Search Results

Search found 11532 results on 462 pages for 'word break'.

Page 85/462 | < Previous Page | 81 82 83 84 85 86 87 88 89 90 91 92  | Next Page >

  • Generate random number histogram using java

    - by Chewart
    Histogram -------------------------------------------------------- 1 ****(4) 2 ******(6) 3 ***********(11) 4 *****************(17) 5 **************************(26) 6 *************************(25) 7 *******(7) 8 ***(3) 9 (0) 10 *(1) -------------------------------------------------------- basically above is what my prgram needs to do.. im missing something somewhere any help would be great :) import java.util.Random; public class Histogram { /*This is a program to generate random number histogram between 1 and 100 and generate a table */ public static void main(String args[]) { int [] randarray = new int [80]; Random random = new Random(); System.out.println("Histogram"); System.out.println("---------"); int i ; for ( i = 0; i<randarray.length;i++) { int temp = random.nextInt(100); //random numbers up to number value 100 randarray[i] = temp; } int [] histo = new int [10]; for ( i = 0; i<10; i++) { /* %03d\t, this generates the random numbers to three decimal places so the numbers are generated with a full number or number with 00's or one 0*/ if (randarray[i] <= 10) { histo[i] = histo[i] + 1; //System.out.println("*"); } else if ( randarray[i] <= 20){ histo[i] = histo[i] + 1; } else if (randarray[i] <= 30){ histo[i] = histo[i] + 1; } else if ( randarray[i] <= 40){ histo[i] = histo[i] + 1; } else if (randarray[i] <= 50){ histo[i] = histo[i] + 1; } else if ( randarray[i] <=60){ histo[i] = histo[i] + 1; } else if ( randarray[i] <=70){ histo[i] = histo[i] + 1; } else if ( randarray[i] <=80){ histo[i] = histo[i] + 1; } else if ( randarray[i] <=90){ histo[i] = histo[i] + 1; } else if ( randarray[i] <=100){ histo[i] = histo[i] + 1; } switch (randarray[i]) { case 1: System.out.print("0-10 | "); break; case 2: System.out.print("11-20 | "); break; case 3: System.out.print("21-30 | "); break; case 4: System.out.print("31-40 | "); break; case 5: System.out.print("41-50 | "); break; case 6: System.out.print("51-60 | "); break; case 7: System.out.print("61-70 | "); break; case 8: System.out.print("71-80 | "); break; case 9: System.out.print("81-90 | "); break; case 10: System.out.print("91-100 | "); } for (int i = 0; i < 80; i++) { randomNumber = random.nextInt(100) index = (randomNumber - 1) / 2; histo[index]++; } } } }

    Read the article

  • Regular Expression; Find whether a line contains any word with more than X characters.

    - by Simpsoid
    Hi, I am trying to use a Validator on a ASP.NET site and need to find whether the Street Address textbox contains a valid entry. Entries with words that are longer than X characters (in this case 25, with no punctuation or spaces) will cause the HTML on a printed A4 page to not wrap properly and therefore not to confrom to certain sizes correctly pushing the margins off. For a street address I want to match that something like "201 Long Road" is valid but "235 ReallyLongAndNarrowWindingRoadBesideTheRiver Street" is invalid. Using a Microsoft .Net Regular Expression Validator I need to know what the RegEx pattern might be. I think if it does find a match the Validator will fire correctly however if there is no match the Validator won't fire and the Update button (in this case) won't fire. Since Street addresses can contain Capital Letters and numbers etc. it will need to accomodate for that and also Spaces, Commas, Semi-Colons and Colons and Hyphens are valid characters too. Any help would be greatly appreciated as I am really stuck with this problem. Thanks, David

    Read the article

  • Code golf: find all anagrams

    - by Charles Ma
    An word is an anagram if the letters in that word can be re-arranged to form a different word. Task: Find all sets of anagrams given a word list Input: a list of words from stdin with each word separated by a new line e.g. A A's AOL AOL's Aachen Aachen's Aaliyah Aaliyah's Aaron Aaron's Abbas Abbasid Abbasid's Output: All sets of anagrams, with each set separated by a separate line Example run: ./anagram < words marcos caroms macros lump's plum's dewar's wader's postman tampons dent tend macho mocha stoker's stroke's hops posh shop chasity scythia ... I have a 149 char perl solution which I'll post as soon as a few more people post :) Have fun!

    Read the article

  • Visual Studio Code Metrics and the Maintainability index of switch case

    - by pee2002
    Hi there! As a person who loves to follow the best practices, If i run code metrics (right click on project name in solution explorer and select "Calculate Code Metrics" - Visual Studio 2010) on: public static string GetFormFactor(int number) { string formFactor = string.Empty; switch (number) { case 1: formFactor = "Other"; break; case 2: formFactor = "SIP"; break; case 3: formFactor = "DIP"; break; case 4: formFactor = "ZIP"; break; case 5: formFactor = "SOJ"; break; } return formFactor; } It Gives me a Maintainability index of 61 (of course this is insignificant if you have only this, but if you use an utility like class whos philosophy is doing stuff like that, your utility class will have the maintainability index much worst..) Whats the solution for this?

    Read the article

  • Java .doc generation

    - by bozo
    Hi, anyone knows an easy method to generate mail merge .doc file from Java? So, I want to create a Word (95/97) document in Word, put some simple placeholders in it (only single value, no iterators and other advanced tags) like the ones used with mailmerge option, and then at runtime replace those placeholders with values from Java. One option is to use Jasperreports, but this would require that I create exact replica of non-trivial Word document in Jasper format, which is not easy and is hard to change later. Is there some method of filling placeholders in Word from Java, which does not require low-level document alteration with positioning and others low-level .doc tags from code, but something like this: docPreparer.fillPlaceholder('placeholder1', 'my real value from runtime'); Some CRMs do this via ActiveX control for internet explorer, and it works great (they use Word's mailmerge) but I need an all-Java solution. Ideas? Thanks, Bozo

    Read the article

  • XMLReader in silverlight <test /> type tag problem

    - by Ummar
    Hi I am parsing XML in silverlight, in my XML I have one tag is like <test attribute1="123" /> <test1 attribute2="345">abc text</test1> I am using XMLReader to parse xml like using (XmlReader reader = XmlReader.Create(new StringReader(xmlString))) { // Parse the file and display each of the nodes. while (reader.Read()) { switch (reader.NodeType) { case XmlNodeType.Element: //process start tag here break; case XmlNodeType.Text: //process text here break; case XmlNodeType.XmlDeclaration: case XmlNodeType.ProcessingInstruction: break; case XmlNodeType.Comment: break; case XmlNodeType.EndElement: //process end tag here break; } } } but the problem is that for test tag no EndElement is received? which is making my whole program logic wrong. (for test1 tag all works fine). Please help me out.

    Read the article

  • PHP Get variable to equal value of switch

    - by user342391
    I am trying to get a variable on my page to equal the result of a switch I have. This is the code: $payment_method = switch ($cardtype) { case "visa" : echo "VSA"; break; case "mastercard" : echo "MSC"; break; case "maestro" : echo "MAE"; break; case "amex" : echo "AMX" ; break; default : echo "Please specify a payment method!"; break; }; How can I get $payment_method to equal the result of this???? So far I recieve an error: Parse error: syntax error, unexpected T_SWITCH in /var/www/account/credits/moneybookers/process.php on line 65

    Read the article

  • read file in C++

    - by Amm Sokun
    I am trying to read a list of words from a file in C++. However, the last word is read twice. I cannot understand why it is so. Can someone help me out? int main () { ifstream fin, finn; vector<string> vin; vector<string> typo; string word; fin.open("F:\\coursework\\pz\\gattaca\\breathanalyzer\\file.in"); if (!fin.is_open()) cout<<"Not open\n"; while (fin) { fin >> word; cout<<word<<endl; vin.push_back(word); } fin.close(); }

    Read the article

  • How to wrap text of html button with fixed width

    - by Peter
    I just noticed that if you give a html button a fixed width, the text inside the button is never wrapped. I've tried it with word-wrap, but that cuts of the word even though there are spaces available to wrap on. How can I make the text of an html button with a fixed width wrap like any tablecell would? <td class="category_column"> <input type="submit" name="ctl00$ContentPlaceHolder1$DataList1$ctl12$ProCat_NameButton" value="Roos Sturingen / Sensors" id="ctl00_ContentPlaceHolder1_DataList1_ctl12_ProCat_NameButton" class="outset" style="height:118px;width:200px;font-size:18px;color:#7F7F7F;width:200px;white-space:pre;" /> </td> the css classes do nothing but adding borders and modify the padding. If I add word-wrap:break-word to this button, it will wrap it like this: Roos Sturingen / Sen sors And I dont want it to cut of in the middle of a word if it is possible to cut it off between words. Thanks

    Read the article

  • split a string based on pattern in java - capital letters and numbers

    - by rookie
    Hi all I have the following string "3/4Ton". I want to split it as -- word[1] = 3/4 and word[2] = Ton. Right now my piece of code looks like this:- Pattern p = Pattern.compile("[A-Z]{1}[a-z]+"); Matcher m = p.matcher(line); while(m.find()){ System.out.println("The word --> "+m.group()); } It carries out the needed task of splitting the string based on capital letters like:- String = MachineryInput word[1] = Machinery , word[2] = Input The only problem is it does not preserve, numbers or abbreviations or sequences of capital letters which are not meant to be separate words. Could some one help me out with my regular expression coding problem. Thanks in advance...

    Read the article

  • Find all words containing characters in UNIX

    - by fahdshariff
    Given a word W, I want to find all words containing the letters in W from /usr/dict/words. For example, "bat" should return "bat" and "tab" (but not "table"). Here is one solution which involves sorting the input word and matching: word=$1 sortedWord=`echo $word | grep -o . | sort | tr -d '\n'` while read line do sortedLine=`echo $line | grep -o . | sort | tr -d '\n'` if [ "$sortedWord" == "$sortedLine" ] then echo $line fi done < /usr/dict/words Is there a better way? I'd prefer using basic commands (instead of perl/awk etc), but all solutions are welcome! To clarify, I want to find all permutations of the original word. Addition or deletion of characters is not allowed.

    Read the article

  • Memento with optional state?

    - by Korey Hinton
    EDIT: As pointed out by Steve Evers and pdr, I am not correctly implementing the Memento pattern, my design is actually State pattern. Menu Program I built a console-based menu program with multiple levels that selects a particular test to run. Each level more precisely describes the operation. At any level you can type back to go back one level (memento). Level 1: Server Type? [1] Server A [2] Server B Level 2: Server environment? [1] test [2] production Level 3: Test type? [1] load [2] unit Level 4: Data Collection? [1] Legal docs [2] Corporate docs Level 4.5 (optional): Load Test Type [2] Multi TIF [2] Single PDF Level 5: Command Type? [1] Move [2] Copy [3] Remove [4] Custom Level 6: Enter a keyword [setup, cleanup, run] Design States PROBLEM: Right now the STATES enum is the determining factor as to what state is BACK and what state is NEXT yet it knows nothing about what the current memento state is. Has anyone experienced a similar issue and found an effective way to handle mementos with optional state? static enum STATES { SERVER, ENVIRONMENT, TEST_TYPE, COLLECTION, COMMAND_TYPE, KEYWORD, FINISHED } Possible Solution (Not-flexible) In reference to my code below, every case statement in the Menu class could check the state of currentMemo and then set the STATE (enum) accordingly to pass to the Builder. However, this doesn't seem flexible very flexible to change and I'm struggling to see an effective way refactor the design. class Menu extends StateConscious { private State state; private Scanner reader; private ServerUtils utility; Menu() { state = new State(); reader = new Scanner(System.in); utility = new ServerUtils(); } // Recurring menu logic public void startPromptingLoop() { List<State> states = new ArrayList<>(); states.add(new State()); boolean redoInput = false; boolean userIsDone = false; while (true) { // get Memento from last loop Memento currentMemento = states.get(states.size() - 1) .saveMemento(); if (currentMemento == null) currentMemento = new Memento.Builder(0).build(); if (!redoInput) System.out.println(currentMemento.prompt); redoInput = false; // prepare Memento for next loop Memento nextMemento = null; STATES state = STATES.values()[states.size() - 1]; // get user input String selection = reader.nextLine(); switch (selection) { case "exit": reader.close(); return; // only escape case "quit": nextMemento = new Memento.Builder(first(), currentMemento, selection).build(); states.clear(); break; case "back": nextMemento = new Memento.Builder(previous(state), currentMemento, selection).build(); if (states.size() <= 1) { states.remove(0); } else { states.remove(states.size() - 1); states.remove(states.size() - 1); } break; case "1": nextMemento = new Memento.Builder(next(state), currentMemento, selection).build(); break; case "2": nextMemento = new Memento.Builder(next(state), currentMemento, selection).build(); break; case "3": nextMemento = new Memento.Builder(next(state), currentMemento, selection).build(); break; case "4": nextMemento = new Memento.Builder(next(state), currentMemento, selection).build(); break; default: if (state.equals(STATES.CATEGORY)) { String command = selection; System.out.println("Executing " + command + " command on: " + currentMemento.type + " " + currentMemento.environment); utility.executeCommand(currentMemento.nickname, command); userIsDone = true; states.clear(); nextMemento = new Memento.Builder(first(), currentMemento, selection).build(); } else if (state.equals(STATES.KEYWORD)) { nextMemento = new Memento.Builder(next(state), currentMemento, selection).build(); states.clear(); nextMemento = new Memento.Builder(first(), currentMemento, selection).build(); } else { redoInput = true; System.out.println("give it another try"); continue; } break; } if (userIsDone) { // start the recurring menu over from the beginning for (int i = 0; i < states.size(); i++) { if (i != 0) { states.remove(i); // remove all except first } } reader = new Scanner(System.in); this.state = new State(); userIsDone = false; } if (!redoInput) { this.state.restoreMemento(nextMemento); states.add(this.state); } } } }

    Read the article

  • Formating a text in a table cell with PHPWord e.g. bold, font, size e.t.c

    - by alphy
    I have the code snippet below //create a new word document $word= new PHPWord(); //create potrait orientation $section=$word->createSection(); $table = $section->addTable(); $word->addFontStyle('rStyle', array('bold'=>true, 'italic'=>true, 'size'=>16)); //header row $table->addRow(400, array('bgColor'=>'dbdbdb')); $table->addCell(2000, array('bgColor'=>'dbdbdb'))->addText('Cell 1','rStyle'); $table->addCell(3500, array('bgColor'=>'dbdbdb'))->addText('Cell 1'); $table->addCell(1500, array('bgColor'=>'dbdbdb'))->addText('Cell 1','rStyle'); $table->addCell(2000, array('bgColor'=>'dbdbdb'))->addText('Cell 1'); // Save File $objWriter = PHPWord_IOFactory::createWriter($word, 'Word2007'); $objWriter->save('Text.docx'); echo 'Text.docx created successfully'; } How can i add text formatting to a cell value to bold, italic, font-size etc, I have tried as shown above but it does not work

    Read the article

< Previous Page | 81 82 83 84 85 86 87 88 89 90 91 92  | Next Page >