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  • Managing .NET External Dependencies

    - by Ben Griswold
    Noah and I continue our screencast series by sharing our approach to managing external dependencies referenced within a .NET solution.  This is another introductory episode but you might find a hidden gem in the short 4-minute clip.  ELMAH (Error Logging Modules and Handlers) is the external dependencies we are focusing on in the presentation.  If you are not familiar with ELMAH, this episode may be worth your time.   YouTube - Managing .NET External Dependencies This is one of our first screencasts.  If you have feedback, I’d love to hear it.

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  • Project Euler 52: Ruby

    - by Ben Griswold
    In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 52.  Compared to Problem 51, this problem was a snap. Brute force and pretty quick… As always, any feedback is welcome. # Euler 52 # http://projecteuler.net/index.php?section=problems&id=52 # It can be seen that the number, 125874, and its double, # 251748, contain exactly the same digits, but in a # different order. # # Find the smallest positive integer, x, such that 2x, 3x, # 4x, 5x, and 6x, contain the same digits. timer_start = Time.now def contains_same_digits?(n) value = (n*2).to_s.split(//).uniq.sort.join 3.upto(6) do |i| return false if (n*i).to_s.split(//).uniq.sort.join != value end true end i = 100_000 answer = 0 while answer == 0 answer = i if contains_same_digits?(i) i+=1 end puts answer puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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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 17: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 17.  As always, any feedback is welcome. # Euler 17 # http://projecteuler.net/index.php?section=problems&id=17 # If the numbers 1 to 5 are written out in words: # one, two, three, four, five, then there are # 3 + 3 + 5 + 4 + 4 = 19 letters used in total. # If all the numbers from 1 to 1000 (one thousand) # inclusive were written out in words, how many letters # would be used? # # NOTE: Do not count spaces or hyphens. For example, 342 # (three hundred and forty-two) contains 23 letters and # 115 (one hundred and fifteen) contains 20 letters. The # use of "and" when writing out numbers is in compliance # with British usage. import time start = time.time() def to_word(n): h = { 1 : "one", 2 : "two", 3 : "three", 4 : "four", 5 : "five", 6 : "six", 7 : "seven", 8 : "eight", 9 : "nine", 10 : "ten", 11 : "eleven", 12 : "twelve", 13 : "thirteen", 14 : "fourteen", 15 : "fifteen", 16 : "sixteen", 17 : "seventeen", 18 : "eighteen", 19 : "nineteen", 20 : "twenty", 30 : "thirty", 40 : "forty", 50 : "fifty", 60 : "sixty", 70 : "seventy", 80 : "eighty", 90 : "ninety", 100 : "hundred", 1000 : "thousand" } word = "" # Reverse the numbers so position (ones, tens, # hundreds,...) can be easily determined a = [int(x) for x in str(n)[::-1]] # Thousands position if (len(a) == 4 and a[3] != 0): # This can only be one thousand based # on the problem/method constraints word = h[a[3]] + " thousand " # Hundreds position if (len(a) >= 3 and a[2] != 0): word += h[a[2]] + " hundred" # Add "and" string if the tens or ones # position is occupied with a non-zero value. # Note: routine is broken up this way for [my] clarity. if (len(a) >= 2 and a[1] != 0): # catch 10 - 99 word += " and" elif len(a) >= 1 and a[0] != 0: # catch 1 - 9 word += " and" # Tens and ones position tens_position_value = 99 if (len(a) >= 2 and a[1] != 0): # Calculate the tens position value per the # first and second element in array # e.g. (8 * 10) + 1 = 81 tens_position_value = int(a[1]) * 10 + a[0] if tens_position_value <= 20: # If the tens position value is 20 or less # there's an entry in the hash. Use it and there's # no need to consider the ones position word += " " + h[tens_position_value] else: # Determine the tens position word by # dividing by 10 first. E.g. 8 * 10 = h[80] # We will pick up the ones position word later in # the next part of the routine word += " " + h[(a[1] * 10)] if (len(a) >= 1 and a[0] != 0 and tens_position_value > 20): # Deal with ones position where tens position is # greater than 20 or we have a single digit number word += " " + h[a[0]] # Trim the empty spaces off both ends of the string return word.replace(" ","") def to_word_length(n): return len(to_word(n)) print sum([to_word_length(i) for i in xrange(1,1001)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 2: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 2.  As always, any feedback is welcome. # Euler 2 # http://projecteuler.net/index.php?section=problems&id=2 # Find the sum of all the even-valued terms in the # Fibonacci sequence which do not exceed four million. # Each new term in the Fibonacci sequence is generated # by adding the previous two terms. By starting with 1 # and 2, the first 10 terms will be: # 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # Find the sum of all the even-valued terms in the # sequence which do not exceed four million. import time start = time.time() total = 0 previous = 0 i = 1 while i <= 4000000: if i % 2 == 0: total +=i # variable swapping removes the need for a temp variable i, previous = previous, previous + i print total 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 Euler 11: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 11.  As always, any feedback is welcome. # Euler 11 # http://projecteuler.net/index.php?section=problems&id=11 # What is the greatest product # of four adjacent numbers in any direction (up, down, left, # right, or diagonally) in the 20 x 20 grid? import time start = time.time() grid = [\ [8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8],\ [49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00],\ [81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65],\ [52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91],\ [22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],\ [24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50],\ [32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],\ [67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],\ [24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],\ [21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95],\ [78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92],\ [16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57],\ [86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58],\ [19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40],\ [04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66],\ [88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69],\ [04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],\ [20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16],\ [20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54],\ [01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48]] # left and right max, product = 0, 0 for x in range(0,17): for y in xrange(0,20): product = grid[y][x] * grid[y][x+1] * \ grid[y][x+2] * grid[y][x+3] if product > max : max = product # up and down for x in range(0,20): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x] * \ grid[y+2][x] * grid[y+3][x] if product > max : max = product # diagonal right for x in range(0,17): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x+1] * \ grid[y+2][x+2] * grid[y+3][x+3] if product > max: max = product # diagonal left for x in range(0,17): for y in xrange(0,17): product = grid[y][x+3] * grid[y+1][x+2] * \ grid[y+2][x+1] * grid[y+3][x] if product > max : max = product print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 16: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 16.  As always, any feedback is welcome. # Euler 16 # http://projecteuler.net/index.php?section=problems&id=16 # 2^15 = 32768 and the sum of its digits is # 3 + 2 + 7 + 6 + 8 = 26. # What is the sum of the digits of the number 2^1000? import time start = time.time() print sum([int(i) for i in str(2**1000)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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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 7: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 7.  As always, any feedback is welcome. # Euler 7 # http://projecteuler.net/index.php?section=problems&id=7 # By listing the first six prime numbers: 2, 3, 5, 7, # 11, and 13, we can see that the 6th prime is 13. What # is the 10001st prime number? import time start = time.time() def nthPrime(nth): primes = [2] number = 3 while len(primes) < nth: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes[nth - 1] print nthPrime(10001) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 4: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 4.  As always, any feedback is welcome. # Euler 4 # http://projecteuler.net/index.php?section=problems&id=4 # Find the largest palindrome made from the product of # two 3-digit numbers. A palindromic number reads the # same both ways. The largest palindrome made from the # product of two 2-digit numbers is 9009 = 91 x 99. # Find the largest palindrome made from the product of # two 3-digit numbers. import time start = time.time() def isPalindrome(s): return s == s[::-1] max = 0 for i in xrange(100, 999): for j in xrange(i, 999): n = i * j; if (isPalindrome(str(n))): if (n > max): max = n print max 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 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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  • 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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  • Numbering grouped data in Excel

    - by Jeff
    I have an Excel spreadsheet (2010) with data similar to this: Dogs Brown Nice Dogs White Nice Dogs White Moody Cats Black Nice Cats Black Mean Cats White Nice Cats White Mean I want to group these animals but I only care about species and color. I don't care about disposition. I want to assign group numbers to the set as shown here. 1 Dogs Brown Nice 2 Dogs White Nice 2 Dogs White Moody 3 Cats Black Nice 3 Cats Black Mean 4 Cats White Nice 4 Cats White Mean I was able to select all the species and colors, then from the data tab select 'advanced', then 'unique records only'. This collapsed the data so that I could number the visible rows. Then when I 'cleared' the filter I could easily just fill the blank areas under the numbers with the number above. The problem is that my real data has far too many rows for this to be practical. Also, the trick about entering 1 in the first cell, 2 in the cell below, selecting both then dragging the corner down to 'auto-number' doesn't seem to work when you're viewing filtered rows. Any way to do this?

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  • Create Downloadable CSV File from PHP Script

    - by Aphex22
    How would I create a formatted version of the following PHP script as a downloadable CSV file from the code below (1.0) At the moment the fputcsv function is currently dumping the unparsed PHP/HTML code into a CSV file. This is incorrect. The downloaded CSV file should contain the columns and rows generated from the code at (1.0) as shown in the image link below. I've tried using the following code at the top of the PHP file: // output headers so that the file is downloaded rather than displayed header('Content-Type: text/csv; charset=utf-8'); header('Content-Disposition: attachment; filename=amazon.csv'); // create a file pointer connected to the output stream $output = fopen('php://output', 'w'); $mysql_hostname = ""; $mysql_user = ""; $mysql_password = ""; $mysql_database = ""; $bd = mysql_connect($mysql_hostname, $mysql_user, $mysql_password) or die("Could not connect database"); mysql_select_db($mysql_database, $bd) or die("Could not select database"); $sql = "select * from product WHERE on_amazon = 'on' AND active = 'on'"; $result = mysql_query($sql) or die ( mysql_error() ); // loop over the rows, outputting them while ($sql_result = mysql_fetch_assoc($sql)) fputcsv($output, $sql_result); 1.0 The start of the code outputs the column headings for the CSV file: // set headers echo " item_sku, external_product_id, external_product_id_type, item_name, brand_name, manufacturer, product_description, feed_product_type, update_delete, part_number, model, standard_price, list_price, currency, quantity, product_tax_code, product_site_launch_date, merchant_release_date, restock_date ... <br>"; And then follows PHP script for the column values // load all stock while ($line = mysql_fetch_assoc($result) ) { ?> <?php $size_suffix = array ("",'_chain','_con_b','_con_c'); $arrayLength = count ($size_suffix); for($y=0;$y<$arrayLength;$y++) { //Possible size array to loop through when checking quantity $con_size = array (36,365,37,375,38,385,39,395,40,405,41,415,42,425,43,435,44,445,45,455,46,465,47,475,48,485); $arrlength=count($con_size); for($x=0;$x<$arrlength;$x++) { // check if size is available if($line['quantity_c_size_'.$con_size[$x].$size_suffix[$y]] > 0 ) { ?> <!-- item sku --> <?=$line['product_id']?>, <!-- external product id --> <?=$line['code_size_'.$con_size[$x].'']?>, <? // external product id type $barcode = $line['code_size_'.$con_size[$x]]; $trim_barcode = trim($barcode); $count = strlen($trim_barcode); if ($count == 12) { echo "UPC"; } if ($count == 13) { echo "EAN"; } elseif ($count < 12) { echo " "; } ?>, <!-- item name --> <?=$line['title']?>, <? // brand_name $brand = $line['jys_brand']; echo ucfirst($brand); ?>, <? // manufacturer $brand = $line['jys_brand']; echo ucfirst($brand); ?>, <!-- product description --> <?=preg_replace('/[^\da-z]/i', ' ', $line['amazon_desc']) ?>, <!-- feed product type --> Shoes, , , , <!-- standard price --> <?=$line['price']?>, , <!-- currency --> GBP, <!-- quantity --> <?=$line['quantity_size_'.$con_size[$x].$size_suffix[$y]]?>, , <!-- product site launch date --> <?=$line['added_y']?>-<?=$line['added_m']?>-<?=$line['added_d']?>, <!-- merchat release date --> <?=$line['added_y']?>-<?=$line['added_m']?>-<?=$line['added_d']?>, , , , , <!-- item package quantity --> 1, , , , , <!-- fulfillment latency --> 2, <!-- max aggregate ship quantity --> 1, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , <!-- main image url, url1, url2, url3 --> http://www.getashoe.co.uk/full/<?=$line['product_id']?>_1.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_2.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_3.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_4.jpg, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , <!-- heel height --> <?=$line['heel']?>, , , , , , , , , , , <!-- colour name --> <?=$line['colour']?>, <!-- colour map --> <? $colour = preg_replace('/[()]/i', ' ', $line['colour']); if (preg_match( '/[\/].*/i', $colour)) { echo 'Multicolour'; } if (preg_match( '/off.*/i', $colour)) { echo 'Off-White'; } elseif( preg_match( '/white.*/i', $colour)) { echo 'White'; } elseif( preg_match( '/moro.*/i', $colour)) { echo 'Brown'; } elseif( preg_match( '/morado.*/i', $colour)) { echo 'Purple'; } elseif( preg_match( '/cream.*/i', $colour)) { echo 'Off-White'; } elseif( preg_match( '/pewter.*/i', $colour)) { echo 'Silver'; } elseif( preg_match( '/yellow.*/i', $colour)) { echo 'Yellow'; } elseif( preg_match( '/camel.*/i', $colour)) { echo 'Beige'; } elseif( preg_match( '/navy.*/i', $colour)) { echo 'Blue'; } elseif( preg_match( '/tan.*/i', $colour)) { echo 'Brown'; } elseif( preg_match( '/rainbow.*/i', $colour)) { echo 'Multicolour'; } elseif( preg_match( '/orange.*/i', $colour)) { echo 'Orange'; } elseif( preg_match( '/leopard.*/i', $colour)) { echo 'Multicolour'; } elseif( preg_match( '/red.*/i', $colour)) { echo 'Red'; } elseif( preg_match( '/pink.*/i', $colour)) { echo 'Pink'; } elseif( preg_match( '/purple.*/i', $colour)) { echo 'Purple'; } elseif( preg_match( '/blue.*/i', $colour)) { echo 'Blue'; } elseif( preg_match( '/green.*/i', $colour)) { echo 'Green'; } elseif( preg_match( '/brown.*/i', $colour)) { echo 'Brown'; } elseif( preg_match( '/grey.*/i', $colour)) { echo 'Grey'; } elseif( preg_match( '/black.*/i', $colour)) { echo 'Black'; } elseif( preg_match( '/gold.*/i', $colour)) { echo 'Gold'; } elseif( preg_match( '/silver.*/i', $colour)) { echo 'Silver'; } elseif( preg_match( '/multi.*/i', $colour)) { echo 'Multicolour'; } elseif( preg_match( '/beige.*/i', $colour)) { echo 'Beige'; } elseif( preg_match( '/nude.*/i', $colour)) { echo 'Beige'; } ?>, <!-- size name --> <? echo $con_size[$x];?>, <!-- size map --> <? if ($con_size[$x] == 36) { echo "3 UK"; } elseif ($con_size[$x] == 37 ) { echo "4 UK"; } elseif ($con_size[$x] == 38) { echo "5 UK"; } elseif ($con_size[$x] == 39 ) { echo "6 UK"; } elseif ($con_size[$x] == 40 ) { echo "7 UK"; } elseif ($con_size[$x] == 41) { echo "8 UK"; } elseif ($con_size[$x] == 42) { echo "9 UK"; } elseif ($con_size[$x] == 43) { echo "10 UK"; } elseif ($con_size[$x] == 44 ) { echo "11 UK"; } elseif ($con_size[$x] == 45 ) { echo "12 UK"; } elseif ($con_size[$x] == 46 ) { echo "13 UK"; } elseif ($con_size[$x] == 47 ) { echo "14 UK"; } elseif ($con_size[$x] == 48 ) { echo "15 UK"; } elseif ($con_size[$x] == 365) { echo "3.5 UK"; } elseif ($con_size[$x] == 375 ) { echo "4.5 UK"; } elseif ($con_size[$x] == 385) { echo "5.5 UK"; } elseif ($con_size[$x] == 395 ) { echo "6.5 UK"; } elseif ($con_size[$x] == 405 ) { echo "7.5 UK"; } elseif ($con_size[$x] == 415) { echo "8.5 UK"; } elseif ($con_size[$x] == 425) { echo "9.5 UK"; } elseif ($con_size[$x] == 435) { echo "10.5 UK"; } elseif ($con_size[$x] == 445 ) { echo "11.5 UK"; } elseif ($con_size[$x] == 455 ) { echo "12.5 UK"; } elseif ($con_size[$x] == 465 ) { echo "13.5 UK"; } elseif ($con_size[$x] == 475 ) { echo "14.5 UK"; } elseif ($con_size[$x] == 485 ) { echo "15.5 UK"; } ?>, <br> <? // finish checking if size is available } } } ?> I've included an image of how the CSV file should appear. https://i.imgur.com/ZU3IFer.png Any help would be great.

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  • multi_index composite_key replace with iterator

    - by Rohit
    Is there anyway to loop through an index in a boost::multi_index and perform a replace? #include <iostream> #include <string> #include <boost/multi_index_container.hpp> #include <boost/multi_index/composite_key.hpp> #include <boost/multi_index/member.hpp> #include <boost/multi_index/ordered_index.hpp> using namespace boost::multi_index; using namespace std; struct name_record { public: name_record(string given_name_,string family_name_,string other_name_) { given_name=given_name_; family_name=family_name_; other_name=other_name_; } string given_name; string family_name; string other_name; string get_name() const { return given_name + " " + family_name + " " + other_name; } void setnew(string chg) { given_name = given_name + chg; family_name = family_name + chg; } }; struct NameIndex{}; typedef multi_index_container< name_record, indexed_by< ordered_non_unique< tag<NameIndex>, composite_key < name_record, BOOST_MULTI_INDEX_MEMBER(name_record,string, name_record::given_name), BOOST_MULTI_INDEX_MEMBER(name_record,string, name_record::family_name) > > > > name_record_set; typedef boost::multi_index::index<name_record_set,NameIndex>::type::iterator IteratorType; typedef boost::multi_index::index<name_record_set,NameIndex>::type NameIndexType; void printContainer(name_record_set & ns) { cout << endl << "PrintContainer" << endl << "-------------" << endl; IteratorType it1 = ns.begin(); IteratorType it2 = ns.end (); while (it1 != it2) { cout<<it1->get_name()<<endl; it1++; } cout << "--------------" << endl << endl; } void modifyContainer(name_record_set & ns) { cout << endl << "ModifyContainer" << endl << "-------------" << endl; IteratorType it3; IteratorType it4; NameIndexType & idx1 = ns.get<NameIndex>(); IteratorType it1 = idx1.begin(); IteratorType it2 = idx1.end(); while (it1 != it2) { cout<<it1->get_name()<<endl; name_record nr = *it1; nr.setnew("_CHG"); bool res = idx1.replace(it1,nr); cout << "result is: " << res << endl; it1++; } cout << "--------------" << endl << endl; } int main() { name_record_set ns; ns.insert( name_record("Joe","Smith","ENTRY1") ); ns.insert( name_record("Robert","Brown","ENTRY2") ); ns.insert( name_record("Robert","Nightingale","ENTRY3") ); ns.insert( name_record("Marc","Tuxedo","ENTRY4") ); printContainer (ns); modifyContainer (ns); printContainer (ns); return 0; } PrintContainer ------------- Joe Smith ENTRY1 Marc Tuxedo ENTRY4 Robert Brown ENTRY2 Robert Nightingale ENTRY3 -------------- ModifyContainer ------------- Joe Smith ENTRY1 result is: 1 Marc Tuxedo ENTRY4 result is: 1 Robert Brown ENTRY2 result is: 1 -------------- PrintContainer ------------- Joe_CHG Smith_CHG ENTRY1 Marc_CHG Tuxedo_CHG ENTRY4 Robert Nightingale ENTRY3 Robert_CHG Brown_CHG ENTRY2 --------------

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  • Changing the background color of a view in real-time.

    - by Tarmon
    Hey Everyone, I am trying to implement a ListView that is composed of rows that contain a View on the left followed by a TextView to the right of that. I want to be able to change the background color of the first View based on it's position in the ListView. Below is what I have at this point but it doesn't seem to due anything. public class Routes extends ListActivity { String[] ROUTES; TextView selection; /** Called when the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); ROUTES = getResources().getStringArray(R.array.routes); setContentView(R.layout.routes); setListAdapter(new IconicAdapter()); selection=(TextView)findViewById(R.id.selection); } public void onListItemClick(ListView parent, View v, int position, long id) { selection.setText(ROUTES[position]); } class IconicAdapter extends ArrayAdapter<String> { IconicAdapter() { super(Routes.this, R.layout.row, R.id.label, ROUTES); } } public View getView(int position, View convertView, ViewGroup parent) { LayoutInflater inflater = getLayoutInflater(); View row = inflater.inflate(R.layout.row, parent, false); TextView label = (TextView) row.findViewById(R.id.label); label.setText(ROUTES[position]); View icon = (View) row.findViewById(R.id.icon); switch(position){ case 0: icon.setBackgroundColor(R.color.Red); break; case 1: icon.setBackgroundColor(R.color.Red); break; case 2: icon.setBackgroundColor(R.color.Green); break; case 3: icon.setBackgroundColor(R.color.Green); break; case 4: icon.setBackgroundColor(R.color.Blue); break; case 5: icon.setBackgroundColor(R.color.Blue); break; case 6: icon.setBackgroundColor(R.color.Gray); break; case 7: icon.setBackgroundColor(R.color.Yellow); break; case 8: icon.setBackgroundColor(R.color.Brown); break; case 9: icon.setBackgroundColor(R.color.Brown); break; case 10: icon.setBackgroundColor(R.color.Brown); break; case 11: icon.setBackgroundColor(R.color.Purple); break; case 12: icon.setBackgroundColor(R.color.Red); break; case 13: icon.setBackgroundColor(R.color.Gold); break; case 14: icon.setBackgroundColor(R.color.Orange); break; } return(row); } } Any input is appreciated and if you have any questions don't hesitate to ask! Thanks, Rob

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  • Having problem loading data from AppDelegate using UITableView into a flip view, loads first view bu

    - by Ms. Ryann
    AppDelegate: @implementation Ripe_ProduceGuideAppDelegate -(void)applicationDidFinishLaunching:(UIApplication *)application { Greens *apricot = [[Greens alloc] init]; apricot.produceName = @"Apricot"; apricot.produceSight = @"Deep orange or yellow orange in appearance, may have red tinge, no marks or bruises. "; apricot.produceTouch = @"Firm to touch and give to gentle pressure, plump."; apricot.produceSmell = @"Should be Fragrant"; apricot.produceHtoP = @"raw, salads, baked, sauces, glazes, desserts, poached, stuffing."; apricot.produceStore = @"Not ripe: place in brown paper bag, at room temperature and out of direct sunlight, close bag for 2 - 3 days. Last for a week. Warning: Only refrigerate ripe apricots."; apricot.produceBest = @"Spring & Summer"; apricot.producePic = [UIImage imageNamed:@"apricot.jpg"]; Greens *artichoke = [[Greens alloc] init]; artichoke.produceName = @"Artichoke"; artichoke.produceSight = @"Slightly glossy dark green color and sheen, tight petals that are not be too open, no marks, no brown petals or dried out look. Stem should not be dark brown or black."; artichoke.produceTouch = @"No soft spots"; artichoke.produceSmell = @" Should not smell"; artichoke.produceHtoP = @"steam, boil, grill, saute, soups"; artichoke.produceStore = @"Stand up in vase of cold water, keeps for 2 -3 days. Or, place in refrigerator loose without plastic bag. May be frozen, if cooked but not raw."; artichoke.produceBest = @"Spring"; artichoke.producePic = [UIImage imageNamed:@"artichoke.jpg"]; self.produce = [[NSMutableArray alloc] initWithObjects:apricot, artichoke, nil]; [apricot release]; [artichoke release]; FirstView: @implementation ProduceView -(id)initWithIndexPath: (NSIndexPath *)indexPath { if (self == [super init] ){ index = indexPath; } return self; } -(void)viewDidLoad { Ripe_ProduceGuideAppDelegate *delegate = (Ripe_ProduceGuideAppDelegate *) [[UIApplication sharedApplication] delegate]; Greens *thisProduce = [delegate.produce objectAtIndex:index.row]; self.title = thisProduce.produceName; sightView.text = thisProduce.produceSight; touchView.text = thisProduce.produceTouch; smellView.text = thisProduce.produceSmell; picView.image = thisProduce.producePic; } FlipView: @implementation FlipsideViewController @synthesize flipDelegate; -(id)initWithIndexPath: (NSIndexPath *)indexPath { if ( self == [super init]) { index = indexPath; } return self; } -(void)viewDidLoad { Ripe_ProduceGuideAppDelegate *delegate = (Ripe_ProduceGuideAppDelegate *) [[UIApplication sharedApplication] delegate]; Greens*thisProduce = [delegate.produce objectAtIndex:index.row]; self.title = thisProduce.produceName; bestView.text = thisProduce.produceBest; htopView.text = thisProduce.produceHtoP; storeView.text = thisProduce.produceStore; picView.image = thisProduce.producePic; } *the app works, the flip view for Artichoke shows the information for Apricot. Been working on it for two days. I have been working with iPhone apps for two months now and would very much appreciate any assistance with this problem. Thank you very much.

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  • What's the best Open Php newsletter manager ?

    - by Bilel
    Hi :) I'm looking for a nice newsletter management solution. I tried CCmail a good script but whaere I can't imort usernames !!! I would like to find a system that is able to import Opt-in lists in the following format : John Smith;[email protected];other paramaeters...;[like] ;Male;Age... I will develop my own module if I could find another emailing manager Are you already satisfied with a similar application with a trusted (spam-prevention) emailer ? Thank you :)

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  • Algorithm to compare people names to detect identicalness

    - by Pentium10
    I am working on address book synchronization algorithm. I would like to reuse some code if there exists, but couldn't find one yet. Does someone know about an algorithm that will tell me in numbers/float/procent how much two names are identical. Levenstein distance is not good in this approach, as names and our adddress books are matching the begining of each of the name sections. John Smith should match Smith Jon, Jonathan Smith, Johnny Smith

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  • SQL & Linq To SQL Help

    - by cre-johnny07
    I have a table which is some thing like below.. Date ID 2009-07-01 1 2009-07-01 2 2009-07-01 3 2009-08-01 4 2009-08-01 5 2009-08-01 6 2009-09-01 7 2009-09-01 8 2009-10-01 9 2009-10-01 10 2009-11-01 11 .... Now I need to write a query which will show a output like below. Date Start End 2009-07 1 3 2009-08 4 6 2009-09 7 8 ... How can I do this.. Any help would be highly appreciated Thanking In Advance Johnny

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  • Efficient way to store tuples in the datastore

    - by Drew Sears
    If I have a pair of floats, is it any more efficient (computationally or storage-wise) to store them as a GeoPtProperty than it would be pickle the tuple and store it as a BlobProperty? If GeoPt is doing something more clever to keep multiple values in a single property, can it be leveraged for arbitrary data? Can I store the tuple ("Johnny", 5) in a single entity property in a similarly efficient manner?

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