To all, 

Thanks Mike's hints on API method, I am able to access Azure Blob Storage from PowerScript.

To share some points which think will help others as well:

1. For Token, it is not either to construct the signature string, so I choosed to use SAS option, which make sense since I am coding for appliction access.

2. To get the SAS Token, access from AzureBlobStorage instance >> Shared access signature >> Shared Access Signature >> Container >> Set the valid period >> Generate SAS and connection string >> copy the SAS token

3. To use httpclient PUT method to upload a file, please replace <...> with your own name/string.

long ll_FileNum
blob blb_file
integer li_rc
HTTPClient lnv_HttpClient

string ls_endpoint, ls_url, ls_account, ls_container
string ls_file,ls_updfile, ls_encoded
integer li_ResponseStatusCode = 0
string ls_ResponseBody = ''
string ls_ResponseStatusMessage = ''
string ls_PathName, ls_FileName

if GetFileOpenName ("Open", ls_PathName, ls_FileName, "PDF") < 1 then
ls_FIle = 'c:\Appn\temp\Tax Invoice format (RM)_Brother.xlsx' //For Testing by default
end if

ls_account = '<...>'
ls_container = 'attachment'
ls_endpoint = 'https://'+ls_account+'.blob.core.windows.net/'+ls_container
ls_updfile = '/DYI/'+ls_FileName

ls_url = ls_endpoint +ls_updfile+ '?sv=2021-06-08&ss=bfqt&srt=sco&sp=rwdlacupiytfx&se=2024-01-30T13:11:46Z&st=2023-01-30T05:11:46Z&spr=https&sig=DLvA7<......>bqntRFW%3D'

ls_file = ls_PathName
ll_FileNum = FileOpen(ls_file, StreamMode!)
if ll_FileNum > 0 then
FileReadEx(ll_FileNum, blb_file)
FileClose(ll_FileNum)
else
MessageBox('File Error', 'Error Opening Filename ' + ls_file)
return
end if

// Send the request
Lnv_HttpClient = create HttpClient
Lnv_HttpClient.SetRequestHeader("Content-Type", "application/json")
Lnv_HttpClient.SetRequestHeader("x-ms-blob-type", "BlockBlob")
Lnv_HttpClient.SetRequestHeader("host", "<...>.blob.core.windows.net")

li_rc = Lnv_HttpClient.sendrequest( "PUT",ls_url,blb_file )

// obtain the response data
if li_rc = 1 then
li_ResponseStatusCode = Lnv_HttpClient.GetResponseStatusCode()
ls_ResponseStatusMessage = Lnv_HttpClient.GetResponseStatusText()

if li_ResponseStatusCode = 201 then
ls_ResponseBody= ls_PathName +' / FileName: '+ls_FileName
MessageBox('Upload Successfull', ls_ResponseBody)
else
MessageBox('Upload Failed', ls_ResponseBody)
end if
else
MessageBox('HTTP Error', 'Error calling ' + ls_endpoint + '. Return Code ' + string(li_rc))
end if

Result as follow:

Regards, JX