I have only one variable in bash script ${PHP_V}
and trying to pass in nginx config file like:
cat <<'EOF' > /etc/nginx/sites-available/default
server {
listen 80 default_server;
listen [::]:80 default_server;
listen 443 ssl default_server;
listen [::]:443 ssl default_server;
root /vagrant/webroot;
index index.php;
server_name _;
ssl_certificate /etc/nginx/certs/vagrantbox.crt;
ssl_certificate_key /etc/nginx/certs/vagrantbox.key;
location / {
try_files $uri $uri/ /index.php?$args;
}
location ~ \.php$ {
try_files $uri =404;
include fastcgi_params;
fastcgi_pass unix:/run/php/php${PHP_V}-fpm.sock;
fastcgi_index index.php;
fastcgi_intercept_errors on;
fastcgi_param SCRIPT_FILENAME
$document_root$fastcgi_script_name;
}
}
EOF
but without success. How to do?