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  • String length difference between ruby 1.8 and 1.9

    - by Raghu
    I have a website thats running on ruby 1.8.7 . I have a validation on an incoming post that checks to make sure that we allow upto max of 12000 characters. The spaces are counted as characters and tab and carriage returns are stripped off before the post is subjected to the validation. Here is the post that is subjected to validation http://pastie.org/5047582 In ruby 1.9 the string length shows up as 11909 which is correct. But when I check the length on ruby 1.8.7 is turns out to be 12044. I used codepad.org to run this ruby code which gives me http://codepad.org/OxgSuKGZ ( which outputs the length as 12044 which is wrong) but when i run this same code in the console at codeacademy.org the string length is 11909. Can anybody explain me why this is happening ??? Thanks

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  • Javascript String Length Differs From PHP mb_strlen

    - by TheOnly92
    I use document.getElementById("text").value.length to get the string length through javascript, and mb_strlen($_POST['text']) to get the string length by PHP and both differs very much. Carriage returns are converted in javascript before getting the string length, but I guess some characters are not being counted. For example, [b]15. Umieszczanie obrazka z logo na stronie zespolu[/b] This block of text is calculated 57 in javascript and 58 in PHP. When the text gets long, the difference increases. Is there any way to overcome this?

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  • Creating a fixed formatted cell in UITableview

    - by Wes
    Hi, I want to have a tableview create rows that look like this: value1 item1 container1 value10 item10 container10 value100 item100 container100 value2 item2 container2 What I am trying to show is that the first word (value) will have a set length of 12 and then the second word (item) will have a set length of 10 and then the last word (container) is just tagged on at the end. I am pulling these from a SQLite database and don't want to use multiple lines, but read in a strictly formatted structure like this.

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  • Efficient way to calculate byte length of a character, depending on the encoding

    - by BalusC
    What's the most efficient way to calculate the byte length of a character, taking the character encoding into account? In UTF-8 for example the characters have a variable byte length, so each character needs to be determined individually. As far now I've come up with this: char c = getItSomehow(); String encoding = "UTF-8"; int length = new String(new char[] { c }).getBytes(encoding).length; But this is clumsy and inefficient in a loop since a new String needs to be created everytime. I can't find other and more efficient ways in the Java API. I imagine that this can be done with bitwise operations like bit shifting, but that's my weak point and I'm unsure how to take the encoding into account here :) If you question the need for this, check this topic.

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  • Try-Catch or Check Length? C# XNA

    - by Shaded
    I was just wondering which would be cheaper, using a try catch block for index out of bounds or checking the length of a multi dimensional array and comparing values? I have a feeling it's the length, since I can store the length in a variable and then just do if's which are relatively cheap. I'm just not sure how expensive try-catch is. Thanks!

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  • Counting down to zero in contrast to counting up to length - 1

    - by Helper Method
    Is it recommended to count in small loops (where possible) down from length - 1 to zero instead of counting up to length - 1? 1.) Counting down for (int i = a.length - 1; i >= 0; i--) { if (a[i] == key) return i; } 2.) Counting up for (int i = 0; i < a.length; i++) { if (a[i] == key) return i; } The first one is slightly faster that the second one (because comparing to zero is faster) but is a little more error-prone in my opinion. Besides, the first one could maybe not be optimized by future improvements of the JVM. Any ideas on that?

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  • Remove leading whitespaces using variable length lookbehind in RegExp

    - by Shizhidi
    Hello, I'm wondering if variable length lookbehind assertions are supported in JavaScript's RegExp engine? For example, I'm trying to match the string "variable length" in the string "[a lot of whitespaces and/or tabs]variable length lookbehind", and I have something like this but it does not go well in various RegExp testers: ^(?<=[ \t]+).+(?= lookbehind) If it's an illegal pattern, what would be a good workaround to it? Thanks!

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  • is memset(ary,0,length) a portable way of inputting zero in double array

    - by monkeyking
    The following code uses memset to set all the bits to zero #include <iostream> #include <cstring> int main(){ int length = 5; double *array = new double[length]; memset(array,0,sizeof(double)*length); for(int i=0;i<length;i++) if(array[i]!=0.0) std::cerr<< "not zero in: " <<i <<std::endl; return 0; } Can I assume that this will work on all platforms? Does the double datatype always correspond to the ieee-754 standard? thanks

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  • Efficient Multiplication of Varying-Length #s [Conceptual]

    - by Milan Patel
    Write the pseudocode of an algorithm that takes in two arbitrary length numbers (provided as strings), and computes the product of these numbers. Use an efficient procedure for multiplication of large numbers of arbitrary length. Analyze the efficiency of your algorithm. I decided to take the (semi) easy way out and use the Russian Peasant Algorithm. It works like this: a * b = a/2 * 2b if a is even a * b = (a-1)/2 * 2b + a if a is odd My pseudocode is: rpa(x, y){ if x is 1 return y if x is even return rpa(x/2, 2y) if x is odd return rpa((x-1)/2, 2y) + y } I have 3 questions: Is this efficient for arbitrary length numbers? I implemented it in C and tried varying length numbers. The run-time in was near-instant in all cases so it's hard to tell empirically... Can I apply the Master's Theorem to understand the complexity...? a = # subproblems in recursion = 1 (max 1 recursive call across all states) n / b = size of each subproblem = n / 1 - b = 1 (problem doesn't change size...?) f(n^d) = work done outside recursive calls = 1 - d = 0 (the addition when a is odd) a = 1, b^d = 1, a = b^d - complexity is in n^d*log(n) = log(n) this makes sense logically since we are halving the problem at each step, right? What might my professor mean by providing arbitrary length numbers "as strings". Why do that? Many thanks in advance

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  • R: Cut and labels/breaks length conflict

    - by AkselO
    I am working with the cut function to prep data for a barplot histogram but keep running into a seeming inconsistency between my labels and breaks: Error in cut.default(sample(1:1e+05, 500, T), breaks = sq, labels = sprintf("$%.0f", : labels/breaks length conflict Here is an example. I pretend that it is income data, using a sequence of 0 to $100,000 in bins of $10,000. I use the same variable to generate both breaks and labels, with minor formating on the label side. I thought they might for some reason have different lengths when comparing to a character vector, but they appear to have the same length, still. > sq<-seq(0,100000,10000) > cut(sample(1:100000, 500, T),breaks=sq,labels=sprintf("$%.0f",sq)) > length(sprintf("$%.0f",sq)) [1] [11] > length(sq) [1] [11]

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  • Length of an HTMLObjectCollection is incorrect in Internet Explorer

    - by Mayank Gupta
    I have three cells in different rows in a table having same name.e.g. <td name = "x"> is present in 3 different rows. I am using document.getElementsByName() to obtain a collection of these cells and trying to calculate the length of this collection. e.g. var obj = doucment.getElementsByName("X"); var length = obj.length; This code works fine in Google Chrome but in IE the length is return as 0(zero). Can anyone tell me how to sove this problem in IE?

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  • Haskell Add Function Return to List Until Certain Length

    - by kienjakenobi
    I want to write a function which takes a list and constructs a subset of that list of a certain length based on the output of a function. If I were simply interested in the first 50 elements of the sorted list xs, then I would use fst (splitAt 50 (sort xs)). However, the problem is that elements in my list rely on other elements in the same list. If I choose element p, then I MUST also choose elements q and r, even if they are not in the first 50 elements of my list. I am using a function finderFunc which takes an element a from the list xs and returns a list with the element a and all of its required elements. finderFunc works fine. Now, the challenge is to write a function which builds a list whose total length is 50 based on multiple outputs of finderFunc. Here is my attempt at this: finish :: [a] -> [a] -> [a] --This is the base case, which adds nothing to the final list finish [] fs = [] --The function is recursive, so the fs variable is necessary so that finish -- can forward the incomplete list to itself. finish ps fs -- If the final list fs is too small, add elements to it | length fs < 50 && length (fs ++ newrs) <= 50 = fs ++ finish newps newrs -- If the length is met, then add nothing to the list and quit | length fs >= 50 = finish [] fs -- These guard statements are currently lacking, not the main problem | otherwise = finish [] fs where --Sort the candidate list sortedps = sort ps --(finderFunc a) returns a list of type [a] containing a and all the -- elements which are required to go with it. This is the interesting -- bit. rs is also a subset of the candidate list ps. rs = finderFunc (head sortedps) --Remove those elements which are already in the final list, because -- there can be overlap newrs = filter (`notElem` fs) rs --Remove the elements we will add to the list from the new list -- of candidates newps = filter (`notElem` rs) ps I realize that the above if statements will, in some cases, not give me a list of exactly 50 elements. This is not the main problem, right now. The problem is that my function finish does not work at all as I would expect it to. Not only does it produce duplicate elements in the output list, but it sometimes goes far above the total number of elements I want to have in the list. The way this is written, I usually call it with an empty list, such as: finish xs [], so that the list it builds on starts as an empty list.

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  • Length of a string in pixels

    - by Rose
    Guys, I'm populating a dropDownList with arrayCollection of strings. I want the width of the drop down list control to match with the size (in pixels) of the longest string in the array collection. The problem I'm facing is: the font width of the strings in the collection are different e.g. 'W' looks wider than 'l'. So I estimated the width of a character to be 8 pixels but that's not pretty neat. If a string that has many 'W' and 'M' is encountered the estimation is wrong. So I want precise pixel width of strings. How can i get the exact length of a string in pixels?? My solution that estimates all character to be 8 pixels wide is given below: public function populateDropDownList():void{ var array:Array = new Array("o","two","three four five six seven eight wwww"); var sampleArrayCollection:ArrayCollection = new ArrayCollection(array); var customDropDownList:DropDownList = new DropDownList(); customDropDownList.dataProvider=sampleArrayCollection; customDropDownList.prompt="Select ..."; customDropDownList.labelField="Answer Options:"; //calculating the max width var componentWidth=10; //default for each(var answerText in array){ Alert.show("Txt size: "+ answerText.length + " pixels: " + answerText.length*9); if(answerText.length * 8 > componentWidth){ componentWidth=answerText.length * 8; } } customDropDownList.width=componentWidth; answers.addChild(customDropDownList); } Any idea or solution is highly valued. Thanks

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  • How to find an audio file's length (in seconds)

    - by mIL3S
    Hi all! (Objective C) Just using simple AudioServicesPlaySystemSoundID and its counterparts, but I can't find in the documentation if there is already a way to find the length of an audio file. I know there is AudioServicesGetPropertyInfo, but that seems to return a byte-buffer - do audio files embed their length in themselves and I can just extract it with this? Or is there perhaps a formula based on bit-rate * fileSize to convert to length-of-time? mIL3S www.milkdrinkingcow.com

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  • Limited-length string class

    - by wood_brian
    Is there a limited-length string class around? I've searched a little on the net and didn't find anything. I'm interested in a class that limits (possibly at compile time) the length to 255, so marshalling the string's length only requires one byte.

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  • wrap all lines that are longer than line length

    - by user1919840
    I am writing a program that limits each line to a certain length. this is what i got so far, i am almost done but i still need to cut each line, but i cant figure it out. def main(): filename = input("Please enter the name of the file to be used: ") openFile = open(filename, 'r+') file = openFile.read() lLength = int(input("enter a number between 10 & 20: ")) while (lLength < 10) or (lLength > 20) : print("Invalid input, please try again...") lLength = int(input("enter a number between 10 & 20: ")) wr = textwrap.TextWrapper() wraped = wr.wrap(file) print("Here is your output formated to a max of", lLength, "characters per line: ") wr.width = lLength wr.expand_tabs = True for lines in wraped: print(lines) an example of what the output SHOULD be is this. If the file specified contains this text: hgytuinghdt #here the length is 11 ughtnjuiknshfyth #here the length is 16 nmjhkaiolgytuhngjuin #here the length is 20 and the lLength is specified to 15 then this should print out: hgytuinghdt ughtnjuiknshfyt h nmjhkaiolgytuhng juin Thanks.

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  • Google Blogger Website CName and/or Text File Issues

    - by Francis Gibbons
    I have a blogger Blog website and I would like to have it show up on my company website. I have read a couple articles out there on how to do it. A hand full of them talk about using FTP which is old and no longer available. However, I am trying to following along with this one: http://www.infinite42.com/small-business/integrate-blogger-blog-website Which seems pretty easy but I am having a problem getting Google to Verify the DNS CName or Text Record that I created on my Windows 2007 Server. Do I need to create this record at the registra level. Right now the domain is setup at the registra to point the www record to my server where on my server I tried the Txt Record and the CName Record with no luck in DNS. Here are the Google instructions for creating a CName file record in DNS: Follow the steps below to create a DNS (Domain Name System) record that proves to Google that you own the domain. Add the CNAME record below to the DNS configuration for abc.com. CNAME Label / Host: CNAME Destination / Target: Click Verify below. When Google finds this DNS record, we'll make you a verified owner of the domain. (Note: DNS changes may take some time. If we don't find the record immediately, we'll check for it periodically.) To stay verified, don't remove the DNS record, even after verification succeeds. Here is the link to do it with a CName: http://googlewebmastercentral.blogspot.com/2012/08/domain-verification-using-cname-records.html When I go to add my CName record on my server's DNS the only two fields available are Alias Name and Fully Qualified Domain Name. How am I suppose to create this record can someone please tell me? Thanks, Frank

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  • Blocking nslookup on A record

    - by msher420
    Need to know which port to be blocked in my local machine so that the nslookup on A record doesnot work? To know the above i need to understand how the lookup on A record how does the request go from the local machine (port) to the nameservers/ rootservers? For example: C: nslookup -type=a google.com Server: MyDslModem.local.lan Address: 192.168.1.1 Non-authoritative answer: Name: google.com Address: 209.85.231.104 Here which from which local port from the local machine does the lookup starts from?

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  • How to record screen and sound

    - by user23950
    Do you know of any application that can record both sound and the screen. Camstudio records only the screen but not the sound. Coming out from the monitor. How do I do this one, because I frequently stream online content. So that I'll just record it and I will have a copy of that stream

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