Remove syncapi debugging output

This commit is contained in:
Neil Alexander 2020-04-24 11:02:34 +01:00
parent e385a5540e
commit ff940c2b08

View file

@ -15,8 +15,6 @@
package sync package sync
import ( import (
"encoding/json"
"fmt"
"net/http" "net/http"
"time" "time"
@ -71,10 +69,6 @@ func (rp *RequestPool) OnIncomingSyncRequest(req *http.Request, device *authtype
logger.WithError(err).Error("rp.currentSyncForUser failed") logger.WithError(err).Error("rp.currentSyncForUser failed")
return jsonerror.InternalServerError() return jsonerror.InternalServerError()
} }
j, _ := json.MarshalIndent(syncData, "", " ")
fmt.Println("Sync response:", string(j))
logger.WithField("next", syncData.NextBatch).Info("Responding immediately") logger.WithField("next", syncData.NextBatch).Info("Responding immediately")
return util.JSONResponse{ return util.JSONResponse{
Code: http.StatusOK, Code: http.StatusOK,
@ -129,9 +123,6 @@ func (rp *RequestPool) OnIncomingSyncRequest(req *http.Request, device *authtype
} }
if !syncData.IsEmpty() || hasTimedOut { if !syncData.IsEmpty() || hasTimedOut {
j, _ := json.MarshalIndent(syncData, "", " ")
fmt.Println("Sync response:", string(j))
logger.WithField("next", syncData.NextBatch).WithField("timed_out", hasTimedOut).Info("Responding") logger.WithField("next", syncData.NextBatch).WithField("timed_out", hasTimedOut).Info("Responding")
return util.JSONResponse{ return util.JSONResponse{
Code: http.StatusOK, Code: http.StatusOK,