Bash substitution cut off end of string - bash

Question

I have bash script which should replace some placeholders in text template by users input.
#!/bin/bash
# Run this script as bash command like: bash create-apache-site.sh
read -p 'Write url without www and http prefixes: ' url
template=$(</etc/apache2/sites-available/000-default.conf)
template2=("${template/1*****/$url}")
echo "$template2" > /home/Camo/template.txt
Template file is multi-line string with placeholders (1*****, 2*****, ...) which looks like
<VirtualHost *:80>
ServerAdmin xxxx.xxxxx#gmail.com
ServerName 1*****
ErrorLog ${APACHE_LOG_DIR}/www/2*****/error.log
CustomLog ${APACHE_LOG_DIR}/www/3*****/access.log combined
DocumentRoot /var/www/html/4*****
<Directory /var/www/html/5*****/>
Options Indexes FollowSymLinks MultiViews
AllowOverride All
Order allow,deny
allow from all
Require all granted
</Directory>
RewriteEngine on
RewriteCond %{SERVER_NAME} = 6*****
RewriteRule ^ https://%{SERVER_NAME}%{REQUEST_URI} [END,NE,R=permanent]
</VirtualHost>
But the result of this script is this damaged file
<VirtualHost *:80>
ServerAdmin xxxx.xxxxx#gmail.com
ServerName ooooooooooooooo.com
As you can see substitution cut off the end of the string. Can somebody tell me please what is wrong with it? Thanks a lot.

Answer 1

You have to escape your * for the replace to work.
Maybe a better way is to not use * as your placeholder? But I leave that up to you.
Try this:
(I also made the replacement global so that you can reuse the same placeholder several times within the file)
#!/bin/bash
# Run this script as bash command like: bash create-apache-site.sh
read -p 'Write url without www and http prefixes: ' url
template=$(</etc/apache2/sites-available/000-default.conf)
template2=("${template//1\*\*\*\*\*/$url}")
echo "$template2" > /home/Camo/template.txt