tableView endUpdates]; For the reverse, we want to reassign the table footer view (which, like the cells, is in the XIB ready and waiting), and remove the last two rows: // Use row02 and row03 from earlier snippet [self.tableView beginUpdates]; state = tableStateOne; self.tableView.tableFooterView = theTableFooterView; [self.tableView deleteRowsAtIndexPaths:[NSArray arrayWithObjects: [NSIndexPath indexPathWithIndexes:row02 length:2], [NSIndexPath indexPathWithIndexes:row03 length:2], nil] withRowAnimation:UITableViewRowAnimationFade]; [self.tableView endUpdates]; Now, when the table asks for rows, it's very cut and dry. The first two cells are the same in both cases. Only the last two appear/disappear depending on the state. The state ivar is consulted when the Table View asks for things like number of rows in a section, height for header/footer in a section, or view for header/footer in a section. This last bit is also where I'm running into trouble. Using the above logic, section 0's header/footer does not disappear. Specifically, the footer stays below the inserted rows, but the header now overlays the topmost row. If we switch back to state one, the section footer is removed, but the section header remains. How about using [self.tableView reloadData] then? Sure, why not. We take care not to use it inside the update block, per Apple's advisement, and simply add it after endUpdates. This time, good news! The section 0 header/footer disappears. :) However ... Toggling back to state one results in a most exquisite mess! The section 0 header returns, only to overlay the first row once again (instead of appear above it). The section 0 footer is placed below the last row just fine, but the overall table footer - now reinstated - overlays the section footer. Waaaaaah … now what? Just to be sure, let's toggle back to state two again. Yep, that looks fine. Coming back to state one? Yecccch. I also tried sprinkling in a few other stunts like using reloadSections:withRowAnimation:, but that only serves to make things worse. NSRange range = {0, 1}; NSIndexSet *indexSet = [NSIndexSet indexSetWithIndexesInRange:range]; ... [self.tableView reloadSections:indexSet withRowAnimation:UITableViewRowAnimationFade]; Case in point: If we invoke reloadSections... just before the end of the update block, changing to state two hides the first two rows from view, even though the space they would otherwise occupy remains. Switching back to state one returns section 0's header/footer to normal, but those first two rows remain invisible. Case two: Moving reloadSections... to just after the update block but before reloadData results in all rows becoming invisible! (I refer to the row as being invisible because, during tracing, tableView:cellForRowAtIndexPath: is returning bona-fide cell objects for those rows.) Case three: Moving reloadSections... after tableView:cellForRowAtIndexPath: brings us a bit closer, but the section 0 header/footer never returns when switching back to state one. Hmm. Perhaps it's a faux pas using both reloadSections... and reloadData, based on what I'm seeing at trace-time, which brings us to: Case four: Replacing reloadData with reloadSections... outright. All cells in state two disappear. All cells in state one remain missing as well (though the space is kept). So much for that theory. :) Tracing through the code, the cell and view objects, as well as the section heights, are all where they should be at the opportune times. They just aren't rendering sanely. So, how to crack this case? Clues welcome/appreciated!

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  • How can I copy Data from one sheet to another Sheet in Excel 07 Through Macro

    - by Mwaseem Alvi
    Hello, I am using MS Office 2007. Please let me know that how can I copy whole data from sheet one to sheet two. I want to copy the whole data from row 5 to onward in sheet two. The whole scenrio is given below in detail. Sheet one: Copy the data from column B and Row 3 Sheet Two: Paste the Copied Data in Column B and Row 3 Sheet One: Copy the whole data from Column B to Column G and Row 5 to onward Sheet Two: Paste whole copied data in sheet two from last filled row to onward Data dont overwrite on any row or column. Every data will be add in sheet two from sheet one when macro will be run. Thanks

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  • Is the Keychain suitable for storing general data, such as strings?

    - by cannyboy
    The Keychain seems to be used a lot for usernames and passwords, but is it a good idea to use it for other sensitive stuff (bank details, ID numbers etc), but with no password? What kind of encryption does the keychain use? The scenario I'm concerned about is a thief acquiring an iPhone (which is screen-locked) and being able to access the file system to get this info. Also, would using the Keychain involve export restrictions due to the use of encryption?

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  • How can I generate a list of words made up of combinations of three word lists in Perl?

    - by Chris Denman
    I have three lists of words. I would like to generate a single list of all the combinations of words from the three lists. List 1: red green blue List 2: one two List 3: apple banana The final list would like like so: red one apple red two apple red one banana red two banana ... and so on Ideally I'd like to pass in three arrays and the routine return one array. I have done a simple loop like so: foreach $word1 (@list1){ foreach $word2 (@list2){ foreach $word3 (@list3){ print "$word1 $word2 $word3\n"; } } } However, this doesn't work if there's nothing in the second or third list (I may only want to iterate between one, two or three lists at a time - in other words, if I only supply two lists it should iterate between those two lists).

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  • which of these modes : cbc,cfb,ctr,ecb,ncfb,nofb,ofb,stream are secure and which are absolute no-no

    - by user393087
    By security I mean that encoded string is indistinguishable from random noise and is different on every encryption of the same text so it is impossible to make a guess on encryption algorithm used or do any dictionary attack on the encoded text. Second: output string length does not correspond to the input string length in easy way, so it is not possible of make guessing on that account. Third: it is possible to verify that the provided password is incorrect so the decoding function could return false instead of supposedly decoded random string.

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