Sorry, I dont understand.
I have a BufferedImage type variable in my .tml can not show, but I can
convert into a variable of type InputStream.
Suppose I call that variable myImage
I was looking at that page but do not quite understand. I'm supposed to use
the last paragraph "Creating a page for streaming files" but I have some
questions.
1. function getuploadfile should return Link instead of a string?
2. I do not understand that I would have to put in the variable "final
String uuid" or in this call ".... getUploadedFile (" SOMEUUID ")"
3. At these points would have to enter some code? return new StreamResponse
() {... }
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