Search Results

Search found 46178 results on 1848 pages for 'java home'.

Page 267/1848 | < Previous Page | 263 264 265 266 267 268 269 270 271 272 273 274  | Next Page >

  • Java: Multithreading & UDP Socket Programming

    - by Ravi
    I am new to multithreading & socket programming in Java. I would like to know what is the best way to implement 2 threads - one for receiving a socket and one for sending a socket. If what I am trying to do sounds absurd, pls let me know why! The code is largely inspired from Sun's tutorials online.I want to use Multicast sockets so that I can work with a multicast group. class server extends Thread { static protected MulticastSocket socket = null; protected BufferedReader in = null; public InetAddress group; private static class receive implements Runnable { public void run() { try { byte[] buf = new byte[256]; DatagramPacket pkt = new DatagramPacket(buf,buf.length); socket.receive(pkt); String received = new String(pkt.getData(),0,pkt.getLength()); System.out.println("From server@" + received); Thread.sleep(1000); } catch (IOException e) { System.out.println("Error:"+e); } catch (InterruptedException e) { System.out.println("Error:"+e); } } } public server() throws IOException { super("server"); socket = new MulticastSocket(4446); group = InetAddress.getByName("239.231.12.3"); socket.joinGroup(group); } public void run() { while(1>0) { try { byte[] buf = new byte[256]; DatagramPacket pkt = new DatagramPacket(buf,buf.length); //String msg = reader.readLine(); String pid = ManagementFactory.getRuntimeMXBean().getName(); buf = pid.getBytes(); pkt = new DatagramPacket(buf,buf.length,group,4446); socket.send(pkt); Thread t = new Thread(new receive()); t.start(); while(t.isAlive()) { t.join(1000); } sleep(1); } catch (IOException e) { System.out.println("Error:"+e); } catch (InterruptedException e) { System.out.println("Error:"+e); } } //socket.close(); } public static void main(String[] args) throws IOException { new server().start(); //System.out.println("Hello"); } }

    Read the article

  • Graph Tour with Uniform Cost Search in Java

    - by user324817
    Hi. I'm new to this site, so hopefully you guys don't mind helping a nub. Anyway, I've been asked to write code to find the shortest cost of a graph tour on a particular graph, whose details are read in from file. The graph is shown below: http://img339.imageshack.us/img339/8907/graphr.jpg This is for an Artificial Intelligence class, so I'm expected to use a decent enough search method (brute force has been allowed, but not for full marks). I've been reading, and I think that what I'm looking for is an A* search with constant heuristic value, which I believe is a uniform cost search. I'm having trouble wrapping my head around how to apply this in Java. Basically, here's what I have: Vertex class - ArrayList<Edge> adjacencies; String name; int costToThis; Edge class - final Vertex target; public final int weight; Now at the moment, I'm struggling to work out how to apply the uniform cost notion to my desired goal path. Basically I have to start on a particular node, visit all other nodes, and end on that same node, with the lowest cost. As I understand it, I could use a PriorityQueue to store all of my travelled paths, but I can't wrap my head around how I show the goal state as the starting node with all other nodes visited. Here's what I have so far, which is pretty far off the mark: public static void visitNode(Vertex vertex) { ArrayList<Edge> firstEdges = vertex.getAdjacencies(); for(Edge e : firstEdges) { e.target.costToThis = e.weight + vertex.costToThis; queue.add(e.target); } Vertex next = queue.remove(); visitNode(next); } Initially this takes the starting node, then recursively visits the first node in the PriorityQueue (the path with the next lowest cost). My problem is basically, how do I stop my program from following a path specified in the queue if that path is at the goal state? The queue currently stores Vertex objects, but in my mind this isn't going to work as I can't store whether other vertices have been visited inside a Vertex object. Help is much appreciated! Josh

    Read the article

  • Java Scanner class reading strings

    - by Max
    I've created a scanner class to read through the text file and get the value what I'm after. Let's assume that I have a text file contains 1 : Fnjiei : ID 7868860 : Age 18 2 : Oipuiieerb : ID 334134 : Age 39 3 : Enekaree : ID 6106274 : Age 31 I'm trying to get a name and id number and age, but everytime I try to run my code it gives me an exception. Here's my code. Any suggestion from java gurus?:) public void readFile(String fileName)throws IOException{ Scanner input = null; input = new Scanner(new BufferedReader(new FileReader(fileName))); try { while (input.hasNextLine()){ int howMany = 3; System.out.println(howMany); String userInput = input.nextLine(); String name = ""; String idS = ""; String ageS = ""; int id; int age; int count=0; for (int j = 0; j <= howMany; j++){ for (int i=0; i < userInput.length(); i++){ if(count < 2){ // for name if(Character.isLetter(userInput.charAt(i))){ name+=userInput.charAt(i); // store the name }else if(userInput.charAt(i)==':'){ count++; i++; } }else if(count == 2){ // for id if(Character.isDigit(userInput.charAt(i))){ idS+=userInput.charAt(i); // store the id } else if(userInput.charAt(i)==':'){ count++; i++; } }else if(count == 3){ // for age if(Character.isDigit(userInput.charAt(i))){ ageS+=userInput.charAt(i); // store the age } } id = Integer.parseInt(idS); // convert id to integer age = Integer.parseInt(ageS); // convert age to integer Fighters newFighters = new Fighters(id, name, age); fighterList.add(newFighters); } userInput = input.nextLine(); } } }finally{ if (input != null){ input.close(); } } } My appology if my mere code begs to be changed.

    Read the article

  • refactoring this function in Java

    - by Joel
    Hi folks, I'm learning Java, and I know one of the big complaints about newbie programmers is that we make really long and involved methods that should be broken into several. Well here is one I wrote and is a perfect example. :-D. public void buildBall(){ /* sets the x and y value for the center of the canvas */ double i = ((getWidth() / 2)); double j = ((getHeight() / 2)); /* randomizes the start speed of the ball */ vy = 3.0; vx = rgen.nextDouble(1.0, 3.0); if (rgen.nextBoolean(.05)) vx = -vx; /* creates the ball */ GOval ball = new GOval(i,j,(2 *BALL_RADIUS),(2 * BALL_RADIUS)); ball.setFilled(true); ball.setFillColor(Color.RED); add(ball); /* animates the ball */ while(true){ i = (i + (vx* 2)); j = (j + (vy* 2)); if (i > APPLICATION_WIDTH-(2 * BALL_RADIUS)){ vx = -vx; } if (j > APPLICATION_HEIGHT-(2 * BALL_RADIUS)){ vy = -vy; } if (i < 0){ vx = -vx; } if (j < 0){ vy = -vy; } ball.move(vx + vx, vy + vy); pause(10); /* checks the edges of the ball to see if it hits an object */ colider = getElementAt(i, j); if (colider == null){ colider = getElementAt(i + (2*BALL_RADIUS), j); } if (colider == null){ colider = getElementAt(i + (2*BALL_RADIUS), j + (2*BALL_RADIUS)); } if (colider == null){ colider = getElementAt(i, j + (2*BALL_RADIUS)); } /* If the ball hits an object it reverses direction */ if (colider != null){ vy = -vy; /* removes bricks when hit but not the paddle */ if (j < (getHeight() -(PADDLE_Y_OFFSET + PADDLE_HEIGHT))){ remove(colider); } } } You can see from the title of the method that I started with good intentions of "building the ball". There are a few issues I ran up against: The problem is that then I needed to move the ball, so I created that while loop. I don't see any other way to do that other than just keep it "true", so that means any other code I create below this loop won't happen. I didn't make the while loop a different function because I was using those variables i and j. So I don't see how I can refactor beyond this loop. So my main question is: How would I pass the values of i and j to a new method: "animateBall" and how would I use ball.move(vx + vx, vy + vy); in that new method if ball has been declared in the buildBall method? I understand this is probably a simple thing of better understanding variable scope and passing arguments, but I'm not quite there yet...

    Read the article

  • Java array of arry [matrix] of an integer partition with fixed term

    - by user335209
    Hello, for my study purpose I need to build an array of array filled with the partitions of an integer with fixed term. That is given an integer, suppose 10 and given the fixed number of terms, suppose 5 I need to populate an array like this 10 0 0 0 0 9 0 0 0 1 8 0 0 0 2 7 0 0 0 3 ............ 9 0 0 1 0 8 0 0 1 1 ............. 7 0 1 1 0 6 0 1 1 1 ............ ........... 0 6 1 1 1 ............. 0 0 0 0 10 am pretty new to Java and am getting confused with all the for loops. Right now my code can do the partition of the integer but unfortunately it is not with fixed term public class Partition { private static int[] riga; private static void printPartition(int[] p, int n) { for (int i= 0; i < n; i++) System.out.print(p[i]+" "); System.out.println(); } private static void partition(int[] p, int n, int m, int i) { if (n == 0) printPartition(p, i); else for (int k= m; k > 0; k--) { p[i]= k; partition(p, n-k, n-k, i+1); } } public static void main(String[] args) { riga = new int[6]; for(int i = 0; i<riga.length; i++){ riga[i] = 0; } partition(riga, 6, 1, 0); } } the output I get it from is like this: 1 5 1 4 1 1 3 2 1 3 1 1 1 2 3 1 2 2 1 1 2 1 2 1 2 1 1 1 what i'm actually trying to understand how to proceed is to have it with a fixed terms which would be the columns of my array. So, am stuck with trying to get a way to make it less dynamic. Any help?

    Read the article

  • java.lang.NumberFormatException: unable to parse '' as integer one more time

    - by Quzziy
    I will take two numbers from user, but this number from EditText must be converted to int. I think it should be working, but I still have problem with compilation code in Android Studio. CatLog show error in line with: int wiek = Integer.parseInt(wiekEditText.getText().toString()); Below is my full Android code: public class MyActivity extends ActionBarActivity { int Wynik; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activity_my); int Tmax, RT; EditText wiekEditText = (EditText) findViewById(R.id.inWiek); EditText tspoczEditText = (EditText) findViewById(R.id.inTspocz); int wiek = Integer.parseInt(wiekEditText.getText().toString()); int tspocz = Integer.parseInt(tspoczEditText.getText().toString()); Tmax = 220 - wiek; RT = Tmax - tspocz; Wynik = 70*RT/100 + tspocz; final EditText tempWiekEdit = wiekEditText; TabHost tabHost = (TabHost) findViewById(R.id.tabHost); //Do TabHost'a z layoutu tabHost.setup(); TabHost.TabSpec tabSpec = tabHost.newTabSpec("Calc"); tabSpec.setContent(R.id.Calc); tabSpec.setIndicator("Calc"); tabHost.addTab(tabSpec); tabSpec = tabHost.newTabSpec("Hints"); tabSpec.setContent(R.id.Hints); tabSpec.setIndicator("Hints"); tabHost.addTab(tabSpec); final Button Btn = (Button) findViewById(R.id.Btn); Btn.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View v) { Toast.makeText(getApplicationContext(),"blablabla"+ "Wynik",Toast.LENGTH_SHORT).show(); } }); wiekEditText.addTextChangedListener(new TextWatcher() { @Override public void beforeTextChanged(CharSequence s, int start, int count, int after) { } @Override public void onTextChanged(CharSequence s, int start, int before, int count) { Btn.setEnabled(!(tempWiekEdit.getText().toString().trim().isEmpty())); } @Override public void afterTextChanged(Editable s) { } }); } @Override public boolean onCreateOptionsMenu(Menu menu) { // Inflate the menu; this adds items to the action bar if it is present. getMenuInflater().inflate(R.menu.my, menu); return true; } @Override public boolean onOptionsItemSelected(MenuItem item) { // Handle action bar item clicks here. The action bar will // automatically handle clicks on the Home/Up button, so long // as you specify a parent activity in AndroidManifest.xml. int id = item.getItemId(); if (id == R.id.action_settings) { return true; } return super.onOptionsItemSelected(item); } }

    Read the article

  • Customs toString in Java not giving desired output and throwing error

    - by user2972048
    I am writing a program in Java to accept and validate dates according to the Gregorian Calendar. My public boolean setDate(String aDate) function for an incorrect entry is suppose to change the boolean goodDate variable to false. That variable is suppose tell the toString function, when called, to output "Invalid Entry" but it does not. My public boolean setDate(int d, int m, int y) function works fine though. I've only included the problem parts as its a long piece of code. Thanks public boolean setDate(int day, int month, int year){ // If 1 <= day <= 31, 1 <= month <= 12, and 0 <= year <= 9999 & the day match with the month // then set object to this date and return true // Otherwise,return false (and do nothing) boolean correct = isTrueDate(day, month, year); if(correct){ this.day = day; this.month = month; this.year = year; return true; }else{ goodDate = false; return false; } //return false; } public boolean setDate(String aDate){ // If aDate is of the form "dd/mm/yyyy" or "d/mm/yyyy" // Then set the object to this date and return true. // Otherwise, return false (and do nothing) Date d = new Date(aDate); boolean correct = isTrueDate(d.day, d.month, d.year); if(correct){ this.day = d.day; this.month = d.month; this.year = d.year; return true; }else{ goodDate = false; return false; } } public String toString(){ // outputs a String of the form "dd/mm/yyyy" // where dd must be 2 digits (with leading zero if needed) // mm must be 2 digits (with leading zero if needed) // yyyy must be 4 digits (with leading zeros if needed) String day1; String month1; String year1; if(day<10){ day1 = "0" + Integer.toString(this.day); } else{ day1 = Integer.toString(this.day); } if(month<10){ month1 = "0" + Integer.toString(this.month); } else{ month1 = Integer.toString(this.month); } if(year<10){ year1 = "00" + Integer.toString(this.year); } else{ year1 = Integer.toString(this.year); } if(goodDate){ return day1 +"/" +month1 +"/" + year1; }else{ goodDate = true; return "Invalid Entry"; } } Thank you

    Read the article

  • on SSH login, get message 'Could not chdir to home directory"

    - by joachim
    I am SSHing into a Mac OS X server running Tiger. When I log in I get put in the root directory and shown this message: Could not chdir to home directory : No such file or directory My $HOME variable seems to be empty. I've googled the problem and found a mailing list thread which suggests using dscl to set up the home directory, but I've done that and the problem still persists even though now dscl correctly reports: $ dscl . -read /users/me NFSHomeDirectory NFSHomeDirectory: /Users/me

    Read the article

  • vsftpd restrict local users to home and group directories

    - by wag2639
    i've got vsftpd install on an ubuntu server 9.10 i can use chroot to restrict users to their own home directories but i also want to give them access to a group shared folder for example, users foo1 and foo2 are local users in the group foos i want foo1 to have access to /home/foo1 and /svr/foos and foo2 to have access to /home/foo2 and /svr/foos other notes: using pam and enforce local user ssl already tried mount --bind but it does weird permissions when you try to mount bind multiple users to the same

    Read the article

  • Change the Safari home page for the OSX Guest account

    - by John Lemberger
    I'd like to setup the guest account with easy access to a particular web site, but cannot figure out how to change the default. In 10.5.8 the parental controls can be used to control the list of bookmarks, but I haven't seen any reference to the home page. And when logged in as Guest, the home page settings are read-only, even if you enter an administrator password. How can the Safari home page be changed (and be made persistent) for the OSX Guest account?

    Read the article

  • Apache to read from /home/user/public_html on CentOS 5.7

    - by C.S.Putra
    this is my first experience using CentOS 5.7 / Linux as my web server OS and I have just finished installing Apache. Then I created a new account using WHM. The account is now created and the domain name can be accessed. I have put the web files under /home/user/public_html/ but when I access the domain assigned for that user which I assigned when creating new account in WHM, it doesn't read the files. In /usr/local/apache/conf/httpd.conf : <VirtualHost 175.103.48.66:80> ServerName domain.com ServerAlias www.domain.com DocumentRoot /home/user/public_html ServerAdmin [email protected] User veevou # Needed for Cpanel::ApacheConf <IfModule mod_suphp.c> suPHP_UserGroup group1 group1 </IfModule> <IfModule !mod_disable_suexec.c> SuexecUserGroup group1 group1 </IfModule> CustomLog /usr/local/apache/domlogs/domain.com-bytes_log "%{%s}t %I .\n%{%s}t %O ." CustomLog /usr/local/apache/domlogs/domain.com combined ScriptAlias /cgi-bin/ /home/user/public_html/cgi-bin/ </VirtualHost> Instead of reading from /home/user/public_html/ apache will read the /var/ww/html/ folder. How to set the apache so that when user access www.domain.com, they will access the files under /home/user/public_html/ ? Please advice. Thanks

    Read the article

  • Changed login for a user with encrypted home, now I can't login

    - by HappyDeveloper
    I changed a user's login by doing this: $ usermod old_login -l new_login I also wanted to move his home to reflect his new username, but it wouldn't let me, so I just rebooted. But now after I login, the screen blinks and I'm redirected back to the login screen. And that's what happens when you cannot access your home, that's why I think it has something to do with his home being encrypted. How do I fix this? I'm on a Ubuntu 12 Virtualbox VM.

    Read the article

  • Linux Mint does not start after renaming home directory

    - by RUBY
    I am new to linux and was just trying to rename the only directory in home from rk to rhk. I messed up the whole thing and the settings. Created some new thing named rhk which I can't remember as it got all messed up and Now I am getting nothing after Linux Mint 10(julia) boots up - no start menu, no panel, no taskbar nothing. I tried to work in the recovery mode and got some(downloaded) 216mb of something(in the repair broken packages) hoping that it might help but didn't help. Moreover whenver I have booted in it shows messages like Could not update ICEauthority file /home/rk./.ICEauthority there is a problem with the configuration server. (usr/lib/libconfig24/gconfsanitycheck2 exited with status 256) The panel encountered a problem while loading "OAFIID: GNOME_mintMenu" The panel encountered a problem while loading "OAFIID: GNOME_IndicatorApplet" Naulitis could not create the following reqiured folders: /home/rk/Desktop, /home/rk/. Naulitis Moreover Alt+F2 gives Run application or run with file and nothing seems to be working.

    Read the article

  • recover files from encrypted home folder

    - by maskiepop
    I can't seem to find the answer to my questions -- hence my posting this here. 1) I have encrypted my home folder in LinuxMint 15 Cinammon x64. If I create images of LM's partitions via fsarchiver, how do I go about selectively restoring home folders, and files from the FS images? 2) I haven't done this yet; but can I restore another users home folder into another user; both unencrypted. Is that a fairly common thing to do in Linux/ubuntu? I mean is the process fairly straightforward? What if the home folder I want to copy over to another user is encrypted? Thx

    Read the article

  • Java issues with Apache 2.0 Agent 2.202 for RHEL5 Linux 64bit

    - by Richard
    In trying to install Apache 2.0 Agent 2.202 for RHEL5 Linux 64bit, the dialogue appears as follows. $ ./setup Error : java is not present in path. Please enter JAVAHOME path to pick up java:/usr/lib/jvm/java-1.6.0-openjdk-1.6.0.0.x86_64/jre/ Launching installer... Attach to native process failed $ ./setup Error : java is not present in path. Please enter JAVAHOME path to pick up java:/usr/lib/jvm/java-1.6.0-openjdk-1.6.0.0.x86_64/jre/lib ./setup: line 80: [: 107:: integer expression expected ./setup: line 83: [: 107:: integer expression expected Error : Incorrect java version (1.2.2 or above is needed). Please enter JAVAHOME path to pick up java: On the server we have the following JREs and I've tried both. $ sudo rpm -qa | egrep "(openjdk|icedtea)" java-1.6.0-openjdk-1.6.0.0-1.27.1.10.8.el5_8 $ find 2>/dev/null | grep -i '/jre/' ./usr/lib/jvm/java-1.4.2-gcj-1.4.2.0/jre/bin/ ... ./usr/lib/jvm/java-1.6.0-openjdk-1.6.0.0.x86_64/jre/ Any suggestions? I know I'm overlooking something. In previous searches I've only found one other posting that comes close but it has no responses (http://forum.parallels.com/showthread.php?t=76556).

    Read the article

  • windows 7 home premium - update to professionel

    - by ben
    Hi there, I bought a new notebook that came with Win7 Home Premium. I know want to sell my home premium license and buy a profesionell one. Is it possible to just change the key in my current running and ready-configuerd win7 home premium, or do I need ti reinstall the system? Thanks for your help!

    Read the article

  • Building a home cluster - hardware and cost analysis

    - by ldigas
    Does anyone know some links / books / anything you can think of, that describe the process of building a little home cluster (when I say home, it doesn't necessarily mean for keeping at home - just means it's relatively cheap and small) for experimental purposes, with a special emphasis on what hardware would be adequate today, and some kind of cost analysis ? Although, if someone here's done it, I'd appreciate all the experience you can share.

    Read the article

  • mod_rewite Rule: root/? root/app/views/home/home.php

    - by Jonathon David Oates
    I am shocking at mod_rewite, here's the scenario: I need a rule that rewrites mydomain.com to mydomain.com/app/views/home/home.php. The rule, or set of rules rather, must also rewite mydomain.com/signin to mydomain.com/app/views/signin/signin.php, and work in a similar fashion for any subdirectory, for example: mydomain.com/subdir must redirect to mydomain.com/app/views/subdir/subdir.php. The rules must also work with or without the trailing slash, for example: ….com or ….com/. Thank you all, your help is much appreciated! If you could outline how and why your solution works or direct me to a good resource that explains it, I'd be exceptionally grateful! Edit: I have got a simple .htaccess file with this: Options +FollowSymLinks RewriteEngine On RewriteRule ^$ http://mydomain.local/~Jay/some_awesome_app/app/views/home/home.php This does the redirect but changes the URL in the address bar too! I've not got a trailing [R] flag so why would this be?

    Read the article

  • Connecting to my home router web interface from work

    - by Joe
    Hi, I'm trying to connect to my home router web interface from work. I use dyndns, because I don't have a static IP at home, and it works perfectly from any other place except my workplace (update: I made a mistake, see edit below). When trying to access the web interface from work I get a "500 Server Error" with the code: SERVER_RESPONSE_RESET. I'm not trying to use any protocols such as remote desktop, I'm only trying to access the web interface. I can access any other web page from my workplace with no problems, and I think my router web interface is like any other web page, isn't it? I thought maybe my work place proxy blocks addresses of services like dyndns, so I also applied another trick. Since I have a web page on my own domain (say www.mydomain.com) which I can access from work, I tried adding a CNAME to my domain which is linked to the dyndns address (router.mydomain.com). This way if anyone enters the address router.mydomain.com from anywhere, they reach my home router web interface, and there's no way of knowing it's a dyndns address (or is there?). However, it still doesn't work from my workplace (I get the same error message). Any ides? Edit: I'm sorry to say I made a mistake earlier. I used to be able to access my home router web interface from my old workplace, and I thought it was still possible since I don't recall making any configuration changes. However, after reading the replies, I went over to my old workplace and checked, and it doesn't work from there either. I'm very sorry for giving out wrong and misleading information about my problem. So to summarize: my problem is that I can't access my home router web interface from anywhere.

    Read the article

  • Laptop connects to other network but not to my home wireless

    - by Nilesh
    My home network's wireless SSID is say "XYZ" I also have an ethernet wire from the same router. I have two laptops A and B Earlier both A and B were able to connect to my home internet through the ethernet and wireless. Suddenly, the laptop B can no longer connect to XYZ or through ethernet. When I do plug the wire, i get the connection icon all green but when I try to access any web page it errors out (page not found) But strangely laptop B connects to my neighbours wireless SSID "ABC". I have also tested laptop B with other networks and it connects fine. Laptop A and many other devices still connect fine with my home wireless "XYZ" Strange thing is when my laptop B connects wireless through XYz, it gets the IP address but then none of the browsers (chrome,firefox, IE) can show any web pages. What settings should I be checking on laptop B that is preventing it to connect to my home internet. Thank you

    Read the article

  • DVD Share on Vista Home Premium Failing

    - by hpunyon
    UPDATE: - I can't find any Local Policy Editor for Vista Home Premium, as suggested. - I did learn about registry keys: allocatecdroms, allocatefloppies, allocatedasd and tried adding these keys (individually and collectively) and setting them to both 0 or 1. There was no positive affect on read access to the DVD root folder - always Access Denied. ORIGINAL POST: Failing read access to the root folder of a DVD drive in Vista Home Premium laptop using the Guest account - Access Denied. The client is an XP Home PC that can see, but not access, the data in the share. I'm only trying to read the data DVD - not trying to write/burn anything. On the Vista laptop, I have: All Firewalls and Antivirus disabled.UAC disabled. Password checking disabled. "Advanced Shared" the DVD drive, with "Everyone" having full-access permissions to the share. Tried adding Guest and Anonymous users having full-access permissions to the share. RestrictAnonymous=0 set in the registry. Both PC's are in the same workgroup (MSHOME) The XP Home client sees the shared DVD in \Vista_Hostname\ but when I double click the drive icon on the client, I get a popup that access is denied, check with the administrator, etc. I can share other folders on the Vista PC and see and READ these from the XP Home client. If I enable password checking on the Vista side, I get a user/password popup, and I can authenticate (using my known Vista account, that happens to have Admin rights) and then I can get to see and read the DVD data. I need to open this up so that the (default) Guest user can see and access the DVD data files.

    Read the article

  • Why can't Windows home editions connect to domains?

    - by TyrionLannister
    The company that I work for continuously hires new people, and I'm the one who has to go and purchase new computers. The majority of them, if not all, come pre-installed with Windows Home editions. I'm noticing that the Windows 7/8 Home editions are unable to connect to domains. I'm having to buy the upgrades to the Pro editions. I'm trying to understand as to why the Home edition of the OS is unable to connect to domains?

    Read the article

< Previous Page | 263 264 265 266 267 268 269 270 271 272 273 274  | Next Page >