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Subject [GitHub] [openwhisk-wskdeploy] alvarolopez commented on a change in pull request #1054: Honor WSK_CONFIG_FILE if variable is set
Date Thu, 05 Sep 2019 09:14:27 GMT
alvarolopez commented on a change in pull request #1054: Honor WSK_CONFIG_FILE if variable
is set
URL: https://github.com/apache/openwhisk-wskdeploy/pull/1054#discussion_r321155044
 
 

 ##########
 File path: cmd/root.go
 ##########
 @@ -102,21 +102,32 @@ func init() {
 func initConfig() {
 	userHome := utils.GetHomeDirectory()
 	defaultPath := path.Join(userHome, whisk.DEFAULT_LOCAL_CONFIG)
-	if utils.Flags.CfgFile != "" {
 
-		// Read the file as a wskprops file, to check if it is valid.
-		_, err := whisk.ReadProps(utils.Flags.CfgFile)
-		if err != nil {
-			utils.Flags.CfgFile = defaultPath
-			warn := wski18n.T(wski18n.ID_WARN_CONFIG_INVALID_X_path_X,
-				map[string]interface{}{
-					wski18n.KEY_PATH: utils.Flags.CfgFile})
-			wskprint.PrintOpenWhiskWarning(warn)
+    // Precedence order for reading the configuration file should be:
+    // 1. --config 
+    // 2. ENV variable WSK_CONFIG_FILE
+    // 3. Default $HOME/.wskprops
+	cfgFiles := []string{
+		utils.Flags.CfgFile,
+		os.Getenv("WSK_CONFIG_FILE"),
+        defaultPath,
+	 }
+
+	for _, cfgFile := range cfgFiles {
 
 Review comment:
   Well, I am not a gopher, therefore I do not know if there is any better method to perform
it. What would you suggest to do?

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