I have a question about an angular application either. If I have a domain that can handle more than one application, and I navigate to each application from a specific url, for example:
www.direccion.com/aplicacion1 - > to access the application 1
www.direccion.com/aplicacion2 - > to access application 2
If I'm working with angular and I publish the project in the application 1. How can I make my route system be handled in there? I'll explain to you:
I publish the application inside the folder "aplicacion1" and I can access it as follows: www.direccion.com/aplicacion1
At this moment, it redirects me by default to the "home" component but it modifies the url and displays it as follows: www.direccion.com/home.
What I want is to be able to work on application 1, and I need the route system to be www.direccion.com/aplicacion1/home
How can I achieve that? I looked for it from several sides, but I did not find what I need, that's why I'm publishing it here.
It's on a server apache2 .
This is my virtual host. I did what they asked me to do and restart the server. But it does not work either:
<VirtualHost *:80>
ServerAdmin [email protected]
ServerName dominio.com
ServerAlias www.dominio.com
DocumentRoot /var/www/dominio.com/html
ErrorLog ${APACHE_LOG_DIR}/error.log
CustomLog ${APACHE_LOG_DIR}/access.log combined
RewriteEngine on
RewriteCond %{SERVER_NAME} =www.dominio.com [OR]
RewriteCond %{SERVER_NAME} =dominio.com
RewriteRule ^ https://%{SERVER_NAME}%{REQUEST_URI} [END,NE,R=permanent]
<Directory "/var/www/dominio.com/html">
Options Indexes FollowSymLinks MultiViews
AllowOverride All
Order allow,deny
allow from all
</Directory>
/VirtualHost> VirtualHost *:80>
ServerAdmin [email protected]
ServerName dominio.com
ServerAlias www.dominio.com
DocumentRoot /var/www/dominio.com/html/app3
ErrorLog ${APACHE_LOG_DIR}/error.log
CustomLog ${APACHE_LOG_DIR}/access.log combined
<Directory "/var/www/dominio.com/html">
Options Indexes FollowSymLinks MultiViews
AllowOverride All
Order allow,deny
allow from all
</Directory>
</VirtualHost>