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  • How to explain to users the advantages of dumb primary key?

    - by Hao
    Primary key attractiveness I have a boss(and also users) that wants primary key to be sophisticated/smart/attractive control number(sort of like Social Security number, or credit card number format) I just padded the primary key(in Views) with zeroes to appease their desire to make the control number sophisticated,smart and attractive. But they wanted it as: first 2 digits as client code, then 4 digits as year year, then last 4 digits as transaction number on that client on a given year, then reset the transaction number of client to 1 when next year flows. Each client's transaction starts with 1. e.g. WM20090001, WM20090002, BB2009001, WM20100001, BB20100001 But as I wanted to make things as simple as possible, I forgo embedding their suggested smartness in primary key, I just keep the primary key auto increments regardless of client and year. But to make it not dull-looking(they really are adamant to make the primary key as smart control number), I made the primary key appears to them smart, on view query, I put the client code and four digit year code on front of the eight-zero padded autoincrement key, i.e. WM200900000001. Sort of slug-like information on autoincremented primary key. Keeping primary key autoincrement regardless of any other information, we are able keep other potential side effects problem when they edit a record, for example, if they made a mistake of entering the transaction on WM, then they edit the client code to BB, if we use smart primary key, the primary keys of WM customer will have gaps in their control number. Or worse yet, instead of letting the control numbers have gaps/holes, the user will request that subsequent records of that gap should shift up to that gap and have their subsequent primary keys re-adjust(decremented). How do you deal with these user requests(reasonable or otherwise)? Do you yield to their request? Or just continue using dumb primary key and explain them the repercussions of having a very smart/sophisticated primary key and educate them the significant advantages of having a dumb primary key? P.S. quotable quote(http://articles.techrepublic.com.com/5100-10878_11-1044961.html): "If you hold your tongue the first time users ask what is for them a reasonable request, things will work a lot better in the end."

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  • Programming R/Sweave for proper \Sexpr output

    - by deoksu
    Hi I'm having a bit of a problem programming R for Sweave, and the #rstats twitter group often points here, so I thought I'd put this question to the SO crowd. I'm an analyst- not a programmer- so go easy on me my first post. Here's the problem: I am drafting a survey report in Sweave with R and would like to report the marginal returns in line using \Sexpr{}. For example, rather than saying: Only 14% of respondents said 'X'. I want to write the report like this: Only \Sexpr{p.mean(variable)}$\%$ of respondents said 'X'. The problem is that Sweave() converts the results of the expression in \Sexpr{} to a character string, which means that the output from expression in R and the output that appears in my document are different. For example, above I use the function 'p.mean': p.mean<- function (x) {options(digits=1) mmm<-weighted.mean(x, weight=weight, na.rm=T) print(100*mmm) } In R, the output looks like this: p.mean(variable) >14 but when I use \Sexpr{p.mean(variable)}, I get an unrounded character string (in this case: 13.5857142857143) in my document. I have tried to limit the output of my function to 'digits=1' in the global environment, in the function itself, and and in various commands. It only seems to contain what R prints, not the character transformation that is the result of the expression and which eventually prints in the LaTeX file. as.character(p.mean(variable)) >[1] 14 >[1] "13.5857142857143" Does anyone know what I can do to limit the digits printed in the LaTeX file, either by reprogramming the R function or with a setting in Sweave or \Sexpr{}? I'd greatly appreciate any help you can give. Thanks, David

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  • codingBat separateThousands using regex (and unit testing how-to)

    - by polygenelubricants
    This question is a combination of regex practice and unit testing practice. Regex part I authored this problem separateThousands for personal practice: Given a number as a string, introduce commas to separate thousands. The number may contain an optional minus sign, and an optional decimal part. There will not be any superfluous leading zeroes. Here's my solution: String separateThousands(String s) { return s.replaceAll( String.format("(?:%s)|(?:%s)", "(?<=\\G\\d{3})(?=\\d)", "(?<=^-?\\d{1,3})(?=(?:\\d{3})+(?!\\d))" ), "," ); } The way it works is that it classifies two types of commas, the first, and the rest. In the above regex, the rest subpattern actually appears before the first. A match will always be zero-length, which will be replaceAll with ",". The rest basically looks behind to see if there was a match followed by 3 digits, and looks ahead to see if there's a digit. It's some sort of a chain reaction mechanism triggered by the previous match. The first basically looks behind for ^ anchor, followed by an optional minus sign, and between 1 to 3 digits. The rest of the string from that point must match triplets of digits, followed by a nondigit (which could either be $ or \.). My question for this part is: Can this regex be simplified? Can it be optimized further? Ordering rest before first is deliberate, since first is only needed once No capturing group Unit testing part As I've mentioned, I'm the author of this problem, so I'm also the one responsible for coming up with testcases for them. Here they are: INPUT, OUTPUT "1000", "1,000" "-12345", "-12,345" "-1234567890.1234567890", "-1,234,567,890.1234567890" "123.456", "123.456" ".666666", ".666666" "0", "0" "123456789", "123,456,789" "1234.5678", "1,234.5678" "-55555.55555", "-55,555.55555" "0.123456789", "0.123456789" "123456.789", "123,456.789" I haven't had much experience with industrial-strength unit testing, so I'm wondering if others can comment whether this is a good coverage, whether I've missed anything important, etc (I can always add more tests if there's a scenario I've missed).

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  • Array: mathematical sequence

    - by VaioIsBorn
    An array of integers A[i] (i 1) is defined in the following way: an element A[k] ( k 1) is the smallest number greater than A[k-1] such that the sum of its digits is equal to the sum of the digits of the number 4* A[k-1] . You need to write a program that calculates the N th number in this array based on the given first element A[1] . INPUT: In one line of standard input there are two numbers seperated with a single space: A[1] (1 <= A[1] <= 100) and N (1 <= N <= 10000). OUTPUT: The standard output should only contain a single integer A[N] , the Nth number of the defined sequence. Input: 7 4 Output: 79 Explanation: Elements of the array are as follows: 7, 19, 49, 79... and the 4th element is solution. I tried solving this by coding a separate function that for a given number A[k] calculates the sum of it's digits and finds the smallest number greater than A[k-1] as it says in the problem, but with no success. The first testing failed because of a memory limit, the second testing failed because of a time limit, and now i don't have any possible idea how to solve this. One friend suggested recursion, but i don't know how to set that. Anyone who can help me in any way please write, also suggest some ideas about using recursion/DP for solving this problem. Thanks.

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  • Using stdint.h and ANSI printf?

    - by nn
    Hi, I'm writing a bignum library, and I want to use efficient data types to represent the digits. Particularly integer for the digit, and long (if strictly double the size of the integer) for intermediate representations when adding and multiplying. I will be using some C99 functionality, but trying to conform to ANSI C. Currently I have the following in my bignum library: #include <stdint.h> #if defined(__LP64__) || defined(__amd64) || defined(__x86_64) || defined(__amd64__) || defined(__amd64__) || defined(_LP64) typedef uint64_t u_w; typedef uint32_t u_hw; #define BIGNUM_DIGITS 2048 #define U_HW_BITS 16 #define U_W_BITS 32 #define U_HW_MAX UINT32_MAX #define U_HW_MIN UINT32_MIN #define U_W_MAX UINT64_MAX #define U_W_MIN UINT64_MIN #else typedef uint32_t u_w; typedef uint16_t u_hw; #define BIGNUM_DIGITS 4096 #define U_HW_BITS 16 #define U_W_BITS 32 #define U_HW_MAX UINT16_MAX #define U_HW_MIN UINT16_MIN #define U_W_MAX UINT32_MAX #define U_W_MIN UINT32_MIN #endif typedef struct bn { int sign; int n_digits; // #digits should exclude carry (digits = limbs) int carry; u_hw tab[BIGNUM_DIGITS]; } bn; As I haven't written a procedure to write the bignum in decimal, I have to analyze the intermediate array and printf the values of each digit. However I don't know which conversion specifier to use with printf. Preferably I would like to write to the terminal the digit encoded in hexadecimal. The underlying issue is, that I want two data types, one that is twice as long as the other, and further use them with printf using standard conversion specifiers. It would be ideal if int is 32bits and long is 64bits but I don't know how to guarantee this using a preprocessor, and when it comes time to use functions such as printf that solely rely on the standard types I no longer know what to use.

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  • Readability and IF-block brackets: best practice

    - by MasterPeter
    I am preparing a short tutorial for level 1 uni students learning JavaScript basics. The task is to validate a phone number. The number must not contain non-digits and must be 14 digits long or less. The following code excerpt is what I came up with and I would like to make it as readable as possible. if ( //set of rules for invalid phone number phoneNumber.length == 0 //empty || phoneNumber.length > 14 //too long || /\D/.test(phoneNumber) //contains non-digits ) { setMessageText(invalid); } else { setMessageText(valid); } A simple question I can not quite answer myself and would like to hear your opinions on: How to position the surrounding (outermost) brackets? It's hard to see the difference between a normal and a curly bracket. Do you usually put the last ) on the same line as the last condition? Do you keep the first opening ( on a line by itself? Do you wrap each individual sub-condition in brackets too? Do you align horizontally the first ( with the last ), or do you place the last ) in the same column as the if? Do you keep ) { on a separate line or you place the last ) on the same line with the last sub-condition and then place the opening { on a new line? Or do you just put the ) { on the same line as the last sub-condition? Community wiki. EDIT Please only post opinions regarding the usage and placement of brackets. The code needs not be re-factored. This is for people who have only been introduced to JavaScript a couple of weeks ago. I am not asking for opinions how to write the code so it's shorter or performs better. I would only like to know how do you place brackets around IF-conditions.

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  • Using $.get with jquery validation

    - by Jimmy McCarthy
    I'm trying to use the Jquery builtin validator on my page. The issue is that I have certain fields that only are required if the JobID (entered into another field) does not already exist in our database. I have a simple service which simply takes JobID and returns True or False based on whether the JobID exists, but I can't seem to get this information where I want it. Some sample code: $("#dep_form").validate({ rules: { JobID: { required: true, digits: true, minlength: 3 }, OrgName: { required: function(element) { //This field is required if the JobID is new. return $("#jobinfo").html().length==15; } } }, messages: { JobID: { required: "Enter a Job Number.", digits: "Only numbers are allowed in Job ID's.", minlength: "Job Number must be at least 3 digits" }, OrgName: { required: "Select a Practice from the dropdown list." } }, errorClass: "ui-state-error-input", errorLabelContainer: "#errorbox", errorElement: 'li', errorContainer: "#validation_errors", onfocusout: false, onkeyup: false, focusinvalid: false }; Currently, I'm using a lazy method to validate (shown above). However, I now have access to a service using the URL: var lookupurl = "/deposits/jobidvalidate/?q=" + $("#id_JobID").val() + "&t=" + new Date().getTime(); which is a page which will contain just the word True or False based on whether that given JobID exists. I've tried half a dozen different ways of setting variables and calling functions within functions and still cannot get a way to simply return the value of that page (which I've been trying to access with $.get() ) to my validator, so that required is set to true when the Job does not exist and false if the job already exists. Any suggestions? Thanks.

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  • Truncate C++ string fields generated by ostringstream, iomanip:setw

    - by Ian Durkan
    In C++ I need string representations of integers with leading zeroes, where the representation has 8 digits and no more than 8 digits, truncating digits on the right side if necessary. I thought I could do this using just ostringstream and iomanip.setw(), like this: int num_1 = 3000; ostringstream out_target; out_target << setw(8) << setfill('0') << num_1; cout << "field: " << out_target.str() << " vs input: " << num_1 << endl; The output here is: field: 00003000 vs input: 3000 Very nice! However if I try a bigger number, setw lets the output grow beyond 8 characters: int num_2 = 2000000000; ostringstream out_target; out_target << setw(8) << setfill('0') << num_2; cout << "field: " << out_target.str() << " vs input: " << num_2 << endl; out_target.str(""); output: field: 2000000000 vs input: 2000000000 The desired output is "20000000". There's nothing stopping me from using a second operation to take only the first 8 characters, but is field truncation truly missing from iomanip? Would the Boost formatting do what I need in one step?

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  • IF-block brackets: best practice

    - by MasterPeter
    I am preparing a short tutorial for level 1 uni students learning JavaScript basics. The task is to validate a phone number. The number must not contain non-digits and must be 14 digits long or less. The following code excerpt is what I came up with and I would like to make it as readable as possible. if ( //set of rules for invalid phone number phoneNumber.length == 0 //empty || phoneNumber.length > 14 //too long || /\D/.test(phoneNumber) //contains non-digits ) { setMessageText(invalid); } else { setMessageText(valid); } A simple question I can not quite answer myself and would like to hear your opinions on: How to position the surrounding (outermost) brackets? It's hard to see the difference between a normal and a curly bracket. Do you usually put the last ) on the same line as the last condition? Do you keep the first opening ( on a line by itself? Do you wrap each individual sub-condition in brackets too? Do you align horizontally the first ( with the last ), or do you place the last ) in the same column as the if? Do you keep ) { on a separate line or you place the last ) on the same line with the last sub-condition and then place the opening { on a new line? Or do you just put the ) { on the same line as the last sub-condition? Community wiki.

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  • Array: Recursive problem cracked me up

    - by VaioIsBorn
    An array of integers A[i] (i 1) is defined in the following way: an element A[k] ( k 1) is the smallest number greater than A[k-1] such that the sum of its digits is equal to the sum of the digits of the number 4* A[k-1] . You need to write a program that calculates the N th number in this array based on the given first element A[1] . INPUT: In one line of standard input there are two numbers seperated with a single space: A[1] (1 <= A[1] <= 100) and N (1 <= N <= 10000). OUTPUT: The standard output should only contain a single integer A[N] , the Nth number of the defined sequence. Input: 7 4 Output: 79 Explanation: Elements of the array are as follows: 7, 19, 49, 79... and the 4th element is solution. I tried solving this by coding a separate function that for a given number A[k] calculates the sum of it's digits and finds the smallest number greater than A[k-1] as it says in the problem, but with no success. The first testing failed because of a memory limit, the second testing failed because of a time limit, and now i don't have any possible idea how to solve this. One friend suggested recursion, but i don't know how to set that. Anyone who can help me in any way please write, also suggest some ideas about using recursion/DP for solving this problem. Thanks.

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  • string maniupulations, oops, how do I replace parts of a string

    - by Joe Gibson
    I am very new to python. Could someone explain how I can manipulate a string like this? This function receives three inputs: complete_fmla: has a string with digits and symbols but has no hyphens ('-') nor spaces. partial_fmla: has a combination of hyphens and possibly some digits or symbols, where the digits and symbols that are in it (other than hyphens) are in the same position as in the complete_formula. symbol: one character The output that should be returned is: If the symbol is not in the complete formula, or if the symbol is already in the partial formula, the function should return the same formula as the input partial_formula. If the symbol is in the complete_formula and not in the partial formula, the function should return the partial_formula with the symbol substituting the hyphens in the positions where the symbol is, in all the occurrences of symbol in the complete_formula. For example: generate_next_fmla (‘abcdeeaa’, ‘- - - - - - - - ’, ‘d’) should return ‘- - - d - - - -’ generate_next_fmla (‘abcdeeaa’, ‘- - - d - - - - ’, ‘e’) should return ‘- - - d e e - -’ generate_next_fmla (‘abcdeeaa’, ‘- - - d e e - - ’, ‘a’) should return ‘a - - d e e a a’ Basically, I'm working with the definition: def generate_next_fmla (complete_fmla, partial_fmla, symbol): Do I turn them into lists? and then append? Also, should I find out the index number for the symbol in the complete_fmla so that I know where to append it in the string with hyphens??

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  • PHP regex for password validation

    - by Fabio Anselmo
    I not getting the desired effect from a script. I want the password to contain A-Z, a-z, 0-9, and special chars. A-Z a-z 0-9 2 special chars 2 string length = 8 So I want to force the user to use at least 2 digits and at least 2 special chars. Ok my script works but forces me to use the digits or chars back to back. I don't want that. e.g. password testABC55$$ is valid - but i don't want that. Instead I want test$ABC5#8 to be valid. So basically the digits/special char can be the same or diff - but must be split up in the string. PHP CODE: $uppercase = preg_match('#[A-Z]#', $password); $lowercase = preg_match('#[a-z]#', $password); $number = preg_match('#[0-9]#', $password); $special = preg_match('#[\W]{2,}#', $password); $length = strlen($password) >= 8; if(!$uppercase || !$lowercase || !$number || !$special || !$length) { $errorpw = 'Bad Password';

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  • Threshold of blurry image - part 2

    - by 1''
    How can I threshold this blurry image to make the digits as clear as possible? In a previous post, I tried adaptively thresholding a blurry image (left), which resulted in distorted and disconnected digits (right): Since then, I've tried using a morphological closing operation as described in this post to make the brightness of the image uniform: If I adaptively threshold this image, I don't get significantly better results. However, because the brightness is approximately uniform, I can now use an ordinary threshold: This is a lot better than before, but I have two problems: I had to manually choose the threshold value. Although the closing operation results in uniform brightness, the level of brightness might be different for other images. Different parts of the image would do better with slight variations in the threshold level. For instance, the 9 and 7 in the top left come out partially faded and should have a lower threshold, while some of the 6s have fused into 8s and should have a higher threshold. I thought that going back to an adaptive threshold, but with a very large block size (1/9th of the image) would solve both problems. Instead, I end up with a weird "halo effect" where the centre of the image is a lot brighter, but the edges are about the same as the normally-thresholded image: Edit: remi suggested morphologically opening the thresholded image at the top right of this post. This doesn't work too well. Using elliptical kernels, only a 3x3 is small enough to avoid obliterating the image entirely, and even then there are significant breakages in the digits:

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  • Project Euler 51: Ruby

    - by Ben Griswold
    In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 51.  I know I started back up with Python this week, but I have three more Ruby solutions in the hopper and I wanted to share. For the record, Project Euler 51 was the second hardest Euler problem for me thus far. Yeah. As always, any feedback is welcome. # Euler 51 # http://projecteuler.net/index.php?section=problems&id=51 # By replacing the 1st digit of *3, it turns out that six # of the nine possible values: 13, 23, 43, 53, 73, and 83, # are all prime. # # By replacing the 3rd and 4th digits of 56**3 with the # same digit, this 5-digit number is the first example # having seven primes among the ten generated numbers, # yielding the family: 56003, 56113, 56333, 56443, # 56663, 56773, and 56993. Consequently 56003, being the # first member of this family, is the smallest prime with # this property. # # Find the smallest prime which, by replacing part of the # number (not necessarily adjacent digits) with the same # digit, is part of an eight prime value family. timer_start = Time.now require 'mathn' def eight_prime_family(prime) 0.upto(9) do |repeating_number| # Assume mask of 3 or more repeating numbers if prime.count(repeating_number.to_s) >= 3 ctr = 1 (repeating_number + 1).upto(9) do |replacement_number| family_candidate = prime.gsub(repeating_number.to_s, replacement_number.to_s) ctr += 1 if (family_candidate.to_i).prime? end return true if ctr >= 8 end end false end # Wanted to loop through primes using Prime.each # but it took too long to get to the starting value. n = 9999 while n += 2 next if !n.prime? break if eight_prime_family(n.to_s) end puts n puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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  • SQL SERVER – Introduction to PERCENTILE_CONT() – Analytic Functions Introduced in SQL Server 2012

    - by pinaldave
    SQL Server 2012 introduces new analytical function PERCENTILE_CONT(). The book online gives following definition of this function: Computes a specific percentile for sorted values in an entire rowset or within distinct partitions of a rowset in Microsoft SQL Server 2012 Release Candidate 0 (RC 0). For a given percentile value P, PERCENTILE_DISC sorts the values of the expression in the ORDER BY clause and returns the value with the smallest CUME_DIST value (with respect to the same sort specification) that is greater than or equal to P. If you are clear with understanding of the function – no need to read further. If you got lost here is the same in simple words – it is lot like finding median with percentile value. Now let’s have fun following query: USE AdventureWorks GO SELECT SalesOrderID, OrderQty, ProductID, PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY ProductID) OVER (PARTITION BY SalesOrderID) AS MedianCont FROM Sales.SalesOrderDetail WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY SalesOrderID DESC GO The above query will give us the following result: You can see that I have used PERCENTILE_COUNT(0.5) in query, which is similar to finding median. Let me explain above diagram with little more explanation. The defination of median is as following: In case of Even Number of elements = In ordered list add the two digits from the middle and devide by 2 In case of Odd Numbers of elements = In ordered list select the digits from the middle I hope this example gives clear idea how PERCENTILE_CONT() works. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Function, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Handling extremely large numbers in a language which can't?

    - by Mallow
    I'm trying to think about how I would go about doing calculations on extremely large numbers (to infinitum - intergers no floats) if the language construct is incapable of handling numbers larger than a certain value. I am sure I am not the first nor the last to ask this question but the search terms I am using aren't giving me an algorithm to handle those situations. Rather most suggestions offer a language change or variable change, or talk about things that seem irrelevant to my search. So I need a little guideance. I would sketch out an algorithm like this: Determine the max length of the integer variable for the language. If a number is more than half the length of the max length of the variable split it in an array. (give a little play room) Array order [0] = the numbers most to the right [n-max] = numbers most to the left Ex. Num: 29392023 Array[0]:23, Array[1]: 20, array[2]: 39, array[3]:29 Since I established half the length of the variable as the mark off point I can then calculate the ones, tenths, hundredths, etc. Place via the halfway mark so that if a variable max length was 10 digits from 0 to 9999999999 then I know that by halfing that to five digits give me some play room. So if I add or multiply I can have a variable checker function that see that the sixth digit (from the right) of array[0] is the same place as the first digit (from the right) of array[1]. Dividing and subtracting have their own issues which I haven't thought about yet. I would like to know about the best implementations of supporting larger numbers than the program can.

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  • remove words containing non-alpha characters

    - by dnkb
    Given a text file with space separated string and a tab separated integer, I'd ;like to get rid of all words that have non-alpha characters but keep words consisting of alpha only characters and the tab plus the integer afterwards. My attempts like the ones below didin't yield any good. What I was trying to express is something like: "replace anything within word boundaries that starts and ends with 0 or more whatever and there is at least one :digits: or :punct: in between". sed 's/\b.[:digits::punct:]+.\b//g' sed 's/\b.[^:alpha:]+.\b//g' What am I missing? See sample input data below. Thank you! asdf 754m 563 a2a 754mm 291 754n 463 754 ppp 1409 754pin 4652 pin pin 462 754pins 652 754 ppp 1409 754pin 4652 pi$n pin 462 754/p ins 652 754 pp+p 1409 754 p=in 4652

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  • Bitmap Font Displays in Center Always Without Coding it Manually (Fix Coordinate Problem onText)

    - by David Dimalanta
    Is there a way on how to stay the texts in center without manually coding it or something, especially when making an update? I'm making a display for the highest score. Let's say that the score is 9. However, if the score is 9,999,999, the text displays still only at the fixed X and Y coordinate. Is there really a way to stay the text in center especially when there is changes when a player beats the new world record? Here's my code inside Sprite Batch: font.setScale(1.5f); font.draw(batch, "HIGHEST SCORE:", (900/10)*1 + 60, (1280/16)*10); font.draw(batch, "" + 9999999 + "", (900/10)*4, (1280/16)*8); batch.draw(grid_guide, 0, 0, 900, 1280); // --> For testing purpose only. // Where 9999999 is a new record score for example. Here's the image shown as example. I add it some red grid so that I could check if the display of score when updated will always display on center no matter how many digits takes place in. However, it is fixed, so I have to figure it out how to display it automatically on center regardless of the number of digits while updating for the new highscore. I have used the LibGDX preferences very well though to save and load records for the highscore.

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  • Excel VLOOKUP using results from a formula as the lookup value [on hold]

    - by Rick Deemer
    I have a cell that I must remove the first 2 characters "RO" for each value in a column on a sheet called RAW DATA and put into a cell on a sheet called ROSS DATA. Some of the values in that cell have 3 digits after the "RO", and some have 5 digits. To do that I used =REPLACE('RAW DATA'!A3,1,2,"") Then I need to use this new resultant string as the lookup value in a VLOOKUP. The VLOOKUP will be looking at a named range called DAP on a sheet called DAP, in column 5 for an exact match, and I need it to return that value to the cell. I have tried using INDIRECT in different ways to no avail, and I'm not sure that I fully understand its usage. So at this point I am Googling for a method to do this and at a standstill.

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  • Working with packed dates in SSIS

    - by Jim Giercyk
    One of the challenges recently thrown my way was to read an EBCDIC flat file, decode packed dates, and insert the dates into a SQL table.  For those unfamiliar with packed data, it is a way to store data at the nibble level (half a byte), and was often used by mainframe programmers to conserve storage space.  In the case of my input file, the dates were 2 bytes long and  represented the number of days that have past since 01/01/1950.  My first thought was, in the words of Scooby, Hmmmmph?  But, I love a good challenge, so I dove in. Reading in the flat file was rather simple.  The only difference between reading an EBCDIC and an ASCII file is the Code Page option in the connection manager.  In my case, I needed to use Code Page 1140 for EBCDIC (I could have also used Code Page 37).       Once the code page is set correctly, SSIS can understand what it is reading and it will convert the output to the default code page, 1252.  However, packed data is either unreadable or produces non-alphabetic characters, as we can see in the preview window.   Column 1 is actually the packed date, columns 0 and 2 are the values in the rest of the file.  We are only interested in Column 1, which is a 2 byte field representing a packed date.  We know that 2 bytes of packed data can be stored in 1 byte of character data, so we are working with 4 packed digits in 2 character bytes.  If you are confused, stay tuned….this will make sense in a minute.   Right-click on your Flat File Source shape and select “Show Advanced Editor”. Here is where the magic begins. By changing the properties of the output columns, we can access the packed digits from each byte. By default, the Output Column data type is DT_STR. Since we want to look at the bytes individually and not the entire string, change the data type to DT_BYTES. Next, and most important, set UseBinaryFormat to TRUE. This will write the HEX VALUES of the output string instead of writing the character values.  Now we are getting somewhere! Next, you will need to use a Data Conversion shape in your Data Flow to transform the 2 position byte stream to a 4 position Unicode string containing the packed data.  You need the string to be 4 bytes long because it will contain the 4 packed digits.  Here is what that should look like in the Data Conversion shape: Direct the output of your data flow to a test table or file to see the results.  In my case, I created a test table.  The results looked like this:     Hold on a second!  That doesn't look like a date at all.  No, of course not.  It is a hex number which represents the days which have passed between 01/01/1950 and the date.  We have to convert the Hex value to a decimal value, and use the DATEADD function to get a date value.  Luckily, I have created a function to convert Hex to Decimal:   -- ============================================= -- Author:        Jim Giercyk -- Create date: March, 2012 -- Description:    Converts a Hex string to a decimal value -- ============================================= CREATE FUNCTION [dbo].[ftn_HexToDec] (     @hexValue NVARCHAR(6) ) RETURNS DECIMAL AS BEGIN     -- Declare the return variable here DECLARE @decValue DECIMAL IF @hexValue LIKE '0x%' SET @hexValue = SUBSTRING(@hexValue,3,4) DECLARE @decTab TABLE ( decPos1 VARCHAR(2), decPos2 VARCHAR(2), decPos3 VARCHAR(2), decPos4 VARCHAR(2) ) DECLARE @pos1 VARCHAR(1) = SUBSTRING(@hexValue,1,1) DECLARE @pos2 VARCHAR(1) = SUBSTRING(@hexValue,2,1) DECLARE @pos3 VARCHAR(1) = SUBSTRING(@hexValue,3,1) DECLARE @pos4 VARCHAR(1) = SUBSTRING(@hexValue,4,1) INSERT @decTab VALUES (CASE               WHEN @pos1 = 'A' THEN '10'                 WHEN @pos1 = 'B' THEN '11'               WHEN @pos1 = 'C' THEN '12'               WHEN @pos1 = 'D' THEN '13'               WHEN @pos1 = 'E' THEN '14'               WHEN @pos1 = 'F' THEN '15'               ELSE @pos1              END, CASE               WHEN @pos2 = 'A' THEN '10'                 WHEN @pos2 = 'B' THEN '11'               WHEN @pos2 = 'C' THEN '12'               WHEN @pos2 = 'D' THEN '13'               WHEN @pos2 = 'E' THEN '14'               WHEN @pos2 = 'F' THEN '15'               ELSE @pos2              END, CASE               WHEN @pos3 = 'A' THEN '10'                 WHEN @pos3 = 'B' THEN '11'               WHEN @pos3 = 'C' THEN '12'               WHEN @pos3 = 'D' THEN '13'               WHEN @pos3 = 'E' THEN '14'               WHEN @pos3 = 'F' THEN '15'               ELSE @pos3              END, CASE               WHEN @pos4 = 'A' THEN '10'                 WHEN @pos4 = 'B' THEN '11'               WHEN @pos4 = 'C' THEN '12'               WHEN @pos4 = 'D' THEN '13'               WHEN @pos4 = 'E' THEN '14'               WHEN @pos4 = 'F' THEN '15'               ELSE @pos4              END) SET @decValue = (CONVERT(INT,(SELECT decPos4 FROM @decTab)))         +                 (CONVERT(INT,(SELECT decPos3 FROM @decTab))*16)      +                 (CONVERT(INT,(SELECT decPos2 FROM @decTab))*(16*16)) +                 (CONVERT(INT,(SELECT decPos1 FROM @decTab))*(16*16*16))     RETURN @decValue END GO     Making use of the function, I found the decimal conversion, added that number of days to 01/01/1950 and FINALLY arrived at my “unpacked relative date”.  Here is the query I used to retrieve the formatted date, and the result set which was returned: SELECT [packedDate] AS 'Hex Value',        dbo.ftn_HexToDec([packedDate]) AS 'Decimal Value',        CONVERT(DATE,DATEADD(day,dbo.ftn_HexToDec([packedDate]),'01/01/1950'),101) AS 'Relative String Date'   FROM [dbo].[Output Table]         This technique can be used any time you need to retrieve the hex value of a character string in SSIS.  The date example may be a bit difficult to understand at first, but with SSIS becoming the preferred tool for enterprise level integration for many companies, there is no doubt that developers will encounter these types of requirements with regularity in the future. Please feel free to contact me if you have any questions.

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  • How can I estimate the entropy of a password?

    - by Wug
    Having read various resources about password strength I'm trying to create an algorithm that will provide a rough estimation of how much entropy a password has. I'm trying to create an algorithm that's as comprehensive as possible. At this point I only have pseudocode, but the algorithm covers the following: password length repeated characters patterns (logical) different character spaces (LC, UC, Numeric, Special, Extended) dictionary attacks It does NOT cover the following, and SHOULD cover it WELL (though not perfectly): ordering (passwords can be strictly ordered by output of this algorithm) patterns (spatial) Can anyone provide some insight on what this algorithm might be weak to? Specifically, can anyone think of situations where feeding a password to the algorithm would OVERESTIMATE its strength? Underestimations are less of an issue. The algorithm: // the password to test password = ? length = length(password) // unique character counts from password (duplicates discarded) uqlca = number of unique lowercase alphabetic characters in password uquca = number of uppercase alphabetic characters uqd = number of unique digits uqsp = number of unique special characters (anything with a key on the keyboard) uqxc = number of unique special special characters (alt codes, extended-ascii stuff) // algorithm parameters, total sizes of alphabet spaces Nlca = total possible number of lowercase letters (26) Nuca = total uppercase letters (26) Nd = total digits (10) Nsp = total special characters (32 or something) Nxc = total extended ascii characters that dont fit into other categorys (idk, 50?) // algorithm parameters, pw strength growth rates as percentages (per character) flca = entropy growth factor for lowercase letters (.25 is probably a good value) fuca = EGF for uppercase letters (.4 is probably good) fd = EGF for digits (.4 is probably good) fsp = EGF for special chars (.5 is probably good) fxc = EGF for extended ascii chars (.75 is probably good) // repetition factors. few unique letters == low factor, many unique == high rflca = (1 - (1 - flca) ^ uqlca) rfuca = (1 - (1 - fuca) ^ uquca) rfd = (1 - (1 - fd ) ^ uqd ) rfsp = (1 - (1 - fsp ) ^ uqsp ) rfxc = (1 - (1 - fxc ) ^ uqxc ) // digit strengths strength = ( rflca * Nlca + rfuca * Nuca + rfd * Nd + rfsp * Nsp + rfxc * Nxc ) ^ length entropybits = log_base_2(strength) A few inputs and their desired and actual entropy_bits outputs: INPUT DESIRED ACTUAL aaa very pathetic 8.1 aaaaaaaaa pathetic 24.7 abcdefghi weak 31.2 H0ley$Mol3y_ strong 72.2 s^fU¬5ü;y34G< wtf 88.9 [a^36]* pathetic 97.2 [a^20]A[a^15]* strong 146.8 xkcd1** medium 79.3 xkcd2** wtf 160.5 * these 2 passwords use shortened notation, where [a^N] expands to N a's. ** xkcd1 = "Tr0ub4dor&3", xkcd2 = "correct horse battery staple" The algorithm does realize (correctly) that increasing the alphabet size (even by one digit) vastly strengthens long passwords, as shown by the difference in entropy_bits for the 6th and 7th passwords, which both consist of 36 a's, but the second's 21st a is capitalized. However, they do not account for the fact that having a password of 36 a's is not a good idea, it's easily broken with a weak password cracker (and anyone who watches you type it will see it) and the algorithm doesn't reflect that. It does, however, reflect the fact that xkcd1 is a weak password compared to xkcd2, despite having greater complexity density (is this even a thing?). How can I improve this algorithm? Addendum 1 Dictionary attacks and pattern based attacks seem to be the big thing, so I'll take a stab at addressing those. I could perform a comprehensive search through the password for words from a word list and replace words with tokens unique to the words they represent. Word-tokens would then be treated as characters and have their own weight system, and would add their own weights to the password. I'd need a few new algorithm parameters (I'll call them lw, Nw ~= 2^11, fw ~= .5, and rfw) and I'd factor the weight into the password as I would any of the other weights. This word search could be specially modified to match both lowercase and uppercase letters as well as common character substitutions, like that of E with 3. If I didn't add extra weight to such matched words, the algorithm would underestimate their strength by a bit or two per word, which is OK. Otherwise, a general rule would be, for each non-perfect character match, give the word a bonus bit. I could then perform simple pattern checks, such as searches for runs of repeated characters and derivative tests (take the difference between each character), which would identify patterns such as 'aaaaa' and '12345', and replace each detected pattern with a pattern token, unique to the pattern and length. The algorithmic parameters (specifically, entropy per pattern) could be generated on the fly based on the pattern. At this point, I'd take the length of the password. Each word token and pattern token would count as one character; each token would replace the characters they symbolically represented. I made up some sort of pattern notation, but it includes the pattern length l, the pattern order o, and the base element b. This information could be used to compute some arbitrary weight for each pattern. I'd do something better in actual code. Modified Example: Password: 1234kitty$$$$$herpderp Tokenized: 1 2 3 4 k i t t y $ $ $ $ $ h e r p d e r p Words Filtered: 1 2 3 4 @W5783 $ $ $ $ $ @W9001 @W9002 Patterns Filtered: @P[l=4,o=1,b='1'] @W5783 @P[l=5,o=0,b='$'] @W9001 @W9002 Breakdown: 3 small, unique words and 2 patterns Entropy: about 45 bits, as per modified algorithm Password: correcthorsebatterystaple Tokenized: c o r r e c t h o r s e b a t t e r y s t a p l e Words Filtered: @W6783 @W7923 @W1535 @W2285 Breakdown: 4 small, unique words and no patterns Entropy: 43 bits, as per modified algorithm The exact semantics of how entropy is calculated from patterns is up for discussion. I was thinking something like: entropy(b) * l * (o + 1) // o will be either zero or one The modified algorithm would find flaws with and reduce the strength of each password in the original table, with the exception of s^fU¬5ü;y34G<, which contains no words or patterns.

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  • Best way to detect integer overflow in C/C++

    - by Chris Johnson
    I was writing a program in C++ to find all solutions of a^b = c (a to the power of b), where a, b and c together use all the digits 0-9 exactly once. The program looped over values of a and b, and ran a digit-counting routine each time on a, b and a^b to check if the digits condition was satisfied. However, spurious solutions can be generated when a^b overflows the integer limit. I ended up checking for this using code like: unsigned long b, c, c_test; ... c_test=c*b; // Possible overflow if (c_test/b != c) {/* There has been an overflow*/} else c=c_test; // No overflow Is there a better way of testing for overflow? I know that some chips have an internal flag that is set when overflow occurs, but I've never seen it accessed through C or C++.

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  • adding jquery validation after form created

    - by CoffeeCode
    im using a jquery validation plugin first i add the validation to the form: $('#EmployeeForm').validate({ rules: { "Employee.FirstName": "required", "Employee.PatronymicName": "required", "Employee.LastName": "required", "Employee.BirthDay": { required: true, date: true } }, messages: { "Employee.FirstName": { required: "*" }, "Employee.PatronymicName": { required: "*" }, "Employee.LastName": { required: "*" }, "Employee.BirthDay": { required: "*", date: "00.00.00 format" } } }); and then latter a need to add validation rules to other form elements: $('#Address_A.Phone1, #Address_A.Phone2, #Address_B.Phone1, #Address_B.Phone2').rules("add", { digits: true, messages: { digits: "?????? ?????" } }); but i get an error: 'form' is null or not an object i check, the form and all the elements in it are created before i add validation to it. cant figer out whats wrong.

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  • Fast modulo 3 or division algorithm?

    - by aaa
    Hello is there a fast algorithm, similar to power of 2, which can be used with 3, i.e. n%3. Perhaps something that uses the fact that if sum of digits is divisible by three, then the number is also divisible. This leads to a next question. What is the fast way to add digits in a number? I.e. 37 - 3 +7 - 10 I am looking for something that does not have conditionals as those tend to inhibit vectorization thanks

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  • migrate method Rand(int) Visual Fox Pro to C#.net

    - by ch2o
    I'm migrating a Visual Fox Pro code to C #. NET What makes the Visual Fox Pro: generates a string of 5 digits ("48963") based on a text string (captured in a textbox), if you always enter the same text string will get that string always 5 digits (no reverse), my code in C #. NET should generate the same string. I want to migrate the following code (Visual Fox Pro 6 to C#) gnLower = 1000 gnUpper = 100000 vcad = 1 For y=gnLower to gnUpper step 52 genClave = **Rand(vcad)** * y vRound = allt(str(int(genclave))) IF Len(vRound) = 3 vDec = Right(allt(str(genClave,10,2)), 2) finClave = vRound+vDec Thisform.txtPass.value = Rand(971); Exit Endif Next y outputs: vcad = 1 return: 99905 vcad = 2 return: 10077 vcad = thanks return: 17200 thks!

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