fix CSRF checking so that localhost always works

main
Brian Picciano 2 years ago
parent f7b4fd644b
commit 98e0c3e89c
  1. 31
      srv/src/http/csrf.go

@ -2,23 +2,38 @@ package http
import (
"errors"
"net"
"net/http"
"net/url"
"github.com/mediocregopher/blog.mediocregopher.com/srv/http/apiutil"
)
func (a *api) checkCSRFMiddleware(h http.Handler) http.Handler {
return http.HandlerFunc(func(rw http.ResponseWriter, r *http.Request) {
func checkCSRF(r *http.Request, publicURL *url.URL) error {
refererURL, err := url.Parse(r.Referer())
if err != nil {
apiutil.BadRequest(rw, r, errors.New("invalid Referer"))
return
if ipStr, _, err := net.SplitHostPort(r.Host); err == nil {
if ip := net.ParseIP(ipStr); ip != nil && ip.IsLoopback() {
return nil
}
}
refererURL, err := url.Parse(r.Referer())
if err != nil {
return errors.New("invalid Referer")
}
if refererURL.Scheme != publicURL.Scheme ||
refererURL.Host != publicURL.Host {
return errors.New("invalid Referer")
}
return nil
}
func (a *api) checkCSRFMiddleware(h http.Handler) http.Handler {
return http.HandlerFunc(func(rw http.ResponseWriter, r *http.Request) {
if refererURL.Scheme != a.params.PublicURL.Scheme ||
refererURL.Host != a.params.PublicURL.Host {
if err := checkCSRF(r, a.params.PublicURL); err != nil {
apiutil.BadRequest(rw, r, errors.New("invalid Referer"))
return
}

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