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  • jQuery: Counter, Tricky problem with effects for brainy people.

    - by Marius
    Hello there! I made this counter that I think is cool because it only makes visible changes to the numbers being changed between each time it is triggered. This is the code // counter $('a').click(function(){ var u = ''; var newStr = ''; $('.counter').each(function(){ var len = $(this).children('span').length; $(this).children('span').each(function(){ u = u + $(this).text(); }); v = parseInt(u) + 1; v = v + ''; for (i=v.length - 1; i >= 0; i--) { if (v.charAt(i) == u.charAt(i)) { break; } } slce = len - (v.length - (i + 1)) updates = $(this).children('span').slice(slce); $(updates).fadeTo(100,0).delay(100).each(function(index){ f = i + 1 + index; $(this).text(v.charAt(f)); }).fadeTo(100,1); }); }); Markup: <span class="counter"> <span></span><span></span><span></span><span></span><span></span><span></span><span style="margin-right:4px;">9</span><span>9</span><span>9</span><span>9</span> </span> <a href="">Add + 1</a> The problem is that I previously used queue() function to delay() $(this).text(v.charAt(f)); by 100ms, (without queue the text()-function would not be delayed because it isnt in the fx catergory) so that the text would be updated before the object had faded to opacity = 0. That would look stupid. When adding multiple counters, only one of the counters would count. When removing the queue function, both counters would work, but as you can imagine, the delay of the text() would be gone (as it isnt fx-category). It is probably a bit tricky to make out how I can have multiple counters, and still delay the text()-function by 100ms, but I was hoping there is somebody here with spare brain capacity for these things ;) You can see a (NSFW) problem demo here: Just look underneath the sharing icons and you will notice that the text changes WHILE the objects fade out. Looking for some help to sove this problem. I would like to call the text() function once the text has faded to opacity 0, then fade in once the text() has executed. Thank you for your time.

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  • Change A Character In A String Using Actionscript

    - by Joshua
    What is the opposite of String.charAt()?? If I Have a string: var Str:String="Hello World"; How do I change the 5th character, for example, from a ' ' to an '_'? I can GET the 5th character like this: var C:String=Str.charAt(5); But how do I SET the 5th character? Thanks in advance.

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  • Java: Preventing array going out of bounds.

    - by Troy
    I'm working on a game of checkers, if you want to read more about you can view it here; http://minnie.tuhs.org/I2P/Assessment/assig2.html When I am doing my test to see if the player is able to get to a certain square on the grid (i.e. +1 +1, +1 -1 .etc) from it's current location, I get an java.lang.ArrayIndexOutOfBoundsException error. This is the code I am using to make the move; public static String makeMove(String move, int playerNumber) { // variables to contain the starting and destination coordinates, subtracting 1 to match array size int colStart = move.charAt(1) - FIRSTCOLREF - 1; int rowStart = move.charAt(0) - FIRSTROWREF - 1; int colEnd = move.charAt(4) - FIRSTCOLREF - 1; int rowEnd = move.charAt(3) - FIRSTROWREF - 1; // variable to contain which player is which char player, enemy; if (playerNumber==1) { player= WHITEPIECE; enemy= BLACKPIECE; } else { player= BLACKPIECE; enemy= WHITEPIECE; } // check that the starting square contains a player piece if (grid [ colStart ] [ rowStart ] == player) { // check that the player is making a diagonal move if (grid [ colEnd ] [ rowEnd ] == grid [ (colStart++) ] [ (rowEnd++) ] && grid [ colEnd ] [ rowEnd ] == grid [ (colStart--) ] [ (rowEnd++) ] && grid [ colEnd ] [ rowEnd ] == grid [ (colStart++) ] [ (rowEnd--) ] && grid [ colEnd ] [ rowEnd ] == grid [ (colStart--) ] [ (rowEnd--) ]) { // check that the destination square is free if (grid [ colEnd ] [ rowEnd ] == BLANK) { grid [ colStart ] [ rowStart ] = BLANK; grid [ colEnd ] [ rowEnd ] = player; } } // check if player is jumping over a piece else if (grid [ colEnd ] [ rowEnd ] == grid [ (colStart+2) ] [ (rowEnd+2) ] && grid [ colEnd ] [ rowEnd ] == grid [ (colStart-2) ] [ (rowEnd+2) ] && grid [ colEnd ] [ rowEnd ] == grid [ (colStart+2) ] [ (rowEnd-2) ] && grid [ colEnd ] [ rowEnd ] == grid [ (colStart-2) ] [ (rowEnd-2) ]) { // check that the piece in between contains an enemy if ((grid [ (colStart++) ] [ (rowEnd++) ] == enemy ) && (grid [ (colStart--) ] [ (rowEnd++) ] == enemy ) && (grid [ (colStart++) ] [ (rowEnd--) ] == enemy ) && (grid [ (colStart--) ] [ (rowEnd--) ] == enemy )) { // check that the destination is free if (grid [ colEnd ] [ rowEnd ] == BLANK) { grid [ colStart ] [ rowStart ] = BLANK; grid [ colEnd ] [ rowEnd ] = player; } } } } I'm not sure how I can prevent the error from happening, what do you recommend?

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  • What does assert do?

    - by peiska
    What does assert do? for example in the function? private static int charAt(String s, int d) { assert d >= 0 && d <= s.length(); if (d == s.length()) return -1; return s.charAt(d); }

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  • File sorting, Comparison method violates its general contract

    - by user2677383
    My file sorting comparator sometimes invoke java.lang.IllegalArgumentException in android app. I cannot figure out the reason. could you explain that? here is my code block: Comparator cc = new Comparator<File>() { @Override public int compare(File f1, File f2) { if (f1.isDirectory() && !f2.isDirectory()) { return 1; } if (f2.isDirectory() && !f1.isDirectory()) { return -1; } if ((f1.isFile() && f2.isFile()) || (f1.isDirectory() && f2.isDirectory())) { if (sort == 1) { return Util.compareFileName(f2.getName(), f1.getName()); } else if (sort == 2) { return (int) (f1.lastModified() - f2.lastModified()); } else if (sort == 3) { return (int) (f2.lastModified() - f1.lastModified()); } else if (sort == 4) { return (int) (f1.length() - f2.length()); } else if (sort == 5) { return (int) (f2.length() - f1.length()); } else return Util.compareFileName(f1.getName(), f2.getName()); } return f1.compareTo(f2); } }; Util.compareFileName is as followings: public static int compareFileName(String s1, String s2) { int thisMarker = 0; int thatMarker = 0; int s1Length = s1.length(); int s2Length = s2.length(); while (thisMarker < s1Length && thatMarker < s2Length) { String thisChunk = getChunk(s1, s1Length, thisMarker); thisMarker += thisChunk.length(); String thatChunk = getChunk(s2, s2Length, thatMarker); thatMarker += thatChunk.length(); // If both chunks contain numeric characters, sort them numerically int result = 0; if (isDigit(thisChunk.charAt(0)) && isDigit(thatChunk.charAt(0))) { // Simple chunk comparison by length. int thisChunkLength = thisChunk.length(); result = thisChunkLength - thatChunk.length(); // If equal, the first different number counts if (result == 0) { for (int i = 0; i < thisChunkLength; i++) { result = thisChunk.charAt(i) - thatChunk.charAt(i); if (result != 0) { return result; } } } } else { result = thisChunk.compareTo(thatChunk); } if (result != 0) return result; } return s1Length - s2Length; }

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  • initialising a 2-dim Array in Scala

    - by Stefan W.
    (Scala 2.7.7:) I don't get used to 2d-Arrays. Arrays are mutable, but how do I specify a 2d-Array which is - let's say of size 3x4. The dimension (2D) is fixed, but the size per dimension shall be initializable. I tried this: class Field (val rows: Int, val cols: Int, sc: java.util.Scanner) { var field = new Array [Char](rows)(cols) for (r <- (1 to rows)) { val line = sc.nextLine () val spl = line.split (" ") field (r) = spl.map (_.charAt (0)) } def put (val rows: Int, val cols: Int, c: Char) = todo () } I get this error: :11: error: value update is not a member of Char field (r) = spl.map (_.charAt (0)) If it would be Java, it would be much more code, but I would know how to do it, so I show what I mean: public class Field { private char[][] field; public Field (int rows, int cols, java.util.Scanner sc) { field = new char [rows][cols]; for (int r = 0; r < rows; ++r) { String line = sc.nextLine (); String[] spl = line.split (" "); for (int c = 0; c < cols; ++c) field [r][c] = spl[c].charAt (0); } } public static void main (String args[]) { new Field (3, 4, new java.util.Scanner ("fraese.fld")); } } and fraese.fld would look, for example, like that: M M M M . M I get some steps wide with val field = new Array Array [Char] but how would I then implement 'put'? Or is there a better way to implement the 2D-Array. Yes, I could use a one-dim-Array, and work with put (y, x, c) = field (y * width + x) = c but I would prefer a notation which looks more 2d-ish.

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  • Javascript: Whitespace Characters being Removed in Chrome (but not Firefox)

    - by Matrym
    Why would the below eliminate the whitespace around matched keyword text when replacing it with an anchor link? Note, this error only occurs in Chrome, and not firefox. For complete context, the file is located at: http://seox.org/lbp/lb-core.js To view the code in action (no errors found yet), the demo page is at http://seox.org/test.html. Copy/Pasting the first paragraph into a rich text editor (ie: dreamweaver, or gmail with rich text editor turned on) will reveal the problem, with words bunched together. Pasting it into a plain text editor will not. // Find page text (not in links) -> doxdesk.com function findPlainTextExceptInLinks(element, substring, callback) { for (var childi= element.childNodes.length; childi-->0;) { var child= element.childNodes[childi]; if (child.nodeType===1) { if (child.tagName.toLowerCase()!=='a') findPlainTextExceptInLinks(child, substring, callback); } else if (child.nodeType===3) { var index= child.data.length; while (true) { index= child.data.lastIndexOf(substring, index); if (index===-1 || limit.indexOf(substring.toLowerCase()) !== -1) break; // don't match an alphanumeric char var dontMatch =/\w/; if(child.nodeValue.charAt(index - 1).match(dontMatch) || child.nodeValue.charAt(index+keyword.length).match(dontMatch)) break; // alert(child.nodeValue.charAt(index+keyword.length + 1)); callback.call(window, child, index) } } } } // Linkup function, call with various type cases (below) function linkup(node, index) { node.splitText(index+keyword.length); var a= document.createElement('a'); a.href= linkUrl; a.appendChild(node.splitText(index)); node.parentNode.insertBefore(a, node.nextSibling); limit.push(keyword.toLowerCase()); // Add the keyword to memory urlMemory.push(linkUrl); // Add the url to memory } // lower case (already applied) findPlainTextExceptInLinks(lbp.vrs.holder, keyword, linkup); Thanks in advance for your help. I'm nearly ready to launch the script, and will gladly comment in kudos to you for your assistance.

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  • jQuery - Compatibility Problem with Internet Explorer 7 and Opera

    - by Marius
    Hello there, I have this counter which counts + 1 every time somebody shares content from the site. When it happens, the social icon that was clicked will bounce. It works in Firefox,Chrome, IE8, and Opera, however the bouncing animation is wrong in opera. $.fn.countExternal = function(animSpeed, num) { // for each counter this.each(function(){ // select all the digit containers var span = $(this).children(); // count the num of digit containers var len = $(span).length; // get the current count u = $(span).text(); // copy variable and add increment(s) v = num + ''; // foreach digit container... for (i=v.length - 1; i >= 0; i--) { // ...check which digits are not affected by the increment(s) if (v.charAt(i) == u.charAt(i)) { break; } } // slice from the total number of digit containers the digits containers which needs updating. slce = len - (v.length - (i + 1)) var updates = $(span).slice(slce); // loop through each digit container and fade out ... $(updates).fadeTo(animSpeed, 0,function(){ $(updates).each(function(index){ f = i + 1 + index; // ...then pick the right digit and update the digit... $(this).text(v.charAt(f)); // ...before fading back in. Cycle complete. $(this).fadeTo(animSpeed, 1); }); }); }); }; }) (jQuery); Demo (NSFW) is here (look underneath the social sharing icons). Any idea how I can solve the IE, and possibly the Opera compatibility problem? Thank you for your time.

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  • one more time about loop that doesn't work

    - by unit
    I have asked a couple of questions about this for loop: String[] book = new String [ISBN_NUM]; bookNum.replaceAll("-",""); if (bookNum.length()!=ISBN_NUM) throw new ISBNException ("ISBN "+ bookNum + " must be 10 characters"); for (int i=0;i<bookNum.length();i++) { if (Character.isDigit(bookNum.charAt(i))) book[j]=bookNum.charAt(i); //this is the problem right here j++; if (book[9].isNotDigit()|| book[9]!="x" || book[9]!="X") throw new ISBNException ("ISBN " + bookNum + " must contain all digits" + "or 'X' in the last position"); } which will not compile. An answer I had from the other question I asked told me that the line where the error occurs is wrong in that bookNum.charAt(i) is an (immutable) string, and I can't get the values into a book array that way. What I need to do on my assignment is check an ISBN number (bookNum) to see that it is all numbers, except the last digit can be an 'x' (valid ISBN). Is this the best way to do it? If so, what the hell am I doing wrong? If not, what method would be a better one to use?

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  • I asked a question about arrays before, but this one won't compile

    - by unit
    I asked about this array a little while ago, and I can't see what the problem is. Too tired. What have I done wrong? Basically, I am taking a string array and trying to check to see if it contains numbers or an x (ISBN number validation). I want to take the number from a given input (bookNum), check the input, and feed any valid input into a new array (book). At the line 'bookNum.charAt[j]==book[i]' I get the 'not a statement error'. What gives? String[] book = new String [ISBN_NUM]; bookNum.replaceAll("-",""); if (bookNum.length()!=ISBN_NUM) throw new ISBNException ("ISBN "+ bookNum + " must be 10 characters"); for (int i=0;i<bookNum.length();i++) { if (Character.isDigit(bookNum.charAt(i))) bookNum.CharAt[j]==book[i]; j++; if (book[9].isNotDigit()|| book[9]!="x" || book[9]!="X") throw new ISBNException ("ISBN " + bookNum + " must contain all digits" + "or 'X' in the last position");

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  • Optimizing Jaro-Winkler algorithm

    - by Pentium10
    I have this code for Jaro-Winkler algorithm taken from this website. I need to run 150,000 times to get distance between differences. It takes a long time, as I run on an Android mobile device. Can it be optimized more? public class Jaro { /** * gets the similarity of the two strings using Jaro distance. * * @param string1 the first input string * @param string2 the second input string * @return a value between 0-1 of the similarity */ public float getSimilarity(final String string1, final String string2) { //get half the length of the string rounded up - (this is the distance used for acceptable transpositions) final int halflen = ((Math.min(string1.length(), string2.length())) / 2) + ((Math.min(string1.length(), string2.length())) % 2); //get common characters final StringBuffer common1 = getCommonCharacters(string1, string2, halflen); final StringBuffer common2 = getCommonCharacters(string2, string1, halflen); //check for zero in common if (common1.length() == 0 || common2.length() == 0) { return 0.0f; } //check for same length common strings returning 0.0f is not the same if (common1.length() != common2.length()) { return 0.0f; } //get the number of transpositions int transpositions = 0; int n=common1.length(); for (int i = 0; i < n; i++) { if (common1.charAt(i) != common2.charAt(i)) transpositions++; } transpositions /= 2.0f; //calculate jaro metric return (common1.length() / ((float) string1.length()) + common2.length() / ((float) string2.length()) + (common1.length() - transpositions) / ((float) common1.length())) / 3.0f; } /** * returns a string buffer of characters from string1 within string2 if they are of a given * distance seperation from the position in string1. * * @param string1 * @param string2 * @param distanceSep * @return a string buffer of characters from string1 within string2 if they are of a given * distance seperation from the position in string1 */ private static StringBuffer getCommonCharacters(final String string1, final String string2, final int distanceSep) { //create a return buffer of characters final StringBuffer returnCommons = new StringBuffer(); //create a copy of string2 for processing final StringBuffer copy = new StringBuffer(string2); //iterate over string1 int n=string1.length(); int m=string2.length(); for (int i = 0; i < n; i++) { final char ch = string1.charAt(i); //set boolean for quick loop exit if found boolean foundIt = false; //compare char with range of characters to either side for (int j = Math.max(0, i - distanceSep); !foundIt && j < Math.min(i + distanceSep, m - 1); j++) { //check if found if (copy.charAt(j) == ch) { foundIt = true; //append character found returnCommons.append(ch); //alter copied string2 for processing copy.setCharAt(j, (char)0); } } } return returnCommons; } } I mention that in the whole process I make just instance of the script, so only once jaro= new Jaro(); If you are going to test and need examples so not break the script, you will find it here, in another thread for python optimization.

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  • 3-way quicksort, question

    - by peiska
    I am trying to understand the 3-way radix Quicksort, and i dont understand why the the CUTOFF variable there? and the insertion method? public class Quick3string { private static final int CUTOFF = 15; // cutoff to insertion sort // sort the array a[] of strings public static void sort(String[] a) { // StdRandom.shuffle(a); sort(a, 0, a.length-1, 0); assert isSorted(a); } // return the dth character of s, -1 if d = length of s private static int charAt(String s, int d) { assert d >= 0 && d <= s.length(); if (d == s.length()) return -1; return s.charAt(d); } // 3-way string quicksort a[lo..hi] starting at dth character private static void sort(String[] a, int lo, int hi, int d) { // cutoff to insertion sort for small subarrays if (hi <= lo + CUTOFF) { insertion(a, lo, hi, d); return; } int lt = lo, gt = hi; int v = charAt(a[lo], d); int i = lo + 1; while (i <= gt) { int t = charAt(a[i], d); if (t < v) exch(a, lt++, i++); else if (t > v) exch(a, i, gt--); else i++; } // a[lo..lt-1] < v = a[lt..gt] < a[gt+1..hi]. sort(a, lo, lt-1, d); if (v >= 0) sort(a, lt, gt, d+1); sort(a, gt+1, hi, d); } // sort from a[lo] to a[hi], starting at the dth character private static void insertion(String[] a, int lo, int hi, int d) { for (int i = lo; i <= hi; i++) for (int j = i; j > lo && less(a[j], a[j-1], d); j--) exch(a, j, j-1); } // exchange a[i] and a[j] private static void exch(String[] a, int i, int j) { String temp = a[i]; a[i] = a[j]; a[j] = temp; } // is v less than w, starting at character d private static boolean less(String v, String w, int d) { assert v.substring(0, d).equals(w.substring(0, d)); return v.substring(d).compareTo(w.substring(d)) < 0; } // is the array sorted private static boolean isSorted(String[] a) { for (int i = 1; i < a.length; i++) if (a[i].compareTo(a[i-1]) < 0) return false; return true; } public static void main(String[] args) { // read in the strings from standard input String[] a = StdIn.readAll().split("\\s+"); int N = a.length; // sort the strings sort(a); // print the results for (int i = 0; i < N; i++) StdOut.println(a[i]); } } from http://www.cs.princeton.edu/algs4/51radix/Quick3string.java.html

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  • Adding view cart function

    - by user228390
    Hey guys need some help in adding a view cart button but I'm stuck not sure how to code it. any help? The way I have coded it is that when a user clicks 'add item' they will get a alert box with info about the total price but I want that to appear in the HTML file but only once I have clicked on 'view cart' and I need it to be in a table format with info about the name, sum, price of the items and total. any ideas how I can do this? here is my javascript var f,d,str,items,qnts,price,bag,total; function cart(){ f=document.forms[0]; d=f.getElementsByTagName('div'); var items=[];var qnts=[];price=[];bag=[] for(i=0,e=0;i<d.length;i++){ items[i]=d[i].getElementsByTagName('b')[0].innerHTML; qnts[i]=d[i].getElementsByTagName('select')[0].value; str=d[i].getElementsByTagName('p')[1].innerHTML; priceStart(str,i); if(qnts[i]!=0){bag.push(new Array()); ib=bag[bag.length-1]; ib.push(items[i]);ib.push(qnts[i]);ib.push(price[i]);ib.push(qnts[i]*price[i]);} } if(bag.length>0){ total=bag[0][3]; if(bag.length>1){for(t=1;t<bag.length;t++){total+=bag[t][3]}} alert(bag.join('\n')+'\n----------------\ntotal='+total) } } function priceStart(str,inx){for(j=0;j<str.length;j++){if(str.charAt(j)!=' ' && !isNaN(str.charAt(j))){priceEnd(j,str,inx);return }}} function priceEnd(j,str,inx){for(k=str.length;k>j;k--){if(str.charAt(k)!=' ' && !isNaN(str.charAt(k))){price[inx]=str.substring(j,k);return }}} and my HTML <script type="text/javascript" src="cart.js" /> </script> <link rel="stylesheet" type="text/css" href="shopping_cart.css" /> <title> A title </title> </head> <body> <form name="form1" method="post" action="data.php" > <div id="product1"> <p id="title1"><b>Star Wars Tie Interceptor</b></p> <img src="images/DS.jpg" /> <p id="price1">Price £39.99</p> <p><b>Qty</b></p> <select name="qty"> <option value="0">0</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> <input type="button" value="Add to cart" onclick="cart()" /> </div>

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  • Why is my program getting slower and slower ?

    - by RedWolf
    I'm using the program to send data from database to the Excel file . It works fine at the beginning and then becomes more and more slowly,finally it run out of the memory and the following error ocurrs: "java.lang.OutOfMemoryError: Java heap space...". The problem can be resolved by adding the jvm heap sapce.But the question is that it spends too much time to run out the program. After several minutes,it finished a loop with 4 seconds which can be finished with 0.5 seconds at the beginning . I can't found a solution to make it always run in a certain speed. Is it my code problem? Any clues on this? Here is the code: public void addAnswerRow(List<FinalUsers> finalUsersList,WritableWorkbook book){ if (finalUsersList.size() >0 ) { try { WritableSheet sheet = book.createSheet("Answer", 0); int colCount = 0; sheet.addCell(new Label(colCount++,0,"Number")); sheet.addCell(new Label(colCount++,0,"SchoolNumber")); sheet.addCell(new Label(colCount++,0,"District")); sheet.addCell(new Label(colCount++,0,"SchoolName")); sheet.setColumnView(1, 15); sheet.setColumnView(3, 25); List<Elements> elementsList = this.elementsManager.getObjectElementsByEduTypeAndQuestionnaireType(finalUsersList.get(0).getEducationType().getId(), this.getQuestionnaireByFinalUsersType(finalUsersList.get(0).getFinalUsersType().getId())); Collections.sort(elementsList, new Comparator<Elements>(){ public int compare(Elements o1, Elements o2) { for(int i=0; i< ( o1.getItemNO().length()>o2.getItemNO().length()? o2.getItemNO().length(): o1.getItemNO().length());i++){ if (CommonFun.isNumberic(o1.getItemNO().substring(0, o1.getItemNO().length()>3? 4: o1.getItemNO().length()-1)) && !CommonFun.isNumberic(o2.getItemNO().substring(0, o2.getItemNO().length()>3? 4: o2.getItemNO().length()-1))){ return 1; } if (!CommonFun.isNumberic(o1.getItemNO().substring(0, o1.getItemNO().length()>3? 4: o1.getItemNO().length()-1)) && CommonFun.isNumberic(o2.getItemNO().substring(0,o2.getItemNO().length()>3? 4:o2.getItemNO().length()-1))){ return -1; } if ( o1.getItemNO().charAt(i)!=o2.getItemNO().charAt(i) ){ return o1.getItemNO().charAt(i)-o2.getItemNO().charAt(i); } } return o1.getItemNO().length()> o2.getItemNO().length()? 1:-1; }}); for (Elements elements : elementsList){ sheet.addCell(new Label(colCount++,0,this.getTitlePre(finalUsersList.get(0).getFinalUsersType().getId(), finalUsersList.get(0).getEducationType().getId())+elements.getItemNO()+elements.getItem().getStem())); } int sheetRowCount =1; int sheetColCount =0; for(FinalUsers finalUsers : finalUsersList){ sheetColCount =0; sheet.addCell(new Label(sheetColCount++,sheetRowCount,String.valueOf(sheetRowCount))); sheet.addCell(new Label(sheetColCount++,sheetRowCount,finalUsers.getSchool().getSchoolNumber())); sheet.addCell(new Label(sheetColCount++,sheetRowCount,finalUsers.getSchool().getDistrict().getDistrictNumber().toString().trim())); sheet.addCell(new Label(sheetColCount++,sheetRowCount,finalUsers.getSchool().getName())); List<AnswerLog> answerLogList = this.answerLogManager.getAnswerLogByFinalUsers(finalUsers.getId()); Map<String,String> answerMap = new HashMap<String,String>(); for(AnswerLog answerLog :answerLogList ){ if (answerLog.getOptionsId() != null) { answerMap.put(answerLog.getElement().getItemNO(), this.getOptionsAnswer(answerLog.getOptionsId())); }else if (answerLog.getBlanks()!= null){ answerMap.put(answerLog.getElement().getItemNO(), answerLog.getBlanks()); }else{ answerMap.put(answerLog.getElement().getItemNO(), answerLog.getSubjectiveItemContent()); } } for (Elements elements : elementsList){ sheet.addCell(new Label(sheetColCount++,sheetRowCount,null==answerMap.get(elements.getItemNO())?"0":answerMap.get(elements.getItemNO()))); } sheetRowCount++; } book.write(); book.close(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } catch (RowsExceededException e) { // TODO Auto-generated catch block e.printStackTrace(); } catch (WriteException e) { // TODO Auto-generated catch block e.printStackTrace(); } } }

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  • When I try to redefine a variable, I get an index out of bounds error

    - by user2770254
    I'm building a program to act as a calculator with memory, so you can give variables and their values. Whenever I'm trying to redefine a variable, a = 5, to a = 6, I get an index out of bounds error. public static void main(String args[]) { LinkedHashMap<String,Integer> map = new LinkedHashMap<String,Integer>(); Scanner scan = new Scanner(System.in); ArrayList<Integer> values = new ArrayList<>(); ArrayList<String> variables = new ArrayList<>(); while(scan.hasNextLine()) { String line = scan.nextLine(); String[] tokens = line.split(" "); if(!Character.isDigit(tokens[0].charAt(0)) && !line.equals("clear") && !line.equals("var")) { int value = 0; for(int i=0; i<tokens.length; i++) { if(tokens.length==3) { value = Integer.parseInt(tokens[2]); System.out.printf("%5d\n",value); if(map.containsKey(tokens[0])) { values.set(values.indexOf(tokens[0]), value); variables.set(variables.indexOf(tokens[0]), tokens[0]); } else { values.add(value); } break; } else if(tokens[i].charAt(0) == '+') { value = addition(tokens, value); System.out.printf("%5d\n",value); variables.add(tokens[0]); if(map.containsKey(tokens[0])) { values.set(values.indexOf(tokens[0]), value); variables.set(variables.indexOf(tokens[0]), tokens[0]); } else { values.add(value); } break; } else if(i==tokens.length-1 && tokens.length != 3) { System.out.println("No operation"); break; } } map.put(tokens[0], value); } if(Character.isDigit(tokens[0].charAt(0))) { int value = 0; if(tokens.length==1) { System.out.printf("%5s\n", tokens[0]); } else { value = addition(tokens, value); System.out.printf("%5d\n", value); } } if(line.equals("clear")) { clear(map); } if(line.equals("var")) { variableList(variables, values); } } } public static int addition(String[] a, int b) { for(String item : a) { if(Character.isDigit(item.charAt(0))) { int add = Integer.parseInt(item); b = b + add; } } return b; } public static void clear(LinkedHashMap<String,Integer> b) { b.clear(); } public static void variableList(ArrayList<String> a, ArrayList<Integer> b) { for(int i=0; i<a.size(); i++) { System.out.printf("%5s: %d\n", a.get(i), b.get(i)); } } I included the whole code because I'm not sure where the error is arising from.

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  • how to think like a computer scientist java edition exercise 7.2 [on hold]

    - by James Canfield
    I cannot figure out how to write this program, can someone please help me?! The purpose of this method is to practice manipulating St rings. Create a new program called Name.java. This program will take a name string consisting of EITHER a first name followed by a last name (nonstandar d format) or a last name followed by a comma then a first name (standard format). Ie . “Joe Smith” vs. “Smith, Joe”. This program will convert the string to standard format if it is not already in standard format. Write a method called hasComma that takes a name as an argument and that returns a boolean indicating whether it contains a comma. If i t does, you can assume that it is in last name first format. You can use the indexOf String m ethod to help you. Write a method called convertName that takes a name as an argument. It should check whether it contains a comma by calling your hasComma method. If it does, it should just return the string. If not, then it should assume th at the name is in first name first format, and it should return a new string that contains the name converted to last name comma first format. Uses charAt, length, substring, and indexOf methods. In your main program, loop, asking the user for a n ame string. If the string is not blank, call convertName and print the results. The loop terminat es when the string is blank. HINTS/SUGGESTIONS: Use the charAt, length, substring, and indexOf Str ing methods. Use scanner for your input. To get the full line, complete with spaces, use reader.nextLine()

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  • Are there any working implementations of the rolling hash function used in the Rabin-Karp string sea

    - by c14ppy
    I'm looking to use a rolling hash function so I can take hashes of n-grams of a very large string. For example: "stackoverflow", broken up into 5 grams would be: "stack", "tacko", "ackov", "ckove", "kover", "overf", "verfl", "erflo", "rflow" This is ideal for a rolling hash function because after I calculate the first n-gram hash, the following ones are relatively cheap to calculate because I simply have to drop the first letter of the first hash and add the new last letter of the second hash. I know that in general this hash function is generated as: H = c1ak - 1 + c2ak - 2 + c3ak - 3 + ... + cka0 where a is a constant and c1,...,ck are the input characters. If you follow this link on the Rabin-Karp string search algorithm , it states that "a" is usually some large prime. I want my hashes to be stored in 32 bit integers, so how large of a prime should "a" be, such that I don't overflow my integer? Does there exist an existing implementation of this hash function somewhere that I could already use? Here is an implementation I created: public class hash2 { public int prime = 101; public int hash(String text) { int hash = 0; for(int i = 0; i < text.length(); i++) { char c = text.charAt(i); hash += c * (int) (Math.pow(prime, text.length() - 1 - i)); } return hash; } public int rollHash(int previousHash, String previousText, String currentText) { char firstChar = previousText.charAt(0); char lastChar = currentText.charAt(currentText.length() - 1); int firstCharHash = firstChar * (int) (Math.pow(prime, previousText.length() - 1)); int hash = (previousHash - firstCharHash) * prime + lastChar; return hash; } public static void main(String[] args) { hash2 hashify = new hash2(); int firstHash = hashify.hash("mydog"); System.out.println(firstHash); System.out.println(hashify.hash("ydogr")); System.out.println(hashify.rollHash(firstHash, "mydog", "ydogr")); } } I'm using 101 as my prime. Does it matter if my hashes will overflow? I think this is desirable but I'm not sure. Does this seem like the right way to go about this?

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  • finding a string of random characters (with possible errors) within a large string of random charact

    - by mike
    I am trying to search a large string w/o spaces for a smaller string of characters. using regex I can easily find perfect matches but I can't figure out how to find partial matches. by partial matches i mean one or two extra characters in the string or one or two characters that have been changed, or one of each. the first and last characters will always match though. this would be similar to a spell checker but there are no spaces and the strings dont contain actual words, just random hex digits. i figured a way to find the string if there are no extra characters using indexOf(string.charAt(0)) and indexOf(charAt(string.length()-1) and looping through the characters between the two indexes. but this can be problematic when dealing with randomized characters because of the possibility of finding the first and last characters at the correct spacing but none of the middle characters matching. i've been scratching my head for hours on this issue. any ideas?

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  • What is the complexity of the below code with respect to memory ?

    - by Cshah
    Hi, I read about Big-O Notation from here and had few questions on calculating the complexity.So for the below code i have calculated the complexity. need your inputs for the same. private void reverse(String strToRevers) { if(strToRevers.length() == 0) { return ; } else { reverse(strToRevers.substring(1)); System.out.print(strToRevers.charAt(0)); } } If the memory factor is considered then the complexity of above code for a string of n characters is O(n^2). The explanation is for a string that consists of n characters, the below function would be called recursively n-1 times and each function call creates a string of single character(stringToReverse.charAT(0)). Hence it is n*(n-1)*2 which translates to o(n^2). Let me know if this is right ?

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  • ActionScript find LineBreak in XML and count them

    - by Pepe Sanchez
    Hi, i have an XML that has line breaks like this: This is a text that has line breaks im reading that xml in action script.. and trying to count how many linebreaks are in the text. Here is my code.. it returns 0 , it should return 5 for the example. function countBreaks(str:String) : Number { var count:Number = 0; for (var i:Number = 0; i < str.length-1; i++) { if(str.charAt(i)+str.charAt(i+1) eq "\n") count++; } return count; } I appreciate any help :)

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  • Recursion problem; completely lost

    - by timeNomad
    So I've been trying to solve this assignment whole day, just can't get it. The following function accepts 2 strings, the 2nd (not 1st) possibly containing *'s (asterisks). An * is a replacement for a string (empty, 1 char or more), it can appear appear (only in s2) once, twice, more or not at all, it cannot be adjacent to another * (ab**c), no need to check that. public static boolean samePattern(String s1, String s2) It returns true if strings are of the same pattern. It must be recursive, not use any loops, static & global variables. Can use local variables & method overloading. Can use only these methods: charAt(i), substring(i), substring(i, j), length(). Examples: 1: TheExamIsEasy; 2: "The*xamIs*y" --- true 1: TheExamIsEasy; 2: "Th*mIsEasy*" --- true 1: TheExamIsEasy; 2: "*" --- true 1: TheExamIsEasy; 2: "TheExamIsEasy" --- true 1: TheExamIsEasy; 2: "The*IsHard" --- FALSE I tried comparing the the chars one by one using charAt until an asterisk is encountered, then check if the asterisk is an empty one by comparing is successive char (i+1) with the char of s1 at position i, if true -- continue recursion with i+1 as counter for s2 & i as counter for s1; if false -- continue recursion with i+1 as counters for both. Continue this until another asterisk is found or end of string. I dunno, my brain loses track of things, can't concentrate, any pointers / hints? Am I in the right direction? Also, it's been told that a backtracking technique is to be used to solve this. My code so far (doesn't do the job, even theoretically): public static boolean samePattern(String s1, String s2) { if (s1.equals(s2) || s2 == "*") { return true; } return samePattern(s1, s2, 1); } public static boolean samePattern(String s1, String s2, int i) { if (s1.equals(s2)) return true; if (i == s2.length() - 1) // No *'s found -- not same pattern. return false; if (s1.substring(0, i).equals(s2.substring(0, i))) samePattern(s1, s2, i+1); else if (s2.charAt(i-1) == '*') samePattern(s1.substring(0, i-1), s2.substring(0, i), 1); // new smaller strings. else samePattern(s1.substring(1), s2, i); }

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  • To remove garbage characters from a string using regex...

    - by Harjit Singh
    Hi I want to remove characters from a string other then a-z, and A-Z. Created following function for the same and it works fine. public String stripGarbage(String s) { String good = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz"; String result = ""; for (int i = 0; i < s.length(); i++) { if (good.indexOf(s.charAt(i)) >= 0) { result += s.charAt(i); } } return result; } Can anyone tell me a better way to achieve the same. Probably regex may be better option. Regards Harry

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  • longest common subsequence

    - by davit-datuashvili
    i have following code public class LCS1 { public static String lcs(String a,String b) { String x; String y; int alen=a.length(); int blen=b.length(); if (alen==0 || blen==0) { return ""; } else if (a.charAt(alen-1)==b.charAt(blen-1)) { return lcs(a.substring(0,alen-1),b.substring(0,blen-1)); } else { x=lcs(a,b.substring(0,blen-1)); y=lcs(a.substring(0,alen-1),b); } return (x.length()>y.length()) ? x : y; } public static void main(String[]args){ String a="computer"; String b="houseboat"; System.out.println(lcs(a,b)); } } it should return "out" but returns nothing what is problem?

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  • Is there a best coding style for indentations (same line, next line)?

    - by Luis Soeiro
    I prefer Pascal-like coding style, where the beginning and ending of a code block are on the same column. I think that it is easier to read and to handle cut and paste than the other kind of coding style. The style I prefer (Pascal-like): void fooBar(String s) { int a; int length=s.length(); for (int i=0;i<length;i++) { if (i>10) { System.out.println(i); System.out.println(s.charAt(i)); } } } The style that was adopted by the Java community: void fooBar(String s) { int a; int length=s.length(); for (int i=0;i<length;i++){ if (i>10){ System.out.println(i); System.out.println(s.charAt(i)); } } } So why do you use one type or the other (please cite an objective reason)?

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  • java decmail string to AS 3.0 conversion procedure

    - by Jack Smith
    Hello, I have a problem with conversion java code to action script 3. Anyone can help with code translation? Thanks. public static short[] decmail_str_to_binary_data(String s) { short[] data = new short[s.length()/2]; for (int i = 0; i < s.length(); i += 2) { char c1 = s.charAt(i); char c2 = s.charAt(i + 1); int comb = Character.digit(c1, 16) & 0xff; comb <<= 4; comb += Character.digit(c2, 16) & 0xff; data[i/2] = (short)comb; } return data; }

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