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  • Project Euler 13: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 6: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 6.  As always, any feedback is welcome. # Euler 6 # http://projecteuler.net/index.php?section=problems&id=6 # Find the difference between the sum of the squares of # the first one hundred natural numbers and the square # of the sum. import time start = time.time() square_of_sums = sum(range(1,101)) ** 2 sum_of_squares = reduce(lambda agg, i: agg+i**2, range(1,101)) print square_of_sums - sum_of_squares print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 20: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 20.  As always, any feedback is welcome. # Euler 20 # http://projecteuler.net/index.php?section=problems&id=20 # n! means n x (n - 1) x ... x 3 x 2 x 1 # Find the sum of digits in 100! import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) print sum([int(i) for i in str(factorial(100))]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 1: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 1.  As always, any feedback is welcome. # Euler 1 # http://projecteuler.net/index.php?section=problems&amp;id=1 # If we list all the natural numbers below 10 that are # multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of # these multiples is 23. Find the sum of all the multiples # of 3 or 5 below 1000. import time start = time.time() print sum([x for x in range(1000) if x % 3== 0 or x % 5== 0]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue') # Also cool def constraint(x): return x % 3 == 0 or x % 5 == 0 print sum(filter(constraint, range(1000)))

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  • Project Euler 3: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 3.  As always, any feedback is welcome. # Euler 3 # http://projecteuler.net/index.php?section=problems&id=3 # The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number # 600851475143? import time start = time.time() def largest_prime_factor(n): max = n divisor = 2 while (n >= divisor ** 2): if n % divisor == 0: max, n = n, n / divisor else: divisor += 1 return max print largest_prime_factor(600851475143) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Motivating developers in a project perceived as **dull** ?

    - by Fanatic23
    As a manager, I can't always end up generating work that'd be cutting edge. Some of the projects do run on maintenance mode, and generate a healthy free cash flow for the company. As a developer what would it take for you to stick around in this project? I have been thinking of re-branding the work, but I could do with a lot of help here. Appreciate a single response per post. Please don't suggest an increased pay-packet, this creates more problems than it solves.

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  • Updating and organizing class diagrams in a growing C++ project

    - by vanna
    I am working on a C++ project that is getting bigger and bigger. I do a lot of UML so it is not really hard to explain my work to co-workers. Lately though I implemented a lot of new features and I gave up updating by hand my Dia UML diagrams. I once used the class diagram of Visual Studio, which is my IDE but didn't get clear results. I need to show my work on a regular basis and I would like to be as clear as possible. Is there any tool that could generate a sort of organized map of my work (namespaces, classes, interactions, etc.) ?

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  • ???: Oracle NoSQL Database??

    - by zhangqm
    ?????????Oracle?????Oracle NoSQL Database,?????NoSQL Database ??????????Oracle NoSQL Database??2???,Community Edition ?Enterprise Edition?????????NoSQL Database 11g R2 (11gR2.1.2.123). ?????????????????: Oracle NoSQL Database OTN portal (includes download facility) Oracle NoSQL Database OTN documentation Oracle NoSQL Database license information ??Oracle NoSQL Database ???????????,????,?????(key-value)???TB????,????????????(???)????,??????????????????????????,????,??????????? ?Oracle NoSQL Database?,???????????key-value???,??key???????:??????????key?(?????string),????????(??????????bytes)??????key-value ??primary key?hash?,????????????????????????????????,???????,????????????????????????? ???????????????????Java API??????Oracle NoSQL Database driver ????????,?????key-value????????????????Oracle NoSQL Database ?????Create, Read, Update and Delete (CRUD)??,???????durability??????????????????????:?????web console???command line??? Oracle Berkeley DB Java Edition Oracle NoSQL Database?? Oracle Berkeley DB Java Edition ????????,??????????????????????????????,?????????????????? ????????????Oracle NoSQL Database Driver?????key-value????????????Oracle NoSQL Database Driver??:?????????hash??????????????????,?????????????????????? ????????Oracle NoSQL Database Oracle NoSQL Database????????????????????????????????????????????????????????????????: ???? ???? ???? ?????? ???? ?????? ????,??,?? ???? ???? ??? (sub-millisecond) ???????? ????? ??????? ????????  ?????Oracle?????? ???? (Oracle Big Data Appliance) ???? ?????????????????????????????????,???“??”???????????,Oracle NoSQL Database???????????Oracle NoSQL Database?????(Cloud)??,????????(TB?PB??)???Oracle NoSQL Database ??????ETL??(??MapReduce, Hadoop)??,??acquire-organize-analyze ?????????? ???????Oracle NoSQL Database?????: • Large schema-less data repositories• Web?? (click-through capture)• ????• ????• ?????????? • Sensor/statistics/network capture (?????, ?????)• ?????????• ???? (MMS, SMS, routing)• ???? Oracle NoSQL Database (Community Edition ??)??????????? Oracle Big Data Appliance???

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  • SQL SERVER – World Shapefile Download and Upload to Database – Spatial Database

    - by pinaldave
    During my recent, training I was asked by a student if I know a place where he can download spatial files for all the countries around the world, as well as if there is a way to upload shape files to a database. Here is a quick tutorial for it. VDS Technologies has all the spatial files for every location for free. You can download the spatial file from here. If you cannot find the spatial file you are looking for, please leave a comment here, and I will send you the necessary details. Unzip the file to a folder and it will have the following content. Then, download Shape2SQL tool from SharpGIS. This is one of the best tools available to convert shapefiles to SQL tables. Afterwards, run the .exe file. When the file is run for the first time, it will ask for the database properties. Provide your database details. Select the appropriate shape files and the tool will fill up the essential details automatically. If you do not want to create the index on the column, uncheck the box beside it. The screenshot below is simply explains the procedure. You also have to be careful regarding your data, whether that is GEOMETRY or GEOGRAPHY. In this example,  it is GEOMETRY data. Click “Upload to Database”. It will show you the uploading process. Once the shape file is uploaded, close the application and open SQL Server Management Studio (SSMS). Run the following code in SSMS Query Editor. USE Spatial GO SELECT * FROM dbo.world GO This will show the complete map of world after you click on Spatial Results in Spatial Tab. In Spatial Results Set, the Zoom feature is available. From the Select label column, choose the country name in order to show the country name overlaying the country borders. Let me know if this tutorial is helpful enough. I am planning to write a few more posts about this later. Note: Please note that the images displayed here do not reflect the original political boundaries. These data are pretty old and can probably draw incorrect maps as well. I have personally spotted several parts of the map where some countries are located a little bit inaccurately. Reference : Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, SQL, SQL Add-On, SQL Authority, SQL Query, SQL Scripts, SQL Server, SQL Spatial, SQL Tips and Tricks, SQL Utility, T SQL, Technology

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  • The Road to Professional Database Development: Database Normalization

    Not only is the process of normalization valuable for increasing data quality and simplifying the process of modifying data, but it actually makes the database perform much faster. To prove the point, Peter Larsson takes a large unnormalised database and subjects it to successive stages of normalisation. Get smart with SQL Backup ProGet faster, smaller backups with integrated verification.Quickly and easily DBCC CHECKDB your backups. Learn more.

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  • ?????????????!??Oracle Exadata??????

    - by Yusuke.Yamamoto
    ????? ??:2010/11/24 ??:?????? ?? Oracle Exadata????????????????????!??????????????? Oracle Exadata ????????!??1??????????????????????????Oracle Exadata Database Machine???????? Oracle Exadata Database Machine ???Oracle Exadata Database Machine ??????????????????????/ For DWH?For OLTPOracle Exadata Database Machine X2 ????/ ?????????????Oracle Exadata Database Machine ????????? ????????? ????????????????? http://www.oracle.com/technology/global/jp/ondemand/otn-seminar/pdf/ord_exadata_public.pdf

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  • Ant: project dependencies in a flat project layout with ivy

    - by MH
    Hello, I have two (Eclipse-) projects. Project A depends on project B, but the projects aren't nested i.e. project A is not a subproject of project B. Apache Ivy is responsible for the dependency management. When I run the compile task in Project A, is there any way to trigger the compile task (in project B) automatically (for example if the jar file of project B doesn't exist)? Thanks a million in advance.

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  • Project Euler 53: Ruby

    - by Ben Griswold
    In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 53.  I first attempted to solve this problem using the Ruby combinations libraries. That didn’t work out so well. With a second look at the problem, the provided formula ended up being just the thing to solve the problem effectively. As always, any feedback is welcome. # Euler 53 # http://projecteuler.net/index.php?section=problems&id=53 # There are exactly ten ways of selecting three from five, # 12345: 123, 124, 125, 134, 135, 145, 234, 235, 245, # and 345 # In combinatorics, we use the notation, 5C3 = 10. # In general, # # nCr = n! / r!(n-r)!,where r <= n, # n! = n(n1)...321, and 0! = 1. # # It is not until n = 23, that a value exceeds # one-million: 23C10 = 1144066. # In general: nCr # How many, not necessarily distinct, values of nCr, # for 1 <= n <= 100, are greater than one-million timer_start = Time.now # There's no factorial method in Ruby, I guess. class Integer # http://rosettacode.org/wiki/Factorial#Ruby def factorial (1..self).reduce(1, :*) end end def combinations(n, r) n.factorial / (r.factorial * (n-r).factorial) end answer = 0 100.downto(3) do |c| (2).upto(c-1) { |r| answer += 1 if combinations(c, r) > 1_000_000 } end puts answer puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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  • Project Euler 14: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 14.  As always, any feedback is welcome. # Euler 14 # http://projecteuler.net/index.php?section=problems&id=14 # The following iterative sequence is defined for the set # of positive integers: # n -> n/2 (n is even) # n -> 3n + 1 (n is odd) # Using the rule above and starting with 13, we generate # the following sequence: # 13 40 20 10 5 16 8 4 2 1 # It can be seen that this sequence (starting at 13 and # finishing at 1) contains 10 terms. Although it has not # been proved yet (Collatz Problem), it is thought that all # starting numbers finish at 1. Which starting number, # under one million, produces the longest chain? # NOTE: Once the chain starts the terms are allowed to go # above one million. import time start = time.time() def collatz_length(n): # 0 and 1 return self as length if n <= 1: return n length = 1 while (n != 1): if (n % 2 == 0): n /= 2 else: n = 3*n + 1 length += 1 return length starting_number, longest_chain = 1, 0 for x in xrange(1, 1000001): l = collatz_length(x) if l > longest_chain: starting_number, longest_chain = x, l print starting_number print longest_chain # Slow 31 seconds print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 12: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 12.  As always, any feedback is welcome. # Euler 12 # http://projecteuler.net/index.php?section=problems&id=12 # The sequence of triangle numbers is generated by adding # the natural numbers. So the 7th triangle number would be # 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms # would be: # 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... # Let us list the factors of the first seven triangle # numbers: # 1: 1 # 3: 1,3 # 6: 1,2,3,6 # 10: 1,2,5,10 # 15: 1,3,5,15 # 21: 1,3,7,21 # 28: 1,2,4,7,14,28 # We can see that 28 is the first triangle number to have # over five divisors. What is the value of the first # triangle number to have over five hundred divisors? import time start = time.time() from math import sqrt def divisor_count(x): count = 2 # itself and 1 for i in xrange(2, int(sqrt(x)) + 1): if ((x % i) == 0): if (i != sqrt(x)): count += 2 else: count += 1 return count def triangle_generator(): i = 1 while True: yield int(0.5 * i * (i + 1)) i += 1 triangles = triangle_generator() answer = 0 while True: num = triangles.next() if (divisor_count(num) >= 501): answer = num break; print answer print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 19: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 19.  As always, any feedback is welcome. # Euler 19 # http://projecteuler.net/index.php?section=problems&id=19 # You are given the following information, but you may # prefer to do some research for yourself. # # - 1 Jan 1900 was a Monday. # - Thirty days has September, # April, June and November. # All the rest have thirty-one, # Saving February alone, # Which has twenty-eight, rain or shine. # And on leap years, twenty-nine. # - A leap year occurs on any year evenly divisible by 4, # but not on a century unless it is divisible by 400. # # How many Sundays fell on the first of the month during # the twentieth century (1 Jan 1901 to 31 Dec 2000)? import time start = time.time() import datetime sundays = 0 for y in range(1901,2001): for m in range(1,13): # monday == 0, sunday == 6 if datetime.datetime(y,m,1).weekday() == 6: sundays += 1 print sundays print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Codenames - Yea or Nay?

    - by rmx
    Where I work, most of our projects have (or at least attempt) descriptive, useful names. However we have a few with names that make no sense: I found that an assembly named WiFi which actually has nothing whatsoever to do with wi-fi, but is a codename. When I asked why, I was told that it's to protect company secrets incase some intern has few too many at the pub on Friday and starts chatting about the brand new 'WiFi' project he's been working on. Its clear that some people find enjoyment in finding silly / amusing codenames for their projects (like in this question). My question is: is it really a good idea to use codenames for your projects or are you better off spending the time to decide upon a descriptive name? My opinion is that in the long-run its better to give your projects relevant names. My reasoning is that if you can't think of a decent name, perhaps you don't really know the requirements well enough. I think there are better ways to 'protect company secrets' and I find it quite confusing when the name does not correlate at all with the content. It's just common sense, surely?! So do you use codenames and what the your reasons for or against this seemingly common, yet annoying (to me at least) practice?

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  • A project idea I've got...

    - by Mr Teeth
    Hi, Next year I will be doing a final year project at Uni. I've already thought of one and was wondering what you guys think of it. I want to create a University Information Search for prospective students who are trying to look for an affordable University to attend. It will depend on the student's family income and the grades they get. They enter in those two parameters (and some more) and it comes up with a list of suitable Unis based on their criteria. This is not about the price of tution fee. It's mostly to do with the cost of living. Stuff like: Rent (if living in a private flat). Student Accomadation. Cost of traveling to your home and back (for holidays). ...and some other stuff stuff I haven't thought about yet. It'll mostly be GUI driven with some textual information. I'm also thinking of using it as a website interface. What do you guys think? Can I program something like this Java? If there's any holes you see in my idea please tell me.

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  • Formatting data from management database

    - by bVector
    I've got some data that goes like this: Config_Name Question Answer Cisco WAN Sensitivity: High Cisco WAN Authorized Users: Brent, Charles Cisco WAN Last Audited: n/a Cisco WAN Next Audit: 3/30/2012 Cisco WAN Audit Signature: Cisco WAN Username: MYCOMPANY Cisco WAN Password: Cisco WAN Encrypted-A ENCRYPTED DATA Cisco WAN Encrypted-B Cisco WAN Encrypted-C vCenter server Sensitivity: High vCenter server Authorized Users: Brent, Charles vCenter server Last Audited: vCenter server Next Audit: 3/30/2012 vCenter server Audit Signature: ENCRYPTED DATA vCenter server Username: administrator vCenter server Password: vCenter server Encrypted-A ENCRYPTED DATA vCenter server Encrypted-B vCenter server Encrypted-C AKSC-NE01 IPMI Sensitivity: High AKSC-NE01 IPMI Authorized Users: Brent, Charles AKSC-NE01 IPMI Last Audited: AKSC-NE01 IPMI Next Audit: 3/30/2012 AKSC-NE01 IPMI Audit Signature: ENCRYPTED DATA AKSC-NE01 IPMI Username: MYCOMPANY AKSC-NE01 IPMI Password: AKSC-NE01 IPMI Encrypted-A ENCRYPTED DATA AKSC-NE01 IPMI Encrypted-B AKSC-NE01 IPMI Encrypted-C and I need it to be in this format: Config_Name Sensitivity: Authorized Users: Last Audited: Next Audit: Audit Signature: Username: Password: Encrypted-A Encrypted-B Encrypted-C AKSC-NE01 IPMI High Brent, Charles 3/30/2012 ENCRYPTED DATA MYCOMPANY ENCRYPTED DATA Cisco ASA5505 WAN High Brent, Charles n/a 3/30/2012 ENCRYPTED DATA MYCOMPANY ENCRYPTED DATA vCenter server High Brent, Charles 3/30/2012 ENCRYPTED DATA administrator ENCRYPTED DATA the tabs get messed up on here but hopefully you get my drift. does anyone know an easy way to do this? I haven't found one with excel just yet.

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  • How can you force the start and end date of a task in Microsoft Project to be on the same day?

    - by Hauke P.
    I have a task called "Interview person A about topic X". The task's duration is set to 2 hours. The start date of the task should automatically be calculated taking dependencies and resource availabilities into account. My question boils down to: How can I force this task to start and end on the same date? Background: In my case, Microsoft Project sets the start date to a Friday at 5pm. As my working hours are set to 8am to 12am and 1pm to 6pm (Mon-Fri), Microsoft Project "splits up" the task at 6pm on Friday and plans to continue it at 8am on the following Monday. However, it does not make any sense to stop the interview on a Friday and restart it on Monday. Therefore the automatic suggestion is not helpful in this case. That's why I'm looking for a way way to force the task to start and end on the very same day. (In my example, I'd like Microsoft Project to delay the start date of the task until Monday 8am as this is the first time slot in which the task "fits in completely".) By the way: I have lots of such cases... for that reason it would be really great if there was a solution that doesn't just deal with this single special case.)

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  • How to force the start and end date of a task in Microsoft Project to be on the same day?

    - by Hauke P.
    I have a task called "Interview person A about topic X". The task's duration is set to 2 hours. The start date of the task should automatically be calculated taking dependencies and resource availabilities into account. My question boils down to: How can I force this task to start and end on the same date? Background: In my case, Microsoft Project sets the start date to a Friday at 5pm. As my working hours are set to 8am to 12am and 1pm to 6pm (Mon-Fri), Microsoft Project "splits up" the task at 6pm on Friday and plans to continue it at 8am on the following Monday. However, it does not make any sense to stop the interview on a Friday and restart it on Monday. Therefore the automatic suggestion is not helpful in this case. That's why I'm looking for a way way to force the task to start and end on the very same day. (In my example, I'd like Microsoft Project to delay the start date of the task until Monday 8am as this is the first time slot in which the task "fits in completely".) By the way: I have lots of such cases... for that reason it would be really great if there was a solution that doesn't just deal with this single special case.

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  • New project created with Flex Mojo's archetype throws Cannot Find Parent Project-Maven Exception

    - by ignorant
    This is probably a silly question but I just cant seem to figure out. I'm completely new to flex and maven. Maven 2.2.1: Maven 2.2.1 unzipped,M2_HOME set and repository altered to point to different drive location in settings.xml Flex 4.0: Installed Created a multi-modular webapp project using flexmojo: mvn archetype:generate -DarchetypeRepository=http://repository.sonatype.org/content/groups/flexgroup -DarchetypeGroupId=org.sonatype.flexmojos -DarchetypeArtifactId=flexmojos-archetypes-modular-webapp -DarchetypeVersion=RELEASE with following options groupId=com.test artifactId=test version=1.0-snapshot package=com.tests * Creates * test |-- pom.xml |--swc -pom.xml |--swf -pom.xml `--war -pom.xml Parent pom has swc, swf, war as modules. Dependency is war-swf-swc. With parent artifactId of swf, swc, war set to swf, swc, test respectively. On executing mvn on test folder(for that matter clean or anything) I get this following error. G:\Projects\testmvn -e + Error stacktraces are turned on. [INFO] Scanning for projects... Downloading: http://repo1.maven.org/maven2/com/test/swc/1.0-snapshot/swc-1.0-snapshot.pom [INFO] Unable to find resource 'com.test:swc:pom:1.0-snapshot' in repository central (http://repo1.maven.org/maven2) [INFO] ------------------------------------------------------------------------ [ERROR] FATAL ERROR [INFO] ------------------------------------------------------------------------ [INFO] Failed to resolve artifact. GroupId: com.test ArtifactId: swc Version: 1.0-snapshot Reason: Unable to download the artifact from any repository com.test:swc:pom:1.0-snapshot from the specified remote repositories: central (http://repo1.maven.org/maven2) [INFO] ------------------------------------------------------------------------ [INFO] Trace org.apache.maven.reactor.MavenExecutionException: Cannot find parent: com.test:swc for project: com.test:swc-swc:swc:1.0-snapshot for project com.test:swc-swc:swc:1.0-snapshot at org.apache.maven.DefaultMaven.getProjects(DefaultMaven.java:404) at org.apache.maven.DefaultMaven.doExecute(DefaultMaven.java:272) at org.apache.maven.DefaultMaven.execute(DefaultMaven.java:138) at org.apache.maven.cli.MavenCli.main(MavenCli.java:362) at org.apache.maven.cli.compat.CompatibleMain.main(CompatibleMain.java:60) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25) at java.lang.reflect.Method.invoke(Method.java:585) at org.codehaus.classworlds.Launcher.launchEnhanced(Launcher.java:315) at org.codehaus.classworlds.Launcher.launch(Launcher.java:255) at org.codehaus.classworlds.Launcher.mainWithExitCode(Launcher.java:430) at org.codehaus.classworlds.Launcher.main(Launcher.java:375) Caused by: org.apache.maven.project.ProjectBuildingException: Cannot find parent: com.test:swc for project: com.test:swc-swc:swc:1.0-snapshot for project com.test:swc-swc:swc:1.0-snapshot at org.apache.maven.project.DefaultMavenProjectBuilder.assembleLineage(DefaultMavenProjectBuilder.java:1396) at org.apache.maven.project.DefaultMavenProjectBuilder.buildInternal(DefaultMavenProjectBuilder.java:823) at org.apache.maven.project.DefaultMavenProjectBuilder.buildFromSourceFileInternal(DefaultMavenProjectBuilder.java:508) at org.apache.maven.project.DefaultMavenProjectBuilder.build(DefaultMavenProjectBuilder.java:200) at org.apache.maven.DefaultMaven.getProject(DefaultMaven.java:604) at org.apache.maven.DefaultMaven.collectProjects(DefaultMaven.java:487) at org.apache.maven.DefaultMaven.collectProjects(DefaultMaven.java:560) at org.apache.maven.DefaultMaven.getProjects(DefaultMaven.java:391) ... 12 more Caused by: org.apache.maven.project.ProjectBuildingException: POM 'com.test:swc' not found in repository: Unable to download the artifact from any repository com.test:swc:pom:1.0-snapshot from the specified remote repositories: central (http://repo1.maven.org/maven2) for project com.test:swc at org.apache.maven.project.DefaultMavenProjectBuilder.findModelFromRepository(DefaultMavenProjectBuilder.java:605) at org.apache.maven.project.DefaultMavenProjectBuilder.assembleLineage(DefaultMavenProjectBuilder.java:1392) ... 19 more Caused by: org.apache.maven.artifact.resolver.ArtifactNotFoundException: Unable to download the artifact from any repository com.test:swc:pom:1.0-snapshot from the specified remote repositories: central (http://repo1.maven.org/maven2) at org.apache.maven.artifact.resolver.DefaultArtifactResolver.resolve(DefaultArtifactResolver.java:228) at org.apache.maven.artifact.resolver.DefaultArtifactResolver.resolve(DefaultArtifactResolver.java:90) at org.apache.maven.project.DefaultMavenProjectBuilder.findModelFromRepository(DefaultMavenProjectBuilder.java:558) ... 20 more Caused by: org.apache.maven.wagon.ResourceDoesNotExistException: Unable to download the artifact from any repository at org.apache.maven.artifact.manager.DefaultWagonManager.getArtifact(DefaultWagonManager.java:404) at org.apache.maven.artifact.resolver.DefaultArtifactResolver.resolve(DefaultArtifactResolver.java:216) ... 22 more [INFO] ------------------------------------------------------------------------ [INFO] Total time: 1 second [INFO] Finished at: Tue Jun 15 19:22:15 GMT+02:00 2010 [INFO] Final Memory: 1M/2M [INFO] ------------------------------------------------------------------------ Looks like its trying to download the project from maven's central repository instead of building it. What am I missing?

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  • Database Design sugessition

    - by Manoj kumar
    I am in the plan of building a information service website similar to http://us.justdial.com/, I was in the confusion of designing the database. The datas stored in the database are List of categories Name of the company and its address, phone number, category, etc... (i am going to use MYSQL Database) how could i design the database that makes the accessing of those data easier ? Thanks in advance

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