我已经尝试过搜索这个问题，但它与我的不同，所以我在这里发布了这个。我正在尝试创建一个使用nginx在子文件夹中托管多个laravel项目的Web服务器。这是我的实验室服务器。所以我希望我的项目是这样的：
- domain.com/project1
- domain.com/project2
- domain.com/project3
我正在为每个项目复制以下nginx location块（我不知道这里发生了什么，我只是从互联网上复制，它起了作用）：
location ^~ /project1/ {
alias /home/web/project1/public;
try_files $uri $uri/ @project1;
location ~ '.php {
fastcgi_pass unix:/var/run/php5-fpm.sock;
fastcgi_index index.php;
include /etc/nginx/fastcgi_params;
fastcgi_param SCRIPT_FILENAME "/home/web/project1/public/index.php";
}
}
location @project1 {
rewrite /avm/(.*)$ /project1/index.php?/$1 last;
}
我的laravel应用程序中的RESTful路由如下：
/*
|--------------------------------------------------------------------------
| Application Routes
|--------------------------------------------------------------------------
|
| Here is where you can register all of the routes for an application.
| It's a breeze. Simply tell Laravel the URIs it should respond to
| and give it the controller to call when that URI is requested.
|
*/
Route::get('/', ['middleware' => 'auth','uses' => 'HomeController@index'])->name('home');
// Authentication
Route::get('auth/login', 'Auth'AuthController@getLogin');
Route::post('auth/login', 'Auth'AuthController@authenticate');
Route::get('auth/logout', 'Auth'AuthController@getLogout');
// Administração
Route::group(['prefix' => 'administracao', 'middleware' => 'auth'], function() {
Route::resource('filiais', 'FiliaisController');
Route::resource('precos', 'PrecosController');
Route::resource('funcionarios', 'FuncionariosController');
Route::resource('cargos', 'CargosController');
Route::resource('vendedores', 'VendedoresController');
});
// Comercial
Route::group(['prefix' => 'comercial', 'middleware' => 'auth'], function() {
Route::resource('clientes', 'ClientesController');
Route::resource('fichas', 'FichasController');
});
// Operacional
Route::group(['prefix' => 'operacional', 'middleware' => 'auth'], function() {
Route::resource('agenda', 'AgendaController');
Route::resource('os', 'OsController');
Route::resource('ambientes', 'AmbientesController');
Route::resource('processos', 'ProcessosController');
Route::get('relatorios', 'RelatoriosController@index');
Route::group(['prefix' => 'processo', 'middleware' => 'auth'], function() {
Route::get('create', 'ProcessoController@create');
Route::get('index', 'ProcessoController@index');
Route::post('{os}/parse', 'ProcessoController@parse');
Route::get('{os}', 'ProcessoController@principal');
Route::match(['get', 'post'], '{os}/detalhe', 'ProcessoController@detalhe');
Route::get('{os}/duplicidades', 'ProcessoController@duplicidades');
Route::get('{os}/restantes', 'ProcessoController@restantes');
Route::match(['get', 'post'], '{os}/auditoria', 'ProcessoController@auditoria');
Route::match(['get', 'post'], '{os}/operadores', 'ProcessoController@operadores');
Route::match(['get', 'post'], '{os}/divergencia', 'ProcessoController@divergencia');
Route::match(['get', 'post'], '{os}/finalizar', 'ProcessoController@finalizar');
Route::get('{os}/excluir/{setor}', 'ProcessoController@destroy');
});
});
尽管当它进入业务逻辑（保存到数据库等）时，它似乎可以工作（页面出现等），但它似乎有很多错误。例如，当我试图在url http://domain.com/project1/administracao/funcionarios中创建一个新员工时，它会给我错误：SQLSTATE[42S22]: Column not found: 1054 Unknown column '/administracao/funcionarios' in（这有点准备了一些url路由）
当我设置像project1.domain.com这样的子域时，一切都很好。但我不想为每个项目创建一个子域，我希望它在子文件夹url中工作。有可能吗？
最近我遇到了完全相同的问题。我想要
- http://customerdemo.example.com/project1/
- http://customerdemo.example.com/project2/
- http://customerdemo.example.com/project3/
但我讨厌每次添加新项目时都要修改nginx-conf。
以下是我的想法：
# Capture $project from /$projectname/controller/action
map $request_uri $project {
~^/(?<captured_project>[a-zA-Z0-9_-]+)/? $captured_project;
default / ;
}
server {
listen 11.22.33.44:80;
server_name customerdemo.example.com www.customerdemo.example.com;
# Use $project/public as root
root /sites/customerdemo.example.com/$project/public;
# Use index.php as directory index
index index.php;
# Include the basic h5bp config set (see https://github.com/h5bp/server-configs-nginx)
include h5bp/basic.conf;
# Process /projectname/the/rest/of/the/url
location ~ ^/([^/]+)/(.*) {
# Save the rest of the URL after project name as $request_url
set $request_url /$2;
# If the saved url refers to a file in public folder (a static file), serve it,
# else redirect to index.php, passing along any ?var=val URL parameters
try_files $request_url /index.php?$is_args$args;
}
# Process any URL containing .php (we arrive here through previous location block)
# If you don't need to serve any other PHP files besides index.php, use location /index.php here
# instead, to prevent possible execution of user uploaded PHP code
location ~ [^/]'.php(/|$) {
# Define $fastcgi_script_name and $fastcgi_path_info
fastcgi_split_path_info ^(.+?'.php)(/.*)$;
# Immediately return 404 when script file does not exist
if (!-f $document_root$fastcgi_script_name) {
return 404;
}
# Mitigate https://httpoxy.org/ vulnerabilities
fastcgi_param HTTP_PROXY "";
# Define PHP backend location (find yours by grepping "listen ="
# from your PHP config folder, e.g. grep -r "listen =" /etc/php/)
fastcgi_pass unix:/run/php/php7.0-fpm.sock;
# Set SCRIPT_FILENAME to execute
fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
# Include the default fastcgi parameters
include fastcgi_params;
# Overwrite REQUEST_URI (default is $request_uri) with $request_url we saved earlier
fastcgi_param REQUEST_URI $request_url;
}
}
结果是，除了在/sites/customerdemo.example.com/文件夹中执行git clone之外，我不必执行任何其他操作，并运行composer install和php artisan migrate来添加新项目。
事实上，我已经为自己开发了一个控制面板，在那里我可以点击"添加项目"并指定项目详细信息和新的比特桶回购，mysql用户和数据库将被创建，customerdemo服务器将被授予对该比特桶回购的部署访问权限，并且在该新回购中设置了一个webhook，每当有人在该回购上提交master时，该webhook将调用customerdemo服务器上的部署脚本，这将触发git克隆或git pull and composer安装和数据库迁移。这就是为什么我需要动态Nginx配置
检查这个Nginx配置我相信它会帮助你
server {
server_name main-app.dev;
root /var/www/projects/main/public;
add_header X-Frame-Options "SAMEORIGIN";
add_header X-XSS-Protection "1; mode=block";
add_header X-Content-Type-Options "nosniff";
index index.html index.htm index.php;
charset utf-8;
# sub_directory
location ^~ /sub-app {
alias /var/www/projects/sub/public;
try_files $uri $uri/ @sub;
location ~ '.php {
fastcgi_split_path_info ^(.+'.php)(/.+)$;
fastcgi_pass 127.0.0.1:9000;
fastcgi_index index.php;
fastcgi_read_timeout 30000;
include fastcgi_params;
fastcgi_param SCRIPT_FILENAME /var/www/projects/sub/public/index.php;
}
access_log off;
error_log /var/www/projects/sub/storage/log/error.log error;
}
location @sub {
rewrite /sub/(.*)$ /sub/index.php?/$1 last;
} # end sub_directory
location / {
try_files $uri $uri/ /index.php?$query_string;
}
location = /favicon.ico { access_log off; log_not_found off; }
location = /robots.txt { access_log off; log_not_found off; }
access_log off;
error_log /var/www/projects/main/storage/log/error.log error;
error_page 404 /index.php;
location ~ '.php$ {
fastcgi_split_path_info ^(.+'.php)(/.+)$;
fastcgi_pass 127.0.0.1:9000;
fastcgi_index index.php;
fastcgi_read_timeout 30000;
include fastcgi_params;
fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
fastcgi_param PATH_INFO $fastcgi_path_info;
}
location ~ /'.(?!well-known).* {
deny all;
}}
我使用一个简单的符号链接在另一个网站的"子文件夹"中成功运行了一个Laravel 5.4项目。
Nginx配置中没有特别的重写规则。没有副本&粘贴项目的各个部分。路线中没有提及子文件夹。只是一个普通的Laravel 5项目整齐地包含在服务器上的某个地方，并从主站点的文档根目录中指向其公共文件夹的符号链接。
/var/www/domain.com/public/project1 --> /var/www/project1/public
所有的路线都很管用！
在编写视图时，您必须在asset()帮助器函数中包装客户端资产的路径，这样HTML中的路径将包含子文件夹，浏览器可以找到它们。
<!-- Styles -->
<link href="{{ asset('css/app.css') }}" rel="stylesheet">
但这样做并不会降低代码的灵活性，因为如果你在一个不能通过子文件夹访问网站的环境中运行网站，它会起作用，因为asset()可以处理地址栏中包含的内容。
我认为问题可能在您的nginx.conf文件中。试试这个：
location ^~ /project1 {
alias /home/web/project1/public;
try_files $uri $uri/ @project1;
location ~ '.php {
fastcgi_pass unix:/var/run/php5-fpm.sock;
fastcgi_index index.php;
include /etc/nginx/fastcgi_params;
}
}
location @project1 {
rewrite /project1/(.*)$ /project1/index.php?/$1 last;
}
此外，请确保/home/web/project1/在您的web根目录之外。
话虽如此，但确实不建议在子文件夹中运行Laravel。在子域中要容易得多。
我从这个要点中得到了这个建议的基本想法。
不完全确定解决方案，但我认为您应该尝试以下方法：
- 首先：在
config/app.php文件中正确设置url - 第二：审查
public/web.config文件。问题可能是这些配置覆盖了你的nginx。考虑更改<action type="Rewrite" url="project1/index.php" />，看看它返回了什么
在最后一个例子中，var_dump是模型，并查找它的来源。
您尝试过这种配置吗？
https://gist.github.com/tsolar/8d45ed05bcff8eb75404
我明天会测试，只要我有时间在我的环境中复制你的情况。
对于"我希望它在子文件夹url中工作。这可能吗？"有一个简单的解决方案。
步骤
1.将主域中的文件夹设为公用文件夹。您需要创建3个文件夹
主页/web/public/project1
主页/web/public/project2
主页/web/public/project3
2.在每个项目文件夹中，你需要粘贴你的laravel应用程序的公共文件夹的内容
（从您的Laravel项目）project1/public/｛contents｝--将其复制到-->（托管服务器）home/web/public/project1/{contents｝
-
将laravel项目的其余部分上传到公共根目录之外，并授予对该文件夹的写访问权限。
-
现在打开（托管服务器）public/project1/index.php更新这两个字段
需要__DIR__'./..//PROJECTONE/bootstrap/autoload.php’；
$app=需要一次__DIR __'./..//PROJECTONE/bootstrap/app.php’；
试试这样的。。。。我的服务器使用以下.conf：
server {
listen 80;
root /vagrant;
index index.html index.htm index.php app.php app_dev.php;
server_name 192.168.33.10.xip.io;
access_log /var/log/nginx/vagrant.com-access.log;
error_log /var/log/nginx/vagrant.com-error.log error;
charset utf-8;
location ~project('d*)/((.*)'.(?:css|cur|js|jpg|jpeg|gif|htc|ico|png|html|xml))$ {
rewrite project('d*)/((.*)'.(?:css|cur|js|jpg|jpeg|gif|htc|ico|png|html|xml))$ /project$1/public/$2 break;
}
location /project1{
rewrite ^/project1/(.*)$ /project1/public/index.php?$1 last;
}
location /project2 {
rewrite ^/project2/(.*)$ /project2/public/index.php?$1 last;
}
location = /favicon.ico { log_not_found off; access_log off; }
location = /robots.txt { access_log off; log_not_found off; }
error_page 404 /index.php;
location ~ '.php$ {
fastcgi_split_path_info ^(.+'.php)(/.+)$;
fastcgi_pass unix:/var/run/php5-fpm.sock;
fastcgi_index index.php;
include fastcgi_params;
set $laravel_uri $request_uri;
if ($laravel_uri ~ project('d*)(/?.*)$) {
set $laravel_uri $2;
}
fastcgi_param REQUEST_URI $laravel_uri;
fastcgi_param LARA_ENV local; # Environment variable for Laravel
fastcgi_param HTTPS off;
}
location ~ /'.ht {
deny all;
}
}