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  • Using StructureMap to create classes by a name?

    - by Bevan
    How can I use StructureMap to resolve to an appropriate implementation of an interface based on a name stored in an attribute? In my project, I have many different kinds of widgets, each descending from IWidget, and each decorated with an attribute specifying the kind of associated element. To illustrate: [Configuration("header")] public class HeaderWidget : IWidget { } [Configuration("linegraph")] public class LineGraphWidget : IWidget { } When processing my (XML) configuration file, I want to obtain an instance of the appropriate concrete class based on the name of the element I'm processing. public IWidget CreateWidget(XElement definition) { var kind = definition.Name.LocalName; var widget = // What goes here? widget.Configure(definition); return widget; } Each definition should result in a different widget being created - I don't need or want the instances to be shared. In the past I've written plenty of code to do this kind of thing manually, including writing a custom "roll-your-own" IoC container for one project. However, one of my goals with this project is to become proficient with StructureMap instead of reinventing the wheel. I think I've already managed to set up automatic scanning of assemblies so that StructureMap knows about all my IWidget implementations: public class WidgetRegistration : Registry { public WidgetRegistration() { Scan( scanner => { scanner.AssembliesFromApplicationBaseDirectory(); scanner.AddAllTypesOf<IWidget>(); }); } } However, this isn't registering the names of my widgets with StructureMap. What do I need to add to make my scenario work? (While I am trying to use StructureMap in this project, an answer showing me how to solve this problem with a different DI/IoC tool would still be valuable.)

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  • Getting registry information using Python

    - by Willy
    I am trying to pull registry info from many servers and put them all into one txt file. I got the code working fine in a .bat file. I hear that there is a way simpler way to do this in Python. I am intrigued and delighted to hear this. Can anyone help finish my code: My working bat file: echo rfsqlcl01app >> foo.txt reg query "\\rfsqlcl01app\HKEY_LOCAL_MACHINE\SOFTWARE\Network Associates\TVD\Shared Components\On Access Scanner\McShield\Configuration\Default" >> foo.txt echo GLADGSQL01 >> foo.txt reg query "\\GLADGSQL01\HKEY_LOCAL_MACHINE\SOFTWARE\Network Associates\TVD\Shared Components\On Access Scanner\McShield\Configuration\Default" >> foo.txt echo GLADGWEB01 >> foo.txt reg query "\\GLADGWEB01\HKEY_LOCAL_MACHINE\SOFTWARE\Network Associates\TVD\Shared Components\On Access Scanner\McShield\Configuration\Default" >> foo.txt echo PAPERVISION >> foo.txt My python code structure: >>> server_list = open('server_test.txt', 'r') >>> for line in server_list: print r'reg query \\%s\blah\blah\blah' % line.strip() reg query \\foo\blah\blah\blah reg query \\moo\blah\blah\blah reg query \\boo\blah\blah\blah >>> server_list.close()

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  • Does everything after my try statement have to be encompassed in that try statement to access variab

    - by Mithrax
    I'm learning java and one thing I've found that I don't like, is generally when I have code like this: import java.util.*; import java.io.*; public class GraphProblem { public static void main(String[] args) { if (args.length < 2) { System.out.println("Error: Please specify a graph file!"); return; } FileReader in = new FileReader(args[1]); Scanner input = new Scanner(in); int size = input.nextInt(); WeightedGraph graph = new WeightedGraph(size); for(int i = 0; i < size; i++) { graph.setLabel(i,Character.toString((char)('A' + i))); } for(int i = 0; i < size; i++) { for(int j = 0; j < size; j++) { graph.addEdge(i, j, input.nextInt()); } } // .. lots more code } } I have an uncaught exception around my FileReader. So, I have to wrap it in a try-catch to catch that specific exception. My question is does that try { } have to encompass everything after that in my method that wants to use either my FileReader (in) or my Scanner (input)? If I don't wrap the whole remainder of the program in that try statement, then anything outside of it can't access the in/input because it may of not been initialized or has been initialized outside of its scope. So I can't isolate the try-catch to just say the portion that intializes the FileReader and close the try statement immediately after that. So, is it the "best practice" to have the try statement wrapping all portions of the code that are going to access variables initialized in it? Thanks!

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  • While loop not reading in the last item

    - by Gandalf StormCrow
    I'm trying to read in a multi line string then split it then print it .. here is the string : 1T1b5T!1T2b1T1b2T!1T1b1T2b2T!1T3b1T1b1T!3T3b1T!1T3b1T1b1T!5T1*1T 11X21b1X 4X1b1X When I split the string with ! I get this without the last line string : 1T1b5T 1T1b5T1T2b1T1b2T 1T2b1T1b2T1T1b1T2b2T 1T1b1T2b2T1T3b1T1b1T 1T3b1T1b1T3T3b1T 3T3b1T1T3b1T1b1T 1T3b1T1b1T5T1*1T 5T1*1T11X21b1X 11X21b1X Here is my code : import java.io.BufferedInputStream; import java.util.Scanner; public class Main { public static void main(String args[]) { Scanner stdin = new Scanner(new BufferedInputStream(System.in)); while (stdin.hasNext()) { for (String line : stdin.next().split("!")) { System.out.println(line); for (int i = 0; i < line.length(); i++) { System.out.print(line.charAt(i)); } } } } } Where did I make the mistake, why is not reading in the last line? After I read in all lines properly I should go trough each line if I encounter number I should print the next char the n times the number I just read, but that is long way ahead first I need help with this. Thank you UPDATE : Here is how the output should look like : 1T1b5T 1T2b1T1b2T 1T1b1T2b2T 1T3b1T1b1T 3T3b1T 1T3b1T1b1T 5T1*1T 11X21b1X 4X1b1X Here is a solution in C(my friend solved it not me), but I'd stil wanted to do it in JAVA : #include <stdio.h> int main (void) { char row[134]; for (;fgets (row,134,stdin)!=NULL;) { int i,j=0; for (i=0;row[i]!='\0';i++) { if (row[i]<='9'&&row[i]>='1') j+=(row[i]-'0'); else if ((row[i]<='Z'&&row[i]>='A')||row[i]=='*') for (;j;j--) printf ("%c",row[i]); else if (row[i]=='b') for (;j;j--) printf (" "); else if (row[i]=='!'||row[i]=='\n') printf ("\n"); } } return 0; }

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  • If statement String trouble

    - by Jeremy Stone
    I'm trying to create a program which takes user input of two words and determines whether or not these words are the same. import java.util.Scanner; public class L7E3 { public static void main(String[] args) { Scanner keyboard = new Scanner(System. in ); String word1, word2; System.out.println("Please enter a word: "); word1 = keyboard.nextLine(); System.out.println("Please enter a word: "); word2 = keyboard.nextLine(); if (word1 == word2) { System.out.println("The words are " + word1 + " and " + word2 + ". These words are the same."); } else { System.out.println("The words are " + word1 + " and " + word2 + ". These words are not the same."); } } } I figured that word1==word2 would have worked to determine whether the two strings were equal, I'm using JGrasp and it goes directly to my else option regardless of input. Am I doing something wrong with strings?

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  • Very simple code for number search gives me infinite loop

    - by Joshua
    Hello, I am a newbie Computer Science high school student and I have trouble with a small snippet of code. Basically, my code should perform a basic CLI search in an array of integers. However, what happens is I get what appears to be an infinite loop (BlueJ, the compiler I'm using, gets stuck and I have to reset the machine). I have set break points but I still don't quite get the problem...(I don't even understand most of the things that it tells me) Here's the offending code (assume that "ArrayUtil" works, because it does): import java.util.Scanner; public class intSearch { public static void main(String[] args) { search(); } public static void search() { int[] randomArray = ArrayUtil.randomIntArray(20, 100); Scanner searchInput = new Scanner(System.in); int searchInt = searchInput.nextInt(); if (findNumber(randomArray, searchInt) == -1) { System.out.println("Error"); }else System.out.println("Searched Number: " + findNumber(randomArray, searchInt)); } private static int findNumber(int[] searchedArray, int searchTerm) { for (int i = 0; searchedArray[i] == searchTerm && i < searchedArray.length; i++) { return i; } return -1; } } This has been bugging me for some time now...please help me identify the problem!

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  • Read large amount of data from file in Java

    - by Crozin
    Hello I've got text file that contains 1 000 002 numbers in following formation: 123 456 1 2 3 4 5 6 .... 999999 100000 Now I need to read that data and allocate it to int variables (the very first two numbers) and all the rest (1 000 000 numbers) to an array int[]. It's not a hard task, but - it's horrible slow. My first attempt was java.util.Scanner: Scanner stdin = new Scanner(new File("./path")); int n = stdin.nextInt(); int t = stdin.nextInt(); int array[] = new array[n]; for (int i = 0; i < n; i++) { array[i] = stdin.nextInt(); } It works as excepted but it takes about 7500 ms to execute. I need to fetch that data in up to several hundred of milliseconds. Then I tried java.io.BufferedReader: Using BufferedReader.readLine() and String.split() I got the same results in about 1700 ms, but it's still too many. How can I read that amount of data in less that 1 second? The final result should be equal to: int n = 123; int t = 456; int array[] = { 1, 2, 3, 4, ..., 999999, 100000 };

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  • String Index Out Of Bound Exception error

    - by Fd Fehfhd
    Im not really sure why a am getting this error. But here is my code it is meant to test palindromes disregarding punctuation. So here is my code import java.util.Scanner; public class PalindromeTester { public static void main(String [] args) { Scanner kb = new Scanner(System.in); String txt = ""; int left; int right; int cntr = 0; do { System.out.println("Enter a word, phrase, or sentence (blank line to stop):"); txt = kb.nextLine(); txt = txt.toLowerCase(); char yP; String noP = ""; for (int i = 0; i < txt.length(); i++) { yP = txt.charAt(i); if (Character.isLetterOrDigit(txt.charAt(yP))) { noP += yP; } } txt = noP; left = 0; right = txt.length() -1; while (txt.charAt(left) == txt.charAt(right) && right > left) { left++; right--; } if (left > right) { System.out.println("Palindrome"); cntr++; } else { System.out.println("Not a palindrome"); } } while (!txt.equals("")); System.out.println("You found " + cntr + " palindromes. Thank you for using palindromeTester."); } } And if i test it and then i put enter so it will tell me how many palindromes you found the error i am getting is javav.lang.StringIndexOutOfBoundException : String index out of range 0 at PalindromeTester.main(PalindromeTester.java:38) and line 28 is while (txt.charAt(left) == txt.charAt(right) && right > left) Thanks for the help in advance

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  • For each loop not reading in the last item

    - by Gandalf StormCrow
    I'm trying to read in a multi line string then split it then print it .. here is the string : 1T1b5T!1T2b1T1b2T!1T1b1T2b2T!1T3b1T1b1T!3T3b1T!1T3b1T1b1T!5T1*1T 11X21b1X 4X1b1X When I split the string with ! I get this without the last line string : 1T1b5T 1T1b5T1T2b1T1b2T 1T2b1T1b2T1T1b1T2b2T 1T1b1T2b2T1T3b1T1b1T 1T3b1T1b1T3T3b1T 3T3b1T1T3b1T1b1T 1T3b1T1b1T5T1*1T 5T1*1T11X21b1X 11X21b1X Here is my code : import java.io.BufferedInputStream; import java.util.Scanner; public class Main { public static void main(String args[]) { Scanner stdin = new Scanner(new BufferedInputStream(System.in)); while (stdin.hasNext()) { for (String line : stdin.next().split("!")) { System.out.println(line); for (int i = 0; i < line.length(); i++) { System.out.print(line.charAt(i)); } } } } } Where did I make the mistake, why is not reading in the last line? After I read in all lines properly I should go trough each line if I encounter number I should print the next char the n times the number I just read, but that is long way ahead first I need help with this. Thank you

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  • How to create a String Array and link it to a Grade array

    - by user1861544
    I have a project that I need to create 2 Arrays, one to hold Student Names and one to hold Student Scores. The user inputs the size of the array, and the array needs to be sorted using BubbleSort (putting the high scores at the top). I have started the project, created the first array for scores, I have successfully done bubble sort and sorted the grades. Now I can't figure out how to make an array for Names, and once I do how do I make the names array correspond to the Grades array BubbleSort? Here is the code I have so far. import java.util.Scanner; public class Grades { public static void main(String[]args){ { Scanner UserIn = new Scanner(System.in); System.out.print( "How many students are there? " ); int[]GradeArray = new int[UserIn.nextInt()]; for( int i=0 ; i<GradeArray.length ; i++ ) { System.out.print( "Enter Grade for Student " + (i+1) + ": " ); GradeArray[i] = UserIn.nextInt(); } bubbleSort(GradeArray); for( int i : GradeArray ) System.out.println( i ); System.out.println(); } } private static void bubbleSort(int[]GradeArray){ int n = GradeArray.length; int temp = 0; String temp2; for(int i=0; i<n; i++){ for(int j=1; j<(n-i);j++){ if(GradeArray[j-1]<GradeArray[j]){ //swap temp=GradeArray[j-1]; GradeArray[j-1]=GradeArray[j]; GradeArray[j]=temp; } } } } } Also how do I change the grades to Double? I started with Int and when I try to change everything to double I get an error saying "Found Double, expected Int".

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  • Is my code a correct implementation of insertion sort?

    - by user1657171
    This code sorts correctly. Is this an insertion sort? import java.util.Scanner; public class InsertionSort { public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.println("Enter the number of elements: "); int count; count = sc.nextInt(); int[] a = new int[count]; System.out.println("Enter elements: "); for(int i = 0 ; i<count;i++){ a[i] = sc.nextInt(); } int j,temp; System.out.println("aftr insertion sort :"); for(int i = 1 ; i<count;i++){ j=i; while(j>0 && a[j-1] > a[j] ){ temp = a[j]; a[j] = a[j-1]; a[j-1] = temp; j--; } } for(int i = 0 ; i<count;i++){ System.out.print(a[i]+" "); } } }

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  • need help with some basic java.

    - by Racket
    Hi, I'm doing the first chapter exercises on my Java book and I have been stuck for a problem for a while now. I'll print the question, prompt/read a double value representing a monetary amount. Then determine the fewest number of each bill and coin needed to represent that amount, starting with the highest (assume that a ten dollar bill is the maximum size needed). For example, if the value entered is 47,63 (forty-seven dollars and sixty-three cents), and the program should print the equivalent amount as: 4 ten dollar bills 1 five dollar bills 2 one dollar bills 2 quarters 1 dimes 0 nickels 3 pennies" etc. I'm doing an example exactly as they said in order to get an idea, as you will see in the code. Nevertheless, I managed to print 4 dollars, and I can't figure out how to get "1 five dollar", only 7 dollars (see code). Please, don't do the whole code for me. I just need some advice in regards to what I said. Thank you. import java.util.Scanner; public class PP29 { public static void main (String[] args) { Scanner sc = new Scanner (System.in); int amount; double value; double test1; double quarter; System.out.println("Enter \"double\" value: "); value = sc.nextDouble(); amount = (int) value / 10; // 47,63 / 10 = 4. int amount2 = (int) value % 10; // 47 - 40 = 7 quarter = value * 100; // 47,63 * 100 = 4736 int sum = (int) quarter % 100; // 4763 / 100 => 4763-4700 = 63. System.out.println(amount); System.out.println(amount2); } }

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  • Identify words with ascending characters from text file

    - by user2914000
    I am having a fair amount of trouble trying to write a program that counts the amount of ascending words (words in which each character is larger than the previous character) in a text file. I have tried a few different methods to solve this but cannot seem to get it working. If anyone could help me revise the code to work properly it would be appreciated. The code will print about 5 of the words from the list of nearly 20000, but none considered are ascending (the file does have many ascending words) and it sometimes prints the same word twice. I am printing theWord to the console simply to see if the code works. import java.util.Scanner; import java.io.*; public class { public static void main (String [] args) throws FileNotFoundException{ String theWord; Scanner inputFile = new Scanner(new File("file.txt")); boolean ascending = true; int i = 1; while(inputFile.hasNextLine()){ theWord = inputFile.nextLine(); if(theWord.length() >= 2){ while(i < theWord.length() - 1){ if(theWord.charAt(i) <= theWord.charAt(i + 1)){ ascending = true; System.out.println("+ " + theWord); totalNum = totalNum + 1; } else{ ascending = false; System.out.println("= " + theWord); } i++; } } }

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  • ArrayIndexOutOfBoundsException double array size

    - by Andy
    I'm going to preface this question with this statement: I know that I can easily handle this problem by reading the amount of lines in a file and making an array that size. I am not allowed to do this. Anyway, here is my question. I need to double my array's size whenever my program encounters an ArrayIndexOutOfBoundsException and then copy all the previous read in information into the larger array. Here is my code public static void main(String[] args) throws IOException { Scanner inScan, fScan = null; int [] A = new int[5]; inScan = new Scanner(System.in); System.out.print("Please enter the file to read from: "); while(true) { try{ String fName = inScan.nextLine(); fScan = new Scanner(new File(fName)); break; } catch (FileNotFoundException ex) { System.out.println("Your file is invalid -- please re-enter"); } } String nextItem; int nextInt = 0; int i = 0; while (fScan.hasNextLine()) { try { nextItem = fScan.nextLine(); nextInt = Integer.parseInt(nextItem); A[i] = nextInt; i++; } catch (NumberFormatException e) { System.out.println("Found an invalid int -- ignored"); } catch (ArrayIndexOutOfBoundsException e) { //double the size of array A until //copy all previous read in information to the larger array } } System.out.println("Here are your " + i + " items:"); for (int j = 0; j < i; j++) { System.out.println(A[j] + " "); } } }

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  • How to scale page size down in Adobe Acrobat X Pro?

    - by WilliamKF
    I scanned a document on a scanner at a friend's house and the pdf ended up being 35 inches by 45 inches in size. I think this the cause of the trouble had for the person I sent it to, they get the error "insufficient image". How can I scale this down in Adobe Acrobat X Pro to a normal 8.5x11 inch sheet so that I can see if that resolves their issue and I can share the document with them. I cannot rescan the document, as I no longer have it. Acrobat is running on Windows 7 OS. The scanner was an HP OfficeJet Pro L7650 All-in-one.

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  • Does not recognize usb sticks and drives

    - by Peter
    When connecting any usb stick to my thinkpad ubuntu 10.10 does not recognize them. I don't see anything on the desktop. the output of "dmesg | tail -n10" gives me: [ 1965.696388] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1965.884537] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.072503] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.260349] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.506227] usb 1-1: new high speed USB device using ehci_hcd and address 9 [ 1966.572375] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.760379] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.948358] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1967.136335] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1967.325423] hub 1-0:1.0: unable to enumerate USB device on port 1 When connecting my usb scanner to the same port: [ 2008.480135] usb 1-1: new high speed USB device using ehci_hcd and address 65 [ 2008.548389] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2008.736786] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2008.924379] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.112348] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.300443] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.488536] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.732180] usb 1-1: new high speed USB device using ehci_hcd and address 71 [ 2014.796299] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2018.000128] usb 2-1: new full speed USB device using uhci_hcd and address 3 And ubuntu 10.10 recognizes that scanner. So: What can i do to see my usb stick? BTW: on my other Thinkpad running fedora 14 it works perfectly... Cheers -Peter

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  • ubuntu 10.10 does not recognize usb sticks and drives

    - by Peter
    When connecting any usb stick to my thinkpad ubuntu 10.10 does not recognize them. I don't see anything on the desktop. the output of "dmesg | tail -n10" gives me: [ 1965.696388] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1965.884537] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.072503] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.260349] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.506227] usb 1-1: new high speed USB device using ehci_hcd and address 9 [ 1966.572375] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.760379] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1966.948358] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1967.136335] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 1967.325423] hub 1-0:1.0: unable to enumerate USB device on port 1 When connecting my usb scanner to the same port: [ 2008.480135] usb 1-1: new high speed USB device using ehci_hcd and address 65 [ 2008.548389] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2008.736786] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2008.924379] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.112348] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.300443] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.488536] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2009.732180] usb 1-1: new high speed USB device using ehci_hcd and address 71 [ 2014.796299] hub 1-0:1.0: unable to enumerate USB device on port 1 [ 2018.000128] usb 2-1: new full speed USB device using uhci_hcd and address 3 And ubuntu 10.10 recognizes that scanner. So: What can i do to see my usb stick? BTW: on my other Thinkpad running fedora 14 it works perfectly... Cheers -Peter

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  • Help to organize game cycle in Java

    - by ASIO22
    I'm pretty new here (as though to a game development). So here's my question. I'm trying to organize a really simple game cycle in my public static main() as follows: Scanner sc = new Scanner(System.in); //Running the game cycle boolean flag=true; while (flag) { int action; System.out.println("Type your action please:"); System.out.println("0: Exit app"); try { action = sc.nextInt(); switch (action) { case 0: flag=false; break; case 1: break; } } catch (InputMismatchException ex) { System.out.println(ex.getClass() + "\n" + "Please type a correct input\n"); //action = sc.nextInt(); continue; } What's wrong with this cycle: I want to catch an exception when user types text instead of number, show a message, warning user, and the continue game cycle, read user input etc. But instead of that, when users types wrong data, it goes into a eternal cycle without even prompting user. What I did wrong?

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  • program logic of printing the prime numbers

    - by Vignesh Vicky
    can any body help to understand this java program it just print prime n.o ,as you enter how many you want and it works good class PrimeNumbers { public static void main(String args[]) { int n, status = 1, num = 3; Scanner in = new Scanner(System.in); System.out.println("Enter the number of prime numbers you want"); n = in.nextInt(); if (n >= 1) { System.out.println("First "+n+" prime numbers are :-"); System.out.println(2); } for ( int count = 2 ; count <=n ; ) { for ( int j = 2 ; j <= Math.sqrt(num) ; j++ ) { if ( num%j == 0 ) { status = 0; break; } } if ( status != 0 ) { System.out.println(num); count++; } status = 1; num++; } } } i dont understand this for loop condition for ( int j = 2 ; j <= Math.sqrt(num) ; j++ ) why we are taking sqrt of num...which is 3....why we assumed it as 3?

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  • Windows system drive letter changed after repair install

    - by taffer
    yesterday I did a repair install, because Windows froze on welcome screen after resizing its partition on a dual boot system. corresponding question Everything worked fine so far, but now I am facing another challenge: The Windows drive letter changed from G: to C:, so that most programs, including drivers, firewall and virus scanner do not work anymore. I tried to reinstall the virus scanner, but the uninstaller said, that drive G: is not available. What to do now, to get all programs running again? Is it possible, to rename C: to G: or will that mess up my system?

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  • HP Cue-Scanning Flow component freezes

    - by Nathan Fellman
    I am trying to scan with an HP network scanner (actually E6500 all-in-one). Whenever I try to scan, it starts up a flash screen with HP Scanning written all over it, which proceeds to do nothing. Digging in, I found that the process that gets stuck is hpqkygrp.exe, aka "HP CUE-Scanning Flow Component". This happened when I tried scanning from onenote or from the HP Solution Center. However, it seems that scanning from Windows' Fax and Scan utility works fine. As a (probably related) side-note, scanning directly from the scanner (using the buttons on its panel) doesn't work either. How can I keep this process from getting stuck?

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  • What is the best way to run ClamAV on Windows Server 2008 R2

    - by gabbsmo
    I'm hosting a Wordpress-site on Windows Server 2008 RS and want to scan all files that are uploaded by users for viruses using this plugin http://wordpress.org/extend/plugins/upload-scanner/. I'm on a really tight budget (no profit) so ClamAV seem like a good choice. What is the best way to run ClamAV under these circumstances? I'm concidering the following options: Just running the raw windows build from http://sourceforge.net/projects/clamav/ an setup definition updates with task scheduler. Any way to automate updates of the scanner (binaries)? Using a "distro" like ClamWin or Immunet (advertised on clamav.net). Any suggestions are welcome.

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  • Prevent machine in a LAN from receiving a remote shutdown

    - by WebDevHobo
    I'm probably just overreacting, but I recently came across a LAN-scanner that showed me the option "remote shutdown", for all found computers on the scanned network. Now, how exactly does this work? If I send such a message, will the shutdown happen no matter what, or is it required to have the password/user-name of the user of that other computer. Mostly I'm wondering: can this be done to me and how do I prevent it? EDIT: what's more, I had the scanner check for shares. The result being this: Double clicking the links opens them in explorer, basically meaning my entire C and F drive(only 2 HD's I have) are completely exposed to anyone in my LAN. Or can I open these because it's my own machine?

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  • keyPressed is not working after adding ActionListener to JButton

    - by Yehonatan
    I have a serious problem while trying to build a menu for my game. I've added two JButton to a main JPanel and added an ActionListener for each of them. The main JPanel also contains the game JPanel which have the keyPressed method inside keyController. That's how it looks - Main -       JPanel -         JButton, JButton,         JPanel which contains the game and keyPressed function inside KeyController class which worked fine before I added the ActionListener for JButton. For some reason after I added an ActionListener for each of the button, the game JPanel is not getting any keyPreseed events nor KeyRealesed. Does anyone know the solution for my situation? Thank you very much! Main window - Scanner in = new Scanner(System.in); JFrame f = new JFrame("Square V.S Circles"); f.setUndecorated(true); f.setResizable(false); f.setDefaultCloseOperation(JFrame.DISPOSE_ON_CLOSE); f.add(new JPanelHandler()); f.pack(); f.setVisible(true); f.setLocationRelativeTo(null); JPanelHandler(main JPanel) - super.setFocusable(true); JButton mybutton = new JButton("Quit"); JButton sayhi = new JButton("Say hi"); sayhi.addActionListener(new ActionListener() { @Override public void actionPerformed(ActionEvent e) { System.out.println("Hi"); } }); mybutton.addActionListener(new ActionListener() { @Override public void actionPerformed(ActionEvent e) { System.exit(0); } }); add(mybutton); add(sayhi); add(new Board(2)); Board KeyController(The code inside is working so it's unnecessary to put it here) - private class KeyController extends KeyAdapter { public KeyController() { ..Code } @Override public void keyPressed(KeyEvent e) { ...Code } @Override public void keyReleased(KeyEvent e){ ...Code } }

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