Question

How remove two specific param from url NGIX

Hello, i have this url xxxxx.com/xxxx/xx-xx-2/xxxx and want remove the “-2” from second parameter in the url and that it stays that way xxxxx.com/xxxx/xx-xx/xxxx, The “-2” is random. i don’t have idea how to do it :( any help would appreciate it.

My apologies for my bad english


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This can be accomplished using an Nginx rewrite rule. These use regular expressions to match different parts of the URL and then “rewite” them.

There will be differences based on whether xxxx is alaways xxxx or if you also want yyyy to match, but here is a minimal working example as a reference point:

server {
    listen 80 default_server;
    listen [::]:80 default_server ipv6only=on;
    server_name xxxxx.com;

    root /var/www/html/;
    index index.html index.htm;

    rewrite (xxxx/.*)-\d+(.*)$ /$1$2 last;
}

Breaking down the regex a bit:

  • (xxxx/.*) captures every thing highlighted in red: xxxx/xx-xx-2/xxxx
  • -\d+ matches on: xxxx/xx-xx-2/xxxx
  • (.*)$ captures every thing that follows: xxxx/xx-xx-2/xxxx

Then /$1$2 takes the first captured part ($1) and appends the second ($2), leaving you with: xxxx/xx-xx/xxxx

regex101.com is a great tool for testing and explaining how different regular expressions work. Check out this one, and modify it as needed for your real use case: