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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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  • 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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  • 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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  • Delegates: A Practical Understanding

    - by samerpaul
    It's been a while since I have written on this blog, and I'm planning on reviving it this summer, since I have more time to do so again.I've also recently started working on the iPhone platform, so I haven't been as busy in .NET as before.In either case, today's blog post applies to both C# and Objective-C, because it's more about a practical understanding of delegates than it is about code. When I was learning coding, I felt like delegates was one of the hardest things to conceptually understand, and a lot of books don't really do a good job (in my opinion) of explaining it. So here's my stab at it.A Real Life Example of DelegatesLet's say there are three of you. You, your friend, and your brother. You're each in a different room in your house so you can't hear each other, even if you shout. 1)You are playing a computer game2) Friend is building a puzzle3) Brother is nappingNow, you three are going to stay in your room but you want to be informed if anything interesting is happening to the one of you. Let's say you (playing the computer game) want to know when your brother wakes up.You could keep walking to the room, checking to see if he's napping, and then walking back to your room. But that would waste a lot of time / resources, and what if you miss when he's awake before he goes back to sleep? That would be bad.Instead, you hand him a 2-way radio that works between your room and his room. And you inform him that when he wakes up, he should press a button on the radio and say "I'm awake". You are going to be listening to that radio, waiting for him to say he's awake. This, in essence, is how a delegate works.You're creating an "object" (the radio) that allows you to listen in on an event you specify. You don't want him to send any other messages to you right now, except when he wakes up. And you want to know immediately when he does, so you can go over to his room and say hi. (the methods that are called when a delegate event fires). You're also currently specifying that only you are listening on his radio.Let's say you want your friend to come into the room at the same time as you, and do something else entirely, like fluff your brother's pillow. You will then give him an identical radio, that also hooks into your brother's radio, and inform him to wait and listen for the "i'm awake" signal.Then, when your brother wakes up, he says "I'm awake!" and both you and your friend walk into the room. You say hi, and your friend fluffs the pillow, then you both exit.Later, if you decide you don't care to say hi anymore, you turn off your radio. Now, you have no idea when your brother is awake or not, because you aren't listening anymore.So again, you are each classes in this example, and each of you have your own methods. You're playing a computer game (PlayComputerGame()), your friend is building a puzzle (BuildPuzzle()) and your brother is napping (Napping()). You create a delegate (ImAwake) that you set your brother to do, when he wakes up. You listen in on that delegate (giving yourself a radio and turning it on), and when you receive the message, you fire a new method called SayHi()). Your friend is also wired up to the same delegate (using an identical radio) and fires the method FluffPillow().Hopefully this makes sense, and helps shed some light on how delegates operate. Let me know! Feel free to drop me a line at Twitter (preferred method of contact) here: samerabousalbi

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  • uploading image & getting back from database

    - by Anup Prakash
    Putting a set of code which is pushing image to database and fetching back from database: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } if ($resultbytes!='') { echo $resultbytes; } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> Above set of code has one problem. The problem is whenever i pressing the "submit" button. It is just displaying the image on a page. But it is leaving all the html codes. even any new line message after the // Printing image on browser echo $resultbytes; //************************// So, for this i put this set of code in html tag: This is other sample code: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <?php if ($resultbytes!='') { // Printing image on browser echo $resultbytes; } ?> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> ** But in this It is showing the image in format of special charaters and digits. 1) So, Please help me to print the image with some HTML code. So that i can print it in my form to display the image. 2) Is there any way to convert the database image into real image, so that i can store it into my hard-disk and call it from tag? Please help me.

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  • android custom dialog imageButton onclicklistener

    - by Asaf Nevo
    this is my custom dialog class: package com.WhosAround.Dialogs; import com.WhosAround.AppVariables; import com.WhosAround.R; import com.WhosAround.AsyncTasks.LoadUserStatus; import com.WhosAround.Facebook.FacebookUser; import android.app.Dialog; import android.content.Context; import android.graphics.drawable.Drawable; import android.view.MotionEvent; import android.view.View; import android.widget.ImageButton; import android.widget.ImageView; import android.widget.TextView; public class MenuFriend extends Dialog{ private FacebookUser friend; private AppVariables app; public MenuFriend(Context context, FacebookUser friend) { super(context, android.R.style.Theme_Translucent_NoTitleBar); this.app = (AppVariables) context.getApplicationContext(); this.friend = friend; } public void setDialog(String userName, Drawable userProfilePicture) { setContentView(R.layout.menu_friend); setCancelable(true); setCanceledOnTouchOutside(true); TextView name = (TextView) findViewById(R.id.menu_user_name); TextView status = (TextView) findViewById(R.id.menu_user_status); ImageView profilePicture = (ImageView) findViewById(R.id.menu_profile_picture); ImageButton closeButton = (ImageButton) findViewById(R.id.menu_close); name.setText(userName); profilePicture.setImageDrawable(userProfilePicture); if (friend.getStatus() != null) status.setText(friend.getStatus()); else new LoadUserStatus(app, friend, status).execute(0); closeButton.setOnClickListener(new OnClickListener() { @Override public void onClick(View v) { dismiss(); } }) } } for some reason eclipse tells me the following errors on closeButton imageButton: The method setOnClickListener(View.OnClickListener) in the type View is not applicable for the arguments (new DialogInterface.OnClickListener(){}) The type new DialogInterface.OnClickListener(){} must implement the inherited abstract method DialogInterface.OnClickListener.onClick(DialogInterface, int) The method onClick(View) of type new DialogInterface.OnClickListener(){} must override or implement a supertype method why is that ?

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  • Runescape Private Server - How does it work?

    - by Friend of Kim
    I've seen a lot of Runescape private servers lately. How do they work? Most of them are based on the old Runescape, but a few look exactly like the real Runescape. How do they make the servers? Has the source code of the game been leaked on several occasions, and is that used to make Runescape servers? Or have some people just replicated Runescape, and tried to make the same game themselves (and "stolen" the 3D objects and texture from Jagex to make it look the same, and written the code to be able to replicate most functions of Runescape)?

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