Fixing top level run outside repo (#1993)

runtime/local/local.go

@@ -99,7 +99,14 @@ func (r *localRuntime) checkoutSourceIfNeeded(s *runtime.Service, secrets map[st
 			fullp := append([]string{uncompressPath}, sourceParts[1:]...)
 			s.Source = filepath.Join(append(fullp, lastFolderPart)...)
 		} else {
-			s.Source = filepath.Join(uncompressPath, tarName)
+			// The tar name is 'helloworld' for both
+			// the case when the code is uploaded from `$REPO/helloworld`
+			// and when it's uploaded from outside a repo ie `~/helloworld`.
+			if _, err := Entrypoint(filepath.Join(s.Source, tarName)); err == nil {
+				s.Source = filepath.Join(uncompressPath, tarName)
+			} else {
+				s.Source = uncompressPath
+			}
 		}
 		return nil
 	}
@@ -276,6 +283,7 @@ func (r *localRuntime) Create(s *runtime.Service, opts ...runtime.CreateOption)
 	for _, o := range opts {
 		o(&options)
 	}
+
 	err := r.checkoutSourceIfNeeded(s, options.Secrets)
 	if err != nil {
 		return err

runtime/local/service.go

@@ -3,7 +3,6 @@ package local
 import (
 	"fmt"
 	"io"
-	"path/filepath"
 	"strconv"
 	"strings"
 	"sync"
@@ -50,25 +49,6 @@ func newService(s *runtime.Service, c runtime.CreateOptions) *service {
 
 	dir := s.Source
 
-	// For uploaded packages, we upload the whole repo
-	// so the correct working directory to do a `go run .`
-	// needs to include the relative path from the repo root
-	// which is the service name.
-	//
-	// Could use a better upload check.
-	if strings.Contains(s.Source, "uploads") {
-		// There are two cases to consider here:
-		// a., if the uploaded code comes from a repo - in this case
-		// the service name is the relative path.
-		// b., if the uploaded code comes from a non repo folder -
-		// in this case the service name is the folder name.
-		// Because of this, we only append the service name to the source in
-		// case `a`
-		if ex, err := exists(filepath.Join(s.Source, s.Name)); err == nil && ex {
-			dir = filepath.Join(s.Source, s.Name)
-		}
-	}
-
 	return &service{
 		Service: s,
 		Process: new(proc.Process),