help request: apisix How to directly return the status code through response rewrite;

[susugo]

Description

The return operation similar to nginx

        location /status/ {
            return 200 ;
        }

Environment

• APISIX version (run apisix version):
• Operating system (run uname -a):
• OpenResty / Nginx version (run openresty -V or nginx -V):
• etcd version, if relevant (run curl http://127.0.0.1:9090/v1/server_info):
• APISIX Dashboard version, if relevant:
• Plugin runner version, for issues related to plugin runners:
• LuaRocks version, for installation issues (run luarocks --version):

[mscb402]

Do you mean you want return status code using response rewrite plugin?
here is a example.

"response-rewrite": {
        "status_code":200,
        "body": ""
}

And this is document you can read https://apisix.apache.org/docs/apisix/plugins/response-rewrite/

[susugo]

• This is how I fulfilled my needs
• You must configure an upstream or you cannot obtain data

    "response-rewrite": {
      "_meta": {
        "disable": false
      },
      "body": "code: ok",
      "headers": {
        "Access-Control-Allow-Headers": "*",
        "Access-Control-Allow-Origin": "*"
      },
      "status_code": 200
    }
  },
  "upstream": {
    "nodes": [{
      "host": "127.0.0.1",
      "port": 80,
      "weight": 100
    }],

[jiangfucheng]

Maybe you can use fault-injection pugin. https://apisix.apache.org/docs/apisix/plugins/fault-injection/

[shreemaan-abhishek]

@jinjianming I am closing this issue, you can reopen it if you have more questions.