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  • sip.conf configuration file - add new line to each record

    - by Flukey
    I have a sip configuration file which looks like this: [1664] username=1664 mailbox=1664@8360 host=192.168.254.3 type=friend subscribemwi=no [1679] username=1679 mailbox=1679@8360 host=192.168.254.3 type=friend subscribemwi=no [1700] username=1700 mailbox=1700@8360 host=192.168.254.3 type=friend subscribemwi=no [1701] username=1701 mailbox=1701@8360 host=192.168.254.3 type=friend subscribemwi=no For each record I need to add another line (vmxten for each record) for example the above becomes: [1664] username=1664 mailbox=1664@8360 host=192.168.254.3 type=friend subscribemwi=no vmexten=1664 [1679] username=1679 mailbox=1679@8360 host=192.168.254.3 type=friend subscribemwi=no vmexten=1679 [1700] username=1700 mailbox=1700@8360 host=192.168.254.3 type=friend subscribemwi=no vmexten=1700 [1701] username=1701 mailbox=1701@8360 host=192.168.254.3 type=friend subscribemwi=no vmexten=1701 What would you say would be the quickest way to do this? there are hundreds of records in the file, therefore modifying all of the records by hand would take a long time. Would you use Regex? Would you use sed? I'm interested to know how you would approach the problem. Thanks

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  • PHP: How to find connections between users so I can create a closed friend circle?

    - by CuSS
    Hi all, First of all, I'm not trying to create a social network, facebook is big enough! (comic) I've chosen this question as example because it fits exactly on what I'm trying to do. Imagine that I have in MySQL a users table and a user_connections table with 'friend requests'. If so, it would be something like this: Users Table: userid username 1 John 2 Amalia 3 Stewie 4 Stuart 5 Ron 6 Harry 7 Joseph 8 Tiago 9 Anselmo 10 Maria User Connections Table: userid_request userid_accepted 2 3 7 2 3 4 7 8 5 6 4 5 8 9 4 7 9 10 6 1 10 7 1 2 Now I want to find circles between friends and create a structure array and put that circle on the database (none of the arrays can include the same friends that another has already). Return Example: // First Circle of Friends Circleid => 1 CircleStructure => Array( 1 => 2, 2 => 3, 3 => 4, 4 => 5, 5 => 6, 6 => 1, ) // Second Circle of Friends Circleid => 2 CircleStructure => Array( 7 => 8, 8 => 9, 9 => 10, 10 => 7, ) I'm trying to think of an algorithm to do that, but I think it will take a lot of processing time because it would randomly search the database until it 'closes' a circle. PS: The minimum structure length of a circle is 3 connections and the limit is 100 (so the daemon doesn't search the entire database) EDIT: I've think on something like this: function browse_user($userget='random',$users_history=array()){ $user = user::get($userget); $users_history[] = $user['userid']; $connections = user::connection::getByUser($user['userid']); foreach($connections as $connection){ $userid = ($connection['userid_request']!=$user['userid']) ? $connection['userid_request'] : $connection['userid_accepted']; // Start the circle array if(in_array($userid,$users_history)) return array($user['userid'] => $userid); $res = browse_user($userid, $users_history); if($res!==false){ // Continue the circle array return $res + array($user['userid'] => $userid); } } return false; } while(true){ $res = browse_user(); // Yuppy, friend circle found! if($res!==false){ user::circle::create($res); } // Start from scratch again! } The problem with this function is that it could search the entire database without finding the biggest circle, or the best match.

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  • How to give friend access to git repository without giving command line access?

    - by Jack Humphries
    I have some git repositories running on my server and I would like to give a friend read/write access to one. That's simple: I add him as a user, give him SSH access, and change the permissions to the repository folder. Everything works fine; I'm able to clone the git repository using Xcode and change things (ssh://www.example.com/repo.git). However, I do not want him to have command line access. If I recall correctly, Github does not give command line access to those who SSH in. I'm using Snow Leopard Server. Is this more of a server issue or a git issue? Do you have any idea where to begin? Setting the user's Login Shell to none (as opposed to /bin/bash) cuts off access to everything.

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  • Where can I collaborate with my friend on source code in real time?

    - by Carson Myers
    I mean, other than a conference room :) Using google docs, I can upload any kind of file and view it with other people, watch them edit it in real time, with a live chat happening in the same window. This is awesome. How can I do the same thing with source code? I'm looking for a web application where I can upload source files that will be displayed in some kind of editor, with syntax highlighting, and allow others to view it and edit it in real time. Preferably with a live chat also, but not necessary. Does anybody know where I can find this?

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  • How can I share files from my Windows 7 machine to my friend's Ubuntu machine?

    - by ProfKaos
    I run a Windows 7 Pro SP1 laptop as my home machine, and my housemate runs an Ubuntu 12.04.1.05 desktop. We share a WLAN. I would like to make certain locations and files available for him to read and maybe write. How can I go about this? Bearing in mind I have very little recent experience with modern Linux, and Ubuntu in particular. My first idea is to share a Windows folder with my Ubuntu VM under VMWare Player, then his Ubuntu machine can connect to my Unbuntu VM, and the two can use whatever magic Ubuntu uses to achieve file sharing. This requires my Ubuntu VM to be always running though, and that may not always be possible. I have also heard that Samba may have a feature to help here, but I know nothing about that. How can I share my Windows files with my mate's Ubuntu machine, preferably with a 1 to 1 connection, i.e. rather not using shim VM's

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  • How do I access abstract private data from derived class without friend or 'getter' functions in C++?

    - by John
    So, I am caught up in a dilemma right now. How am I suppose to access a pure abstract base class private member variable from a derived class? I have heard from a friend that it is possible to access it through the base constructor, but he didn't explain. How is it possible? There are some inherited classes from base class. Is there any way to gain access to the private variables ? class Base_button { private: bool is_vis; Rect rButton; public: // Constructors Base_button(); Base_button( const Point &corner, double height, double width ); // Destructor virtual ~ Base_button(); // Accessors virtual void draw() const = 0; bool clicked( const Point &click ) const; bool is_visible() const; // Mutators virtual void show(); virtual void hide(); void move( const Point &loc ); }; class Button : public Base_button { private: Message mButton; public: // Constructors Button(); Button( const Point &corner, const string &label ); // Acessors virtual void draw() const; // Mutators virtual void show(); virtual void hide(); }; I want to be able access Rect and bool in the base class from the subclass

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  • org.hibernate.NonUniqueObjectException Within GWT application using hibernate through gilead

    - by molleman
    Hello Guys, i am working on a project for college that uses GWT,Hibernate and Gilead. Basically for the moment users should be able to add friends and remove them. also a user can see if his or her friends are online or not. my trouble is that when i add a friend that is already related to another friend i get this error org.hibernate.NonUniqueObjectException: a different object with the same identifier value was already associated with the session: [com.example.client.YFUser#4] i have a service class public class TestServiceImpl extends PersistentRemoteService implements TestService { this is my service class for my gwt application. my toruble is here with my implmentation class of my serivce in this method that is called when a user presses add friend button on the client-side public void addYFUserFriend(String userName){ //this retrieves the current user YFUser user = (YFUser)getSession().getAttribute(SESSION_USER); Session session = com.example.server.HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); YFUser friend = (YFUser) session.createQuery("select u FROM YFUser u where u.username = :username").setParameter("username", userName).uniqueResult(); System.out.println("user " + friend.getUsername() + " Found"); user.getFriends().add(friend); friend.getBefriended().add(user); session.update(user); session.update(friend); session.getTransaction().commit(); } a scenerio : user1 adds user2 as a friend. this works fine. then user3 adds user2 and the exeception is thrown. any ideas why and where my logic is going wrong Ok so i have changed my code, and i have removed all the getCurrentASession() calls and replaced with openSession() call which are closed at the appropiate point, now the error i am getting is com.google.gwt.user.server.rpc.UnexpectedException: Service method 'public abstract void com.example.client.TestService.addYFUserFriend(java.lang.String)' threw an unexpected exception: org.hibernate.NonUniqueResultException: query did not return a unique result: 3

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  • What Can I Do To One Of My Team Number (Good Friend As Well) Who Lost His Passion.

    - by skyflyer
    It seems this question is not program related, but there are lot of similar questions. So please bear with me! By the way, I am programmer and my team is also charging a software project. And SO is the only place which solved me lot of thorny troubles!THANK YOU GUYS! I joined my company with him years ago. At that time he was quite passionate on his job which is a front-end development. He gave us lot of useful suggestions concerning his work like design. And I believed he was a smart guy. I believe he still is smart too by the way. One years later, however, he seemed lost his passion and fooling around every day, did not care about his work any more and produced poorwork. Even worse he literally stopped learning new skills and honing his work related skills. For me it is horrible, we got to keep abreast with new technology development, otherwise we will be throw out. Since we were just coworkers, I did not care about it too much except mentioned my thoughts several times. But last month, we resembled a new group and assigned very important project. And I am the team leader, sadly! My boss gave me lot of support and expectation as well. I did a pretty good job before and I am very optimism to our future. But as a team, if my team does not work hard, we will be doomed to failure no matter how hard I work and push. In order to revitalize his passion, I tried couple of ways like talking to him about my concern and my boss's angry. I offered his new task which is quite new to him. I even persuaded my boss to give him new incentive package. But all of them knocked wall. His reaction was just he did not care. Even worse he did not want to talk about his situation. I want to be hard on him, but since we are friends and coworkers, I really can not see it will work. Even it works, I can not so quickly change my self from friend and coworker into manager. As a novice in management, I am really overwhelmed! I do not want get him fired, we are friends and I do not see him fired as my team number. What can I do? Thank you guys!

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  • Deadlock Analysis in NetBeans 8

    - by Geertjan
    Lock contention profiling is very important in multi-core environments. Lock contention occurs when a thread tries to acquire a lock while another thread is holding it, forcing it to wait. Lock contentions result in deadlocks. Multi-core environments have even more threads to deal with, causing an increased likelihood of lock contentions. In NetBeans 8, the NetBeans Profiler has new support for displaying detailed information about lock contention, i.e., the relationship between the threads that are locked. After all, whenever there's a deadlock, in any aspect of interaction, e.g., a political deadlock, it helps to be able to point to the responsible party or, at least, the order in which events happened resulting in the deadlock. As an example, let's take the handy Deadlock sample code from the Java Tutorial and look at the tools in NetBeans IDE for identifying and analyzing the code. The description of the deadlock is nice: Alphonse and Gaston are friends, and great believers in courtesy. A strict rule of courtesy is that when you bow to a friend, you must remain bowed until your friend has a chance to return the bow. Unfortunately, this rule does not account for the possibility that two friends might bow to each other at the same time. To help identify who bowed first or, at least, the order in which bowing took place, right-click the file and choose "Profile File". In the Profile Task Manager, make the choices below: When you have clicked Run, the Threads window shows the two threads are blocked, i.e., the red "Monitor" lines tell you that the related threads are blocked while trying to enter a synchronized method or block: But which thread is holding the lock? Which one is blocked by the other? The above visualization does not answer these questions. New in NetBeans 8 is that you can analyze the deadlock in the new Lock Contention window to determine which of the threads is responsible for the lock: Here is the code that simulates the lock, very slightly tweaked at the end, where I use "setName" on the threads, so that it's even easier to analyze the threads in the relevant NetBeans tools. Also, I converted the anonymous inner Runnables to lambda expressions. package org.demo; public class Deadlock { static class Friend { private final String name; public Friend(String name) { this.name = name; } public String getName() { return this.name; } public synchronized void bow(Friend bower) { System.out.format("%s: %s" + " has bowed to me!%n", this.name, bower.getName()); bower.bowBack(this); } public synchronized void bowBack(Friend bower) { System.out.format("%s: %s" + " has bowed back to me!%n", this.name, bower.getName()); } } public static void main(String[] args) { final Friend alphonse = new Friend("Alphonse"); final Friend gaston = new Friend("Gaston"); Thread t1 = new Thread(() -> { alphonse.bow(gaston); }); t1.setName("Alphonse bows to Gaston"); t1.start(); Thread t2 = new Thread(() -> { gaston.bow(alphonse); }); t2.setName("Gaston bows to Alphonse"); t2.start(); } } In the above code, it's extremely likely that both threads will block when they attempt to invoke bowBack. Neither block will ever end, because each thread is waiting for the other to exit bow. Note: As you can see, it really helps to use "Thread.setName", everywhere, wherever you're creating a Thread in your code, since the tools in the IDE become a lot more meaningful when you've defined the name of the thread because otherwise the Profiler will be forced to use thread names like "thread-5" and "thread-6", i.e., based on the order of the threads, which is kind of meaningless. (Normally, except in a simple demo scenario like the above, you're not starting the threads in the same class, so you have no idea at all what "thread-5" and "thread-6" mean because you don't know the order in which the threads were started.) Slightly more compact: Thread t1 = new Thread(() -> { alphonse.bow(gaston); },"Alphonse bows to Gaston"); t1.start(); Thread t2 = new Thread(() -> { gaston.bow(alphonse); },"Gaston bows to Alphonse"); t2.start();

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  • Inviting friends in facebook application

    - by Ahmy
    I have a facebook application that is published at facebook platform and i used facebook API to invite friends and i have succeeded in creating invitation form but the problem is that when u invite friend and send invitation and the invitation request sent to the user and the user accept it this friend appears again in the friend list that can be invited again For example : i have friend in my friend list named X and when i send invitation to him the invitation is sent and and X accept the invitation and when i try to send invitation again the friend X appears again in the list that i can select from to send invitation this means that may i send an invitation to this user (X) and he is already playing the game i need to know how to fix this problem so friends appear in the friend list (for invitation )only friends that not use the application. My application at the following link My Game application visit it and see the problem exactly after inviting friends they will appear again is this normal in any game application? thanks in advance for any reply

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  • how do i search from php file ?

    - by Tum Bin
    Dear Friends, im totally new in php. Just learning. I got 2 Assingment with php and html. Assignment 01: I have to mansion some pplz name and all of them some friends name. and I have to print common friend if there have common friend. Bt there have a prob that I got also msg which dnt have any common friend like “Rana has 0 friends in common with Roni.” I want to stop this and how can i? Assignment 02: I made a html form to search a person from that php file. Like: when I will search for Rana php form will b open and and print : Rana have 4 friends and he has a common friend with Nandini and Mamun. when I will search for Tanmoy the page will be open and print: Tonmoy is Rana’s friend who have 4 friend and common friends with Nandini and Mamun. for this I have to use the function “post/get/request” Plz plz plzzzzzzzzzz help me! Here im posting my codes; <?php # Function: finfCommon function findCommon($current, $arr) { $cUser = $arr[$current]; unset($arr[$current]); foreach ($arr As $user => $friends) { $common = array(); $total = array(); foreach ($friends As $friend) { if (in_array($friend, $cUser)) { $common[] = $friend; } } $total = count($common); $add = ($total != 1) ? 's' : ''; $final[] = "<i>{$current} has {$total} friend{$add} in common with {$user}.</i>"; } return implode('<br />', $final); } # Array of users and friends $Friends = array( "Rana" => array("Pothik", "Zaman", "Tanmoy", "Ishita"), "Nandini" => array("Bonna", "Shakib", "Kamal", "Minhaj", "Ishita"), "Roni" => array("Akbar", "Anwar", "Khakan", "Pavel"), "Liton" => array("Mahadi", "Pavel"), "Mamun" => array("Meheli", "Tarek", "Zaman") ); # Creating the output value $output = "<ul>"; foreach ($Friends As $user => $friends) { $total = count($friends); $common = findCommon($user, $Friends); $output .= "<li><u>{$user} has {$total} friends.</u><br /><strong>Friends:</strong>"; if (is_array($friends) && !empty($friends[0])) { $output .= "<ul>"; foreach ($friends As $friend) { $output .= "<li>{$friend}</li>"; } $output .= "</ul>"; } $output .= "{$common}<br /><br /></li>"; } $output .= "</ul>"; # Printing the output value print $output; ?>

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  • ExpandoObject (dynamics) my greatest friend or my new greatest foe?

    - by WeNeedAnswers
    Yes I know that it shouldn't be abused and that C# is primariy used as a static language. But seriously folks if you could just dirty up some code, in the python style, or create some dynamic do hicky, would you? My mind is working overtime on this having spent a while loving the dynamics of python, is c# going over to the dark side through the back door? Is the argument for static typing a dead one with this obvious addition? Is the argument for less Unit testing a bit silly when we are all grown ups? Or has the addition of dynamics ruined a strongly static typed and well designed language?

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  • fatal error C1034: windows.h: no include path set

    - by nathan
    OS Windows Vista Ultimate trying to run a program called minimal.c when i type at command line C:\Users\nathan\Desktopcl minimal.c Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 14.00.50727.762 for 80x86 Copyright (C) Microsoft Corporation. All rights reserved. minimal.c minimal.c(5) : fatal error C1034: windows.h: no include path set i have set all the paths: C:\Users\nathan\Desktoppath PATH=C:\Program Files (x86)\Microsoft Visual Studio 8\VC\bin;C:\Windows\system3 ;C:\Windows;C:\Windows\System32\Wbem;C:\Program Files (x86)\ATI Technologies\AT .ACE\Core-Static;C:\Program Files\Intel\DMIX;c:\Program Files (x86)\Microsoft S L Server\100\Tools\Binn\;c:\Program Files (x86)\Microsoft SQL Server\100\DTS\Bi n\;C:\Program Files (x86)\QuickTime\QTSystem\;C:\Program Files (x86)\Java\jdk1. .0_13\bin;C:\Program Files (x86)\Autodesk\Backburner\;C:\Program Files (x86)\Co mon Files\Autodesk Shared\;C:\Program Files (x86)\Microsoft DirectX SDK (March 009)\Include;C:\Users\nathan\Desktop\glut-3.7.6-bin\glut-3.7.6-bin;C:\Program F les (x86)\Microsoft Visual Studio 8\Common7\IDE;C:\Program Files (x86)\Microsof Visual Studio 8\VC\PlatformSDK\Include;C:\Program Files (x86)\Microsoft Visual Studio 8\VC\PlatformSDK\Include\gl i have gone and made sure windows.h is in the directory im setting the path too. its in C:\Program Files (x86)\Microsoft Visual Studio 8\VC\PlatformSDK\Include. i have visual studio 2005 i have exhausted all possiblies any ideas

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  • No Program Entry Point TASM Error

    - by Nathan Campos
    I'm trying to develop a simple kernel using TASM, using this code: ; beroset.asm ; ; This is a primitive operating system. ; ;********************************************************************** code segment para public use16 '_CODE' .386 assume cs:code, ds:code, es:code, ss:code org 0 Start: mov ax,cs mov ds,ax mov es,ax mov si,offset err_msg call DisplayMsg spin: jmp spin ;**************************************************************************** ; DisplayMsg ; ; displays the ASCIIZ message to the screen using int 10h calls ; ; Entry: ; ds:si ==> ASCII string ; ; Exit: ; ; Destroyed: ; none ; ; ;**************************************************************************** DisplayMsg proc push ax bx si cld nextchar: lodsb or al,al jz alldone mov bx,0007h mov ah,0eh int 10h jmp nextchar alldone: pop si bx ax ret DisplayMsg endp err_msg db "Operating system found and loaded.",0 code ends END Then I compile it like this: C:\DOCUME~1\Nathan\Desktop tasm /la /m2 beroset.asm Turbo Assembler Version 4.1 Copyright (c) 1988, 1996 Borland International Assembling file: beroset.asm Error messages: None Warning messages: None Passes: 2 Remaining memory: 406k C:\DOCUME~1\Nathan\Desktop tlink beroset, loader.bin Turbo Link Version 7.1.30.1. Copyright (c) 1987, 1996 Borland International Fatal: No program entry point C:\DOCUME~1\Nathan\Desktop What can I to correct this error?

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  • Rails: show some examples of code from controllers, models and views

    - by Totty
    Hy, my controller example: class FriendsController < ApplicationController before_filter :authorize, :except => [:friends] ############## ############## ## REQUESTS ## ############## ############## ################## # GET MY FRIENDS # ################## # Get my friends. def friends @friends = @my_profile.friends.paginate({:page => params[:page], :per_page => 3}) @profile = @my_profile end ################### # REMOVED FRIENDS # ################### # Get my deleted friends. def removed_friends @removed_friends = @my_profile.friends('removed_friends', params[:page]) end ################### # PENDING FRIENDS # ################### # Friend requests made by other profiles to me. def pending_friends @pending_friends = @my_profile.friends('pending_friends', params[:page]) end ############################ # REJECTED PENDING FRIENDS # ############################ # Rejected friend requests made by other profiles to me. def rejected_pending_friends @rejected_pending_friends = @my_profile.friends('rejected_pending_friends', params[:page]) end ##################### # REQUESTED FRIENDS # ##################### # The friend requests I've sent to others profiles. def requested_friends @requested_friends = @my_profile.friends('requested_friends', params[:page]) end ############################# # DELETED REQUESTED FRIENDS # ############################# # The requests I've sent to others # profiles and then canceled. def deleted_requested_friends @deleted_requested_friends = @my_profile.friends('deleted_requested_friends', params[:page]) end ############# ############# ## ACTIONS ## ############# ############# ########################## # ADD FRIENDSHIP REQUEST # ########################## # Add a friendship request. def add_friendship_request friendship = @my_profile.add_friendship_request(params[:profile_id]) render :json => friendship end ############################# # REMOVE FRIENDSHIP REQUEST # ############################# # Removes a friendship request I've done. def remove_friendship_request friendship = @my_profile.remove_friendship_request(params[:profile_id]) render :json => friendship end ###################### # PROCESS FRIENDSHIP # ###################### # Process friendship: accept or reject a friend. # This will make a new friend or # will make a new rejected pending friend. def process_friendship friendship = @my_profile.process_friendship(params[:profile_id].to_i, params[:accepted].to_i) render :json => friendship end ################### # REMOVE A FRIEND # ################### # Remove a friend from my friends by id. def remove_friend friendship = @my_profile.remove_friend(params[:profile_id]) render :json => friendship end end

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  • Binding, Prefixes and generated HTML

    - by Vman
    MVC newbie question re binders. Supposing I have two strongly typed partial actions that happen to have a model attributes with the same name, and are rendered in the same containing page i.e.: Class Friend {string Name {get; set ;} DateTime DOB {get; set ;}} Class Foe {string Name {get; set ;} string ReasonForDislike {get; set ;}} Both partials will have a line: <%= Html.TextBoxFor(model => model.Name) %> And associated controller actions: public ActionResult SaveFriend(Friend friend) public ActionResult SaveFoe(Foe foe) My problem is that both will render on my containing page with the same id (of course, bad for lots of reasons). I’m aware of the [Bind] attribute that allows me add a prefix, resulting in code: public ActionResult SaveFriend([Bind(Prefix = “friend”)] Friend friend) <%= Html.TextBox("friend.Name", Model. Name) %> //Boo, no TextBoxFor :( But this still doesn’t cut it. I can just about tolerate the loss of the strongly typed TextBoxFor helpers but I’ve yet to get clientside validation to work with prefixes: I’ve tried: <%= Html.ValidationMessage("friend.Name") %> ...and every other variant I can think of. I seem to need the model to be aware of the prefix in both directions but bind only applies when mapping the inbound request. It seems (to me) a common scenario but I’m struggling to find examples out there. What am I missing! Thanks in advance.

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  • A friend told me Python is garbage, I'm taking web design classes in the Spring and I have a textbook on C++. What should I do? [on hold]

    - by user107165
    I dont know if I should start digging into Python beforehand just to get acquanited with programming and "whet my appetite" or if I should work on the C++ book... Python definitely has more resources around town and I like the beginner friendly approach that seems to go along with every site that appeals to it. Or should I just wait for my assignments that start in 4 months? Any tips for an aspiring programmer?

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  • I cannot read second internal HD

    - by Nathan
    I reformatted my Win "C" drive and loaded Ubuntu 11.04. The second internal HD of course is not read. It is shown in 'disk utility'. I try to mount it in disk utility but I get an error message error mounting: mount exited with exit code 1: helper failed with: mount: according to mtab, /dev/sda1 is already mounted on / One problem I see is the error relates to sda1, but I am trying to mount sdb. Unfortunately, I am windows literate and ignorant of the coding for linux. Can someone help me with the code to mount my second drive (sdb) Nathan

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  • Improving performance of fuzzy string matching against a dictionary [closed]

    - by Nathan Harmston
    Hi, So I'm currently working for with using SecondString for fuzzy string matching, where I have a large dictionary to compare to (with each entry in the dictionary has an associated non-unique identifier). I am currently using a hashMap to store this dictionary. When I want to do fuzzy string matching, I first check to see if the string is in the hashMap and then I iterate through all of the other potential keys, calculating the string similarity and storing the k,v pair/s with the highest similarity. Depending on which dictionary I am using this can take a long time ( 12330 - 1800035 entries ). Is there any way to speed this up or make it faster? I am currently writing a memoization function/table as a way of speeding this up, but can anyone else think of a better way to improve the speed of this? Maybe a different structure or something else I'm missing. Many thanks in advance, Nathan

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  • View Script Over SSH?

    - by user74781
    A friend, using a remote machine, ran a script that SSHed to my machine, and ran the following python script that resides on my machine: while (1): ....print "hello world" (this script simply prints 'hello world' continuously). I am now logged in to my machine. How can I see the output of the script my friend was running? if it helps, I can 'spot' the script my friend is using: me@home:~$ ps aux | grep justprint.py **friend 7494 12.8 0.3 7260 3300 ? Ss 17:24 0:06 python TEST_AREA/justprint.py** friend 7640 0.0 0.0 3320 800 pts/3 S+ 17:25 0:00 grep --color=auto just what steps should I take in order to view the "hello world" messages on my screen?

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  • iOS: game with facebook challenges

    - by nazz_areno
    I created a game for iPad and I want to challenge my facebook friends. I follow the iOS tutorial in "facebook dev docs", with the "Smash game", but it doesn't explain how to challenge a friend directly to a game. I will explain with an example: I want to start a new match and I want challenge a friend on facebook. Then I send him a request to install the app and when I detect that its app is installed I send him a request to play vs me. Then, when I finish the match I sent him my result and my friend do the same thing. But if I and my friend don't finish the match it is not possible to send another challenge. This scenario is not explained by facebook sdk. Is it necessary to use another instrument to do this situation?

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