I am creating a picture using jpeg.Encode and want to send it. How can I avoid creating an intermediate file?
Create and save file.
// ...
outFile, err := os.Create("./images/" + name + ".jpg")
if err != nil {
log.Fatal(err)
os.Exit(-1)
}
defer outFile.Close()
buff := bufio.NewWriter(outFile)
err = jpeg.Encode(buff, background, &jpeg.Options{Quality: 90})
if err != nil {
log.Fatal(err)
os.Exit(-1)
}
err = buff.Flush()
if err != nil {
log.Fatal(err)
os.Exit(-1)
}
Send file.
file, err := os.Open("./images/" + name + ".jpg")
if err != nil {
log.Fatal(err)
}
body := &bytes.Buffer{}
writer := multipart.NewWriter(body)
part, err := writer.CreateFormFile("photo", file.Name())
if err != nil {
return nil, err
}
io.Copy(part, file)
if err = writer.Close(); err != nil {
return nil, err
}
resp, err := http.Post(uploadURL, writer.FormDataContentType(), body)
if err != nil {
return nil, err
}
defer resp.Body.Close()
How can I send a photo without saving it on the server?
Simply pass the part
target to jpeg.Encode()
:
// ...
part, err := writer.CreateFormFile("photo", file.Name())
if err != nil {
return nil, err
}
err = jpeg.Encode(part, background, &jpeg.Options{Quality: 90})
if err != nil {
return nil, err
}
if err = writer.Close(); err != nil {
return nil, err
}
resp, err := http.Post(uploadURL, writer.FormDataContentType(), body)
if err != nil {
return nil, err
}
defer resp.Body.Close()