File.OpenOutput 'too many parameters'

khaleel

Member
Licensed User
Longtime User
Why do I get this error when I use jOkHttpUtils2 for File.OpenOutput Function ????
While It's documentation states it needs three parameters !

upload_2017-4-9_10-22-30.png
 
Last edited:

khaleel

Member
Licensed User
Longtime User
Whops.
I missed that it's a method.
thanks anyway !!!
 

Attachments

  • upload_2017-4-9_10-20-53.png
    upload_2017-4-9_10-20-53.png
    151.4 KB · Views: 307
Upvote 0
Top