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  • UltraSurf not working, used to work on same network (and is working fine on friend's computer on same network)

    - by Kush
    I'm using UltraSurf (10.17), it used to work fine but now its not working on my computer (Windows 7) and the window keeps showing "Connecting Server..." and after sometime, it fails to establish the connection. While, its working perfectly on the friend's computer on the same network. Any help will be appreciated. (Please note that using Proxifier would be my last option if I can't get UltraSurf working anyway).

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  • Is it still true, to make cross broswer layouts for desktop browsers using table+css is easier then

    - by metal-gear-solid
    My one of web designer friend still making sites with table but he use css very nicely and I also use css nicely but with <div> and i face cross browser problem in layout more than my friend. and i given some reason to my friend about cons of <table>. read my whole discussion with friend? I - you site will be problematic with screen reader My friend - OK, but i never got any call from any client regarding this. I - you will devote more time to make any changes in layout, if changes comes from client My friend - I don't think so, but if it is then show me how can i save time with <div>? I - your sites will not work well with search engine. My friend - it's not true. I've made many site and no problem with any site or client regarding this I - layout is old way, non w3c and non standard way. My friend - what is old and what is new, Who is W3C i don't know, What is standard? Whatever i make works in all browsers, it's enough for me and my client will not pay for standard and W3C guidelines rules I - Your site will not work in mobile browsers My friend - No problem for me, my client don't care about mobile phone I - Your sites are not accessible? My Friend - What do u mean not accessible? Whatever i make works in all browsers. my any client never asked about accessibility I - You will not get more work in future, with table? My friend - OK, no problem when clients will not accept site with table then i will learn about div based layouts in future. My questions? Is it still true, to make cross browser layouts for desktop browsers using table+css is easier then div+css? What is the benefit for developer to use DIV+CSS layout in place of <table> layouts if client would not mind if i use ?

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  • I lent my 3 USB modem to a friend because her Meteor USB modem had run out of credit.

    - by oddbyte7
    I lent my 3 USB modem to a friend because her Meteor USB modem had run out of credit. The 3 modem worked on her machine fine once the driver was installed. She bought Meteor credit a couple of days later, uninstalled the 3 driver and she only now seems to get on the Meteor server, when you try IE8 or Firefox you get the message “Page could not be found”. I uninstalled the Meteor driver and reinstalled…no change. The OS is Win_7 home edition. Any help would be much appreciated. Regards to all, oddbyte7.

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  • Why this friend function can't access a private member of the class?

    - by Alceu Costa
    I am getting the following error when I try to access bins private member of the GHistogram class from within the extractHistogram() implementation: error: 'QVector<double> MyNamespace::GHistogram::bins' is private error: within this context Where the 'within this context' error points to the extractHistogram() implementation. Does anyone knows what's wrong with my friend function declaration? Here's the code: namespace MyNamespace{ class GHistogram { public: GHistogram(qint32 numberOfBins); qint32 getNumberOfBins(); /** * Returns the frequency of the value i. */ double getValueAt(qint32 i); friend GHistogram * Gbdi::extractHistogram(GImage *image, qint32 numberOfBins); private: QVector<double> bins; }; GHistogram * extractHistogram(GImage * image, qint32 numberOfBins); }

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  • Why do I get the error "X is not a member of Y" even though X is a friend of Y?

    - by user1232138
    I am trying to write a binary tree. Why does the following code report error C2039, "'<<' : is not a member of 'btree<T'" even though the << operator has been declared as a friend function in the btree class? #include<iostream> using namespace std; template<class T> class btree { public: friend ostream& operator<<(ostream &,T); }; template<class T> ostream& btree<T>::operator<<(ostream &o,T s) { o<<s.i<<'\t'<<s.n; return o; }

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  • SQL SERVER – Solution – User Not Able to See Any User Created Object in Tables – Security and Permissions Issue

    - by pinaldave
    There is an old quote “A Picture is Worth a Thousand Words”. I believe this quote immensely. Quite often I get phone calls that something is not working if I can help. My reaction is in most of the cases, I need to know more, send me exact error or a screenshot. Until and unless I see the error or reproduce the scenario myself I prefer not to comment. Yesterday I got a similar phone call from an old friend, where he was not sure what is going on. Here is what he said. “When I try to connect to SQL Server, it lets me connect just fine as well let me open and explore the database. I noticed that I do not see any user created instances but when my colleague attempts to connect to the server, he is able to explore the database as well see all the user created tables and other objects. Can you help me fix it? “ My immediate reaction was he was facing security and permission issue. However, to make the same recommendation I suggested that he send me a screenshot of his own SSMS and his friend’s SSMS. After carefully looking at both the screenshots, I was very confident about the issue and we were able to resolve the issue. Let us reproduce the same scenario and many there is some learning for us. Issue: User not able to see user created objects First let us see the image of my friend’s SSMS screen. (Recreated on my machine) Now let us see my friend’s colleague SSMS screen. (Recreated on my machine) You can see that my friend could not see the user tables but his colleague was able to do the same for sure. Now I believed it was a permissions issue. Further to this I asked him to send me another image where I can see the various permissions of the user in the database. My friends screen My friends colleagues screen This indeed proved that my friend did not have access to the AdventureWorks database and because of the same he was not able to access the database. He did have public access which means he will have similar rights as guest access. However, their SQL Server had followed my earlier advise on having limited access for guest access, which means he was not able to see any user created objects. My next question was to validate what kind of access my friend’s colleague had. He replied that the colleague is the admin of the server. I suggested that if my friend was suppose to have admin access to the database, he should request of having admin access to his colleague. My friend promptly asked for the same to his colleague and on following screen he added him as an admin. You can do the same using following T-SQL script as well. USE [AdventureWorks2012] GO ALTER ROLE [db_owner] ADD MEMBER [testguest] GO Once my friend was admin he was able to access all the user objects just like he was expecting. Please note, this complete exercise was done on a development server. One should not play around with security on live or production server. Security is such an issue, which should be left with only senior administrator of the server. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Security, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Making one of a group of similar form fields required in CakePHP

    - by Pickledegg
    I have a bunch of name/email fields in my form like this: data[Friend][0][name] data[Friend][1][name] data[Friend][2][name] etc. and data[Friend][0][email] data[Friend][1][email] data[Friend][2][email] etc. I have a custom validation rule on each one that checks to see if the corresponding field is filled in. Ie. if data[Friend][2][name] then data[Friend][2][email] MUST be filled in. FYI, heres what one of the two rules look like: My form validation rule: ( I have an email validation too but that's irrelevant here) 'name' => array( 'checkEmail' => array( 'rule' => 'hasEmail', 'message' => 'You must fill in the name field', 'last' => true ) ) My custom rule code: function hasEmail($data){ $name = array_values($data); $name = $name[0]; if(strlen($name) == 0){ return empty($this->data['Friend']['email']); } return true; } I need to make it so that one of the pairs should be filled in as a minimum. It can be any as long as the indexes correspond. I can't figure a way, as if I set the form rule to be required or allowEmpty false, it fails on ALL empty fields. How can I check for the existence of 1 pair and if present, carry on? Also, I need to strip out all of the remaining empty [Friend] fields, so my saveAll() doesn't save a load of empty rows, but I think I can handle that part using extract in my controller. The main problem is this validation. Thanks.

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  • Can I do this in only one query ?

    - by Paté
    Merry christmas everyone, I Know my way around SQL but I'm having a hard time figuring this one out. First here are my tables (examples) User id name friend from //userid to //userid If user 1 is friend with user 10 then you a row with 1,10. User 1 cannot be friend with user 10 if user 10 is not friend with user 1 so you have 1,10 10,1 It may look weird but I need those two rows per relations. Now I'm trying to make a query to select the users that have the most mutual friend with a given user. For example User 1 is friend with user 10,9 and 7 and user 8 is friend with 10,9 and 7 too ,I want to suggest user 1 to invite him (like facebook). I want to get like the 10 first people with the most mutual friend. The output would be like User,NumOfMutualFriends I dont know if that can be done in a single query ? Thanks in advance for any help.

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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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  • Customer Support Spotlight: Clemson University

    - by cwarticki
    I've begun a Customer Support Spotlight series that highlights our wonderful customers and Oracle loyalists.  A week ago I visited Clemson University.  As I travel to visit and educate our customers, I provide many useful tips/tricks and support best practices (as found on my blog and twitter). Most of all, I always discover an Oracle gem who deserves recognition for their hard work and advocacy. Meet George Manley.  George is a Storage Engineer who has worked in Clemson's Data Center all through college, partially in the Hardware Architecture group and partially in the Storage group. George and the rest of the Storage Team work with most all of the storage technologies that they have here at Clemson. This includes a wide array of different vendors' disk arrays, with the most of them being Oracle/Sun 2540's.  He also works with SAM/QFS, ACSLS, and our SL8500 Tape Libraries (all three Oracle/Sun products). (pictured L to R, Matt Schoger (Oracle), Mark Flores (Oracle) and George Manley) George was kind enough to take us for a data center tour.  It was amazing.  I rarely get to see the inside of data centers, and this one was massive. Clemson Computing and Information Technology’s physical resources include the main data center located in the Information Technology Center at the Innovation Campus and Technology Park. The core of Clemson’s computing infrastructure, the data center has 21,000 sq ft of raised floor and is powered by a 14MW substation. The ITC power capacity is 4.5MW.  The data center is the home of both enterprise and HPC systems, and is staffed by CCIT staff on a 24 hour basis from a state of the art network operations center within the ITC. A smaller business continuance data center is located on the main campus.  The data center serves a wide variety of purposes including HPC (supercomputing) resources which are shared with other Universities throughout the state, the state's medicaid processing system, and nearly all other needs for Clemson University. Yes, that's no typo (14,256 cores and 37TB of memory!!! Thanks for the tour George and thank you very much for your time.  The tour was fantastic. I enjoyed getting to know your team and I look forward to many successes from Clemson using Oracle products. -Chris WartickiGlobal Customer Management

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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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  • Start Firefox from Terminal on Mac OS X (Snow Leopard)

    - by George Francis
    This is what I get when I try to start the executable: Mac-Pro:MacOS george$ /Applications/Firefox.app/Contents/MacOS/firefox dyld: Library not loaded: /usr/lib/libsqlite3.dylib Referenced from: /System/Library/Frameworks/Security.framework/Versions/A/Security Reason: Incompatible library version: Security requires version 9.0.0 or later, but libsqlite3.dylib provides version 1.0.0 /Applications/Firefox.app/Contents/MacOS/run-mozilla.sh: line 131: 9870 Trace/BPT trap "$prog" ${1+"$@"} I also tried: MacOS george$ open -a /Applications/Firefox FSPathMakeRef(/Applications/Firefox) failed with error -43. Is there a particular way to start the application from the command line?

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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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