如何在Golang中将multipart.File类型转换为* os.File

Question

I am trying to upload video to CloudFlare, but in docs they use os.Open but in my situation user uploads file via html form

Golang CloudFlare Docs: https://developers.cloudflare.com/stream/getting-started/uploading-golang/

file, err := c.FormFile("file")
if err != nil {
    log.Errorf("get file error: %s", err)
    return c.JSONStatus(http.StatusBadRequest)
}

sourceFile, err := file.Open()
if err != nil {
    log.Errorf("open file error: %s", err)
    return c.JSONStatus(http.StatusInternalServerError)
}

headers := make(http.Header)
headers.Add("X-Auth-Email", "***")
headers.Add("X-Auth-Key", "***")

config := &tus.Config{
    ChunkSize:           5 * 1024 * 1024, // Cloudflare Stream requires a minimum chunk size of 5MB.
    Resume:              false,
    OverridePatchMethod: false,
    Store:               nil,
    Header:              headers,
}

client, _ := tus.NewClient("https://api.cloudflare.com/client/v4/accounts/"+ accountID +"/media", config)

upload, _ := tus.NewUploadFromFile(sourceFile)

uploader, _ := client.CreateUpload(upload)

uploader.Upload()

Answer 1

This is actually related to the go-tus client.

Cloudflare's example creates a tus.Upload from an *os.File, but rather than trying to "convert" your multipart.File to an *os.File, I would consider the other functions go-tus provides for getting a tus.Upload.

Looking at the docs, you should consider these two:

• func NewUpload(reader io.Reader, size int64, metadata Metadata, fingerprint string) *Upload
• func NewUploadFromBytes(b []byte) *Upload

Considering multipart.File implements the io.Reader interface, you can go with both of these. Which is best depends on your use case, but if the files being uploaded have a size of more than several tens of KB's, you should really NewUpload. NewUploadFromBytes forces you to read the entire file into memory first.

In case you need some inspiration for what the size, metadata and fingerprint arguments should contain, look at the implementation of NewUploadFromFile.