Search Results

Search found 53 results on 3 pages for 'laravel 4'.

Page 1/3 | 1 2 3  | Next Page >

• Laravel Passing variable not working

  - by Friend

  Hello People here is my code i have used in controller... public function bulk() { return View::make('bulk')->with('message','hii there'); } my route file contains... Route::get('bulk',array('uses'=>'HomeController@bulk'))->before('auth'); In my view Iam testing it by ... @if(Session::has('message')) Present @else not Present @endif The page is making a view with the message 'not Present' why is it?? I even tried return Redirect::to('bulk')->with('message','hii there'); I get an erro mesage on Console mypro/public/bulk net::ERR_TOO_MANY_REDIRECTS What could be the problem?? is there any issues with name?? I tried this method earlier which worked fine for me.... :( Iam using Blade Template..

  Read the article

• laravel multiple where clauses within a loop

  - by user1424508

  Pretty much I want the query to select all records of users that are 25 years old AND are either between 150-170cm OR 190-200cm. I have this query written down below. However the problem is it keeps getting 25 year olds OR people who are 190-200cm instead of 25 year olds that are 150-170 OR 25 year olds that 190-200cm tall. How can I fix this? thanks $heightarray=array(array(150,170),array(190,200)); $user->where('age',25); for($i=0;$i<count($heightarray);i++){ if($i==0){ $user->whereBetween('height',$heightarray[$i]) }else{ $user->orWhereBetween('height',$heightarray[$i]) } } $user->get(); Edit: I tried advanced wheres (http://laravel.com/docs/queries#advanced-wheres) and it doesn't work for me as I cannot pass the $heightarray parameter into the closure. from laravel documentation DB::table('users') ->where('name', '=', 'John') ->orWhere(function($query) { $query->where('votes', '>', 100) ->where('title', '<>', 'Admin'); }) ->get();

  Read the article

• Pattern filters in Laravel 4

  - by ali A

  I want to make a pattern route, to redirect users to login page when they are not logged in. I searched but couldn't find a solution. as always Laravel's Documentation is useless! I have this in my filter.php Route::filter('auth', function() { if (Auth::guest()) return Redirect::guest('login'); }); Route::filter('auth.basic', function() { return Auth::basic(); }); And this route in my routes.php Route::when('/*', 'auth' ); but It's not working. How can I do that?

  Read the article

• Laravel 4 Aliases in custom classes

  - by Leonel Franchelli

  i want to use the alias classes on laravel 4 "facades" like App::method , Config::method. Well the thing is that i create a custom class and i have to import the namespaces like <?php namespace Face\SocialHandlers; use Illuminate\Support\Facades\App; use Illuminate\Support\Facades\Config; class FacebookHandler implements SocialHandlerInterface { public function registrar($perfil) { Config::get('facebook'); } } is there any way to use those classes like in controllers or routes files of the framework ? like <?php namespace Face\SocialHandlers; //use Illuminate\Support\Facades\App; //use Illuminate\Support\Facades\Config; class FacebookHandler implements SocialHandlerInterface { public function registrar($perfil) { Config::get('facebook'); } } Cya ps: sry for my english

  Read the article

• Laravel - Paginate and get()

  - by Bajongskie

  With the code below, what I wanted was paginate the query I created. But, when I try to add paginate after get, it throws an error. I wanted to remain get since I want to limit to columns that was set on $fields. What would should be the better idea to paginate this thing? or what's a good substitute for get and limit the columns? What I tried: ->get($this->fields)->paginate($this->limit) Part of my controller: class PhonesController extends BaseController { protected $limit = 5; protected $fields = array('Phones.*','manufacturers.name as manufacturer'); /** * Display a listing of the resource. * * @return Response */ public function index() { if (Request::query("str")) { $phones = Phone::where("model", 'LIKE', '%'. Request::query('str') . '%') ->join('manufacturers', 'manufacturers_id', '=', 'manufacturers.id') ->get($this->fields); } else { $phones = Phone::join('manufacturers', 'manufacturers_id', '=', 'manufacturers.id') ->get($this->fields); } return View::make('phones.index')->with('phones', $phones); } }

  Read the article

• Laravel: validation for unique username issue

  - by T2theC

  I have a noodle scratcher that I can't seem to figure out. I am validating for unique usernames and emails addresses. I have the following rules: protected $rules = array( 'username' => 'required|min:4|unique:users,username', 'email' => 'email|unique:users,email', 'password' => 'min:6', ); If I update a form that has the username (which is the same as it currently is), I get a The username has already been taken. error. If I do the same with the email, I get no error at all. I know about adding the last param to the unique validation: 'unique:users,username,{{$id}}' or 'unique:users,username,'.$id This doesn't work on the username. I get the same issue. I can't figure out why this works for the email, but not the username. Anyone had a similar issue or know where I am going wrong? Many thanks,

  Read the article

• Laravel check if id exists?

  - by devt204

  I've two columns in contents tables 1. id 2. content now this is what i'm trying to do Route::post('save', function() { $editor_content=Input::get('editor_content'); $rules = array('editor_content' => 'required'); $validator= Validator::make(Input::all(), $rules); if($validator->passes()) { //1. check if id is submitted? //2. if id exists update content table //3. else insert new content //create new instance $content= new Content; // insert the content to content column $content->content = $editor_content; //save the content $content->save(); // check if content has id $id=$content->id; return Response::json(array('success' => 'sucessfully saved', 'id' => $id)); } if($validator->fails()) { return $validator->messages() ; } }); i wanted to check if id has been already submit or checked i'm processing the request via ajax, and if id exists i wanted update the content column and if it doesn't i wanted to create new instance how do i do it ?

  Read the article

• Configure clean URLs using Laravel using a rewrite rule to index.php

  - by yannis hristofakis

  Recently I've started learning Laravel , I have none experience with framework before. I'm encountering the following problem .I'm trying to configure the .htaccess file so I can have clean URLs but the only thing I get are 404 Not Found error pages. I have created a virtual host - you can see below the configuration file - and changed the .htaccesss file on the public directory. /etc/apache2/sites-available <VirtualHost *:80> ServerAdmin [email protected] ServerName laravel.lar DocumentRoot "/home/giannis/Desktop/laravel/public" <Directory "/home/giannis/Desktop/laravel/public"> Options Indexes FollowSymLinks MultiViews AllowOverride All </Directory> <Directory /var/www/> Options Indexes FollowSymLinks MultiViews AllowOverride None Order allow,deny allow from all </Directory> ScriptAlias /cgi-bin/ /usr/lib/cgi-bin/ <Directory "/usr/lib/cgi-bin"> AllowOverride None Options +ExecCGI -MultiViews +SymLinksIfOwnerMatch Order allow,deny Allow from all </Directory> ErrorLog ${APACHE_LOG_DIR}/error.log # Possible values include: debug, info, notice, warn, error, crit, # alert, emerg. LogLevel warn CustomLog ${APACHE_LOG_DIR}/access.log combined Alias /doc/ "/usr/share/doc/" <Directory "/usr/share/doc/"> Options Indexes MultiViews FollowSymLinks AllowOverride None Order deny,allow Deny from all Allow from 127.0.0.0/255.0.0.0 ::1/128 </Directory> </VirtualHost> .htaccesss file: laravel/public # Apache configuration file # http://httpd.apache.org/docs/current/mod/quickreference.html # Note: ".htaccess" files are an overhead for each request. This logic should # be placed in your Apache config whenever possible. # http://httpd.apache.org/docs/current/howto/htaccess.html # Turning on the rewrite engine is necessary for the following rules and # features. "+FollowSymLinks" must be enabled for this to work symbolically. <IfModule mod_rewrite.c> Options +FollowSymLinks RewriteEngine On </IfModule> # For all files not found in the file system, reroute the request to the # "index.php" front controller, keeping the query string intact <IfModule mod_rewrite.c> RewriteEngine On RewriteBase / RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule ^(.*)$ index.php?/$1 [L] </IfModule> In order to test it, I have created a view named about and made the proper routing. If I link to http://laravel.lar/index.php/about/ I'm routing to the about page instead if I link to http://laravel.lar/about/ I get a 404 Not Found error. I'm using a Debian based system.

  Read the article

• Laravel with Homestead

  - by Ahmed el-Gendy

  I new with virtual box and vagrant , Now I using Homestead image and every thing is run well but when i create my project named laravel on virtual machine it supposed that i see this new folder named laravel on my machine but i didn't get any thing on my machine , The synchronization is not working. NOTE: I'm using ubuntu 14.04 This is my homestead.yaml ip: "192.168.10.10" memory: 2048 cpus: 1 authorize: ~/.ssh/id_rsa.pub keys: - ~/.ssh/id_rsa folders: - map: /var/projects/ to: /home/vagrant/projects/ sites: - map: homestead.app to: /home/vagrant/projects/laravel/public variables: - key: APP_ENV value: local thanks advance

  Read the article

• Laravel - public layout not needed in every function

  - by fischer

  I have just recently started working with Laravel. Great framework so far! However I have a question. I am using a layout template like this: public $layout = 'layouts.private'; This is set in my Base_Controller: public function __construct(){ //Styles Asset::add('reset', 'css/reset.css'); Asset::add('main', 'css/main.css'); //Scripts Asset::add('jQuery', 'http://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js'); //Switch layout template according to the users auth credentials. if (Auth::check()) { $this -> layout = 'layouts.private'; } else { $this -> layout = 'layouts.public'; } parent::__construct(); } However I get an error exception now when I try to access functions in my diffrent controllers, which should not call any view, i.e. when a user is going to login: class Login_Controller extends Base_Controller { public $restful = true; public function post_index() { $user = new User(); $credentials = array('username' => Input::get('email'), 'password' => Input::get('password')); if (Auth::attempt($credentials)) { } else { } } } The error I get, is that I do not set the content of the different variables in my public $layout. But since no view is needed in this function, how do I tell Laravel not to include the layout in this function? The best solution that I my self have come a cross (don't know if this is a bad way?) is to unset($this -> layout); from function post_index()... To sum up my question: how do I tell Laravel not to include public $layout in certain functions, where a view is not needed? Thanks in advance, fischer

  Read the article

• Laravel 4 - Using multiple paginators on one page

  - by Fry Farnsworth

  I've been googling for this question but can't find any good answers. I found that there is a method called setPageName() for the paginator but it only works when used like Paginator::setPageName('page'). I looked up the paginator in the Laravel API (http://laravel.com/api/class-Illuminate.Pagination.Environment.html) and the method is a public function just like setBaseUrl which I am currently using like $albums->setBaseUrl('/album/'); for instance I want to have a Photo paginator (?photoPage=n) and a Comment paginator (?commentPage=n). Is there any way that I can use two paginators that use different page names in the URL on one page, or at least call the setPageName() method like setBaseUrl()?

  Read the article

• Setting up Eloquent relationships in Laravel for existing InnoDB relationships

  - by adam

  I have an initial migration that sets up two tables (users and projects), with a relationship (innoDB). $table->foreign('user_id')->references('id')->on('users'); I have two Eloquent models set up, blank except for the relationship: return $this->has_many('Project'); Do i definitely need to tell eloquent about the relationship in the models and the database? I'd assumed something as comprehensive as Laravel would infer it from the Schema? Is there something I'm missing?

  Read the article

• Avoiding exceptions when uploading files in laravel

  - by occam98

  I've got a file upload field (attachment1) in a form that may or may not have a file uploaded in it when I process the form in laravel. When I'm trying to process the page, this line generates an exception: Input::upload('attachment1',path('storage').'attachments/'.$name); Here is the text of the exception: Message: Call to a member function move() on a non-object it seems that I need to check in advance to see if 'attachment1' has a file, and I found that the function Input::has_file('attachment1') is supposed to tell me whether or not 'attachment1' has a file, but even when I submit an empty form, it returns true. Also, from reading documentation, it seems that Input::upload is supposed to just return false when trying to upload a non-existant file, so why does it produce this exception instead, and how can I fix it?

  Read the article

• best way to add route under resource in Laravel 4

  - by passingby

  I would like know if there is a better way to add additional route aside from the default of resource in Laravel 4. I have this code below which is no problem with regard to the functionality, it's just that it seems to be long: <?php Route::group(array('before' => 'auth'), function() { # API Route::group(array('prefix' => 'api'), function() { Route::resource('projects', 'ProjectsController'); Route::resource('projects.groups', 'GroupsController'); Route::post('/projects/{projects}/groups/{groups}/reorder', 'GroupsController@reorder'); }); }); If in Rails Rails.application.routes.draw do # API namespace :api, defaults: { format: 'json' } do scope module: :v1 do resources :projects do resources :groups do member do post :reorder end end end end end end

  Read the article

• Route URL laravel 4

  - by nabil l.

  How can I do this in Laravel 4 URL::route('auth.reset', array('kh34KHh4K342')) Instead of http://domain.tld/auth/reset?kh34KHh4K342 I want to get this : http://domain.tld/auth/reset/kh34KHh4K342 EDIT This my routes.php : The problem is because I have two routes with the same name? How do you explain that ? Should i set a different name event is the same kind, and different methods // Set locale $locale = Request::segment(1); if(in_array($locale, Config::get('app.languages'))) { \App::setLocale($locale); } else { $locale = null; } Route::group( array( 'prefix' => $locale ), function () { Route::get('/'), array( 'uses' => 'HomeController@getIndex', 'as' => '/' )); // Auth Route::group( array( 'prefix' => 'auth' ), function () { // Login Route::get( 'login', array( 'before' => 'guest', 'uses' => 'UserController@getLogin', 'as' => 'auth.login' ) ); Route::post( 'login', array( 'before' => 'guest|csrf', 'uses' => 'UserController@postLogin', 'as' => 'auth.login' ) ); Route::get( 'reset/{$token}', array( 'uses' => 'UserController@getReset', 'as' => 'auth.reset' ) ); Route::post( 'reset', array( 'uses' => 'UserController@postReset', 'as' => 'auth.reset' ) ); } );

  Read the article

• Passing parameter to controller from route in laravel

  - by bretterer

  Given the following route Route::get('groups/(:any)', array('as' => 'group', 'uses' => 'groups@show')); And the URL I would like to use, http://www.example.com/groups/1 I would like to be able to use the (:any) value in my controller. My controller looks like class Groups_Controller extends Base_Controller { public $restful = true; public function get_show($groupID) { return 'I am group id ' . $groupID; } } How is this possible to do? I have tried a few things including the following Route::get('groups/(:any)', array('as' => 'group', 'uses' => 'groups@show((:1))')); but it did not work. UPDATE Anytime I try to pass in the arguments as show above i get a 404 error. Thanks for the help!

  Read the article

• Laravel 4 showing all field from one to one relationship in

  - by rivai04

  I'm trying to show all field of the chosen row from one to one relationship... Route Route::get('relasi-pasien', function() { $pasien = PasienIri::where('no_ipd', '=', '100')->first(); foreach($pasien->keadaanumum as $temp) { echo'<li> Name : '.$temp->name. 'Tekdar: '.$temp->tekdar. 'Nadi : '.$temp->nadi. '</li>'; } }); Relation at PasienIri's model public function keadaanumum() { return $this->hasOne('KeadaanUmum', 'no_ipd'); } Relation at KeadaanUmum's Model public function pasieniri() { return $this->belongsTo('PasienIri', 'no_ipd'); } When I used that way, it showed error: 'Trying to get property of non-object' But if I just trying to show only one of the field, it works, showing one field: Route::get('relasi-pasien', function() { $pasien = PasienIri::where('no_ipd', '=', '100')->first(); return $pasien->keadaanumum->name; }); anyone could help me to show all field with one to one relationship or I really have to change it to one to many relationship? cause if I change it to one to many relationship, it works

  Read the article

• Creating search functionality with Laravel 4

  - by Mitch Glenn

  I am trying to create a way for users to search through all the products on a website. When they search for "burton snowboards", I only want the snowboards with the brand burton to appear in the results. But if they searched only "burton", then all products with the brand burton should appear. This is what I have attempted to write but isn't working for multiple reasons. Controller: public function search(){ $input = Input::all(); $v= Validator::make($input, Product::$rules); if($v->passes()) { $searchTerms = explode(' ', $input); $searchTermBits = array(); foreach ($searchTerms as $term) { $term = trim($term); if (!empty($term)){ $searchTermBits[] = "search LIKE '%$term%'"; } } $result = DB::table('products') ->select('*') ->whereRaw(". implode(' AND ', $searchTermBits) . ") ->get(); return View::make('layouts/search', compact('result')); } return Redirect::route('/'); } I am trying to recreate the first solution given for this stackoverflow.com problem The first problem I have identified is that i'm trying to explode the $input, but it's already an array. So i'm not sure how to go about fixing that. And the way I have written the ->whereRaw(". implode(' AND ', $searchTermBits) . "), i'm sure isn't correct. I'm not sure how to fix these problems though, any insights or solutions will be greatly appreciated.

  Read the article

• Laravel - Mail class Exception

  - by Christian Giupponi

  I need to send email within my app and this is my code: if( $agent->save() ) { //Preparo la mail da inviare con i dati di login $data = [ 'nome' => $input['nome'], 'cognome' => $input['cognome'], 'email' => $input['email'], 'password' => $input['password'] ]; //ATTENZIONE //Questo è da rimuovere in produzione, finge di inviare la mail Mail::pretend(); //Recuero il template e passo alla funzione i dati Mail::send('emails.agents.registration', $data, function($message) use ($data) { $message->to( $data['email'], $data['nome'].' '.$data['cognome'] )->subject('Benvenuto!'); }); return Redirect::action('admin.agents.index')->with('positive_flash_message', 'Agente inserito correttamente.'); } As you can see I have use the Mail::pretend to avoid the email send in development, the problem is that I get this error every time I try to send an email: Undefined property: Illuminate\Mail\Message::$email (View: /var/www/progetti/app/views/emails/agents/registration.blade.php) nd this is my blade view: Email: {{ $message->email }} Password: {{ $message->password }} What's wrong with $message?

  Read the article

• Pass database data to multiples views-Laravel

  - by user3696018

  I have a database with details of daily sales. To query a database, I have a form in a view with parameters that will query as date of admission, client and others. The result is shown in another view with the daily details of income, and below is a summary of the article do all entered. The summary I wish to transfer to another view, try to view :: composer but only transfer the empty query (I saw it with debug bar). Just appeared an empty view. How I can transfer data from the database without the latter view is empty? The second html view is totaly diferent , only the data is the same.

  Read the article

• Existing laravel 4 project gives 404 in browser

  - by Richard A

  I'm trying to set up a development environment on a virtual machine running Ubuntu 14.04 LTS using Nginx and HHVM. To do this, I followed the tutorial here. This goes well with a new installation of Laravel. But when I import an existing Laravel 4 project and try to open that on my actual machine (which will serve as the client running Windows 7), I'm getting a 404 File Not Found error on the screen while connecting to http://sav.savrichard.dev. I did add this to the hosts file with the correct IP Address. The virtual machine is receiving the request and responds with a 404 error. How do I solve this error? I'm pretty new to Ubuntu so I'm not exactly sure what's wrong. The project is located at /var/www/sav.savrichard.net The server configuration is as follow: server { listen 80 default_server; root /var/www/sav.savrichard.net/public; index index.html index.htm index.php; server_name sav.savrichard.dev; access_log /var/log/nginx/localhost.sav.savrichard.dev-access.log; error_log /var/log/nginx/localhost.sav.savrichard.dev-error.log error; charset utf-8; location / { try_files \$uri \$uri/ /index.php?\$query_string; } location = /favicon.ico { log_not_found off; access_log off; } location = /robots.txt { log_not_found off; access_log off; } error_page 404 /index.php; include hhvm.conf; # Deny .htaccess file access location ~ /\.ht { deny all; } } And the hhvm.conf file is: location ~ \.(hh|php)$ { fastcgi_keep_conn on; fastcgi_pass 127.0.0.1:9000; fastcgi_index index.php; fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name; include fastcgi_params; }

  Read the article

• Routing not working correctly using the laravel framework

  - by samayres1992

  I'm using the book wrote by one of the guys that created laravel, so I'd like to think for the most part this code isn't horribly wrong. My server is setup with nginx serving all static files and apache2 serving php. My config for each are the following: apache2: <VirtualHost *> # Host that will serve this project. ServerName litl.it # The location of our projects public directory. DocumentRoot /var/www/litl.it/laravel/public # Useful logs for debug. CustomLog /var/log/apache.access.log common ErrorLog /var/log/apache.error.log # Rewrites for pretty URLs, better not to rely on .htaccess. <Directory /var/www/litl.it/laravel/public> <IfModule mod_rewrite.c> Options -MultiViews RewriteEngine On RewriteCond %{REQUEST_FILENAME} !-f RewriteRule ^ index.php [L] </IfModule> </Directory> nginx: server { # Port that the web server will listen on. listen 80; # Host that will serve this project. server_name litl.it *.litl.it; # Useful logs for debug. access_log /var/log/nginx.access.log; error_log /var/log/nginx.error.log; rewrite_log on; # The location of our projects public directory. root /var/www/litl.it/laravel/public; # Point index to the Laravel front controller. index index.php; location / { # URLs to attempt, including pretty ones. try_files $uri $uri/ /index.php?$query_string; } # Remove trailing slash to please routing system. if (!-d $request_filename) { rewrite ^/(.+)/$ /$1 permanent; } # PHP FPM configuration. location ~* \.php$ { proxy_pass http://127.0.0.1:8080; include /etc/nginx/proxy_params; try_files index index.php $uri =404; include /etc/nginx/fastcgi_params; fastcgi_index index.php; fastcgi_param SCRIPT_FILENAME $document_root/php/$fastcgi_script_name; } # We don't need .ht files with nginx. location ~ /\.ht { deny all; } location @proxy { proxy_pass http://127.0.0.1:8080; include /etc/nginx/proxy_params; } error_page 403 /error/403.html; error_page 404 /error/404.html; error_page 405 /error/405.html; error_page 500 501 502 503 504 /error/5xx.html; location ^~ /error/ { internal; root /var/www/litl.it/lavarel/public/error; } } I'm including these server configs, as I feel this maybe the issue? Here is my incredibly basic routing file that should return "routing is working" on domain.com/test but instead it just returns the homepage. <?php Route::get('/', function() { return View::make('hello'); }); Route::get('/test', function() { return "routing is working"; }); Any ideas where I'm going wrong, I'm following this tutorial very closely and I'm confused why there is issue. Thanks!

  Read the article

• Artisan unable to access environment variables from $_ENV

  - by hansn

  Any artisan command I enter into the command line throws this error: $ php artisan <? return array( 'DB_HOSTNAME' => 'localhost', 'DB_USERNAME' => 'root', 'DB_NAME' => 'pc_booking', 'DB_PASSWORD' => 'secret', ); PHP Warning: Invalid argument supplied for foreach() in /home/martin/code/www/pc_backend/vendor/laravel/framework/src/Illuminate/Config/EnvironmentVariables.php on line 35 {"error":{"type":"ErrorException","message":"Undefined index: DB_HOSTNAME","file":"\/home\/martin\/code\/www\/pc_backend\/app\/config\/database.php","line":57}} This is only on my local development system, where I recently installed apache and php. On my production system on a shared host artisan commands work just fine. The prod system has it's own .env.php, but other than that the code should be identical. Relevant files: .env.local.php <? return array( 'DB_HOSTNAME' => 'localhost', 'DB_USERNAME' => 'root', 'DB_NAME' => 'pc_booking', 'DB_PASSWORD' => 'secret', ); app/config/database.php <?php return array( 'fetch' => PDO::FETCH_CLASS, 'default' => 'mysql', 'connections' => array( 'mysql' => array( 'driver' => 'mysql', 'host' => $_ENV['DB_HOSTNAME'], 'database' => $_ENV['DB_NAME'], 'username' => $_ENV['DB_USERNAME'], 'password' => $_ENV['DB_PASSWORD'], 'charset' => 'utf8', 'collation' => 'utf8_unicode_ci', 'prefix' => '', ), ), 'migrations' => 'migrations', ), ); The $_ENV array is populated as expected on the website - the problem appears to be with artisan only.

  Read the article

• Laravel4: Call to a member function on a non-object

  - by s0hno

  The following code will throw an error Call to a member function `links()` on a non-object routes.php: Route::get('videos', function(){ $data = DB::table('video_data_r')->paginate(5); return View::make('video',$data); }); Corresponding video view: <?php foreach($data as $item): ?> <div class="video_entry"> <a href="<?php echo $item -> url; ?>" target="_blank"><img src="<?php echo $item -> thumb; ?>" /></a> <a href="<?php echo $item -> url; ?>" target="_blank"><?php echo $item -> title; ?>"</a> </div> <?php endforeach; ?> <?php echo $data->links();?> Could you give me a good hint on what looks like a trivial error?

  Read the article

• Laravel Forge SSH on windows not recognized

  - by alysdal

  I decided to try out the Laravel Forge, but I'm having trouble with the SSH keys. I use digitalocean.com for the VMs and have used SSH keys with success many times before. However when I paste my public key into Forge (which adds it to the server) I'm not able to connect to it. I use Putty and I get the following error: Disconnected: No supported authentication methods available (server sent: publickey) Key was generated using puttygen and was added to pageant. I have tried using the very same key files as my other VM and tried creating a new one, but all without luck. I'm hoping someone can help me out here. Edit: here is my putty log http://pastebin.com/ZGmm8Mcv

  Read the article

1 2 3  | Next Page >