Do not make any assumption on the file name when parsing

The parser should act generically, and be able to parse
any file it is provided, regardless of name and file extension.

Signed-off-by: Armel Soro <[email protected]>

Author: rm3l

pkg/devfile/parser/context/context.go

@@ -16,11 +16,7 @@
 package parser
 
 import (
-	"fmt"
 	"net/url"
-	"os"
-	"path/filepath"
-	"strings"
 
 	"github.com/devfile/library/v2/pkg/testingutil/filesystem"
 	"github.com/devfile/library/v2/pkg/util"
@@ -96,17 +92,6 @@ func (d *DevfileCtx) populateDevfile() (err error) {
 
 // Populate fills the DevfileCtx struct with relevant context info
 func (d *DevfileCtx) Populate() (err error) {
-	if !strings.HasSuffix(d.relPath, ".yaml") {
-		if _, err := os.Stat(filepath.Join(d.relPath, "devfile.yaml")); os.IsNotExist(err) {
-			if _, err := os.Stat(filepath.Join(d.relPath, ".devfile.yaml")); os.IsNotExist(err) {
-				return fmt.Errorf("the provided path is not a valid yaml filepath, and devfile.yaml or .devfile.yaml not found in the provided path : %s", d.relPath)
-			} else {
-				d.relPath = filepath.Join(d.relPath, ".devfile.yaml")
-			}
-		} else {
-			d.relPath = filepath.Join(d.relPath, "devfile.yaml")
-		}
-	}
 	if err := d.SetAbsPath(); err != nil {
 		return err
 	}