S3 MultipartUpload pass contentType to meta

This commit is contained in:
Konstantin Lebedev 2021-10-12 17:14:54 +05:00
parent de1ec050e8
commit 458145425e
2 changed files with 10 additions and 0 deletions

View file

@ -38,6 +38,9 @@ func (s3a *S3ApiServer) createMultipartUpload(input *s3.CreateMultipartUploadInp
for k, v := range input.Metadata {
entry.Extended[k] = []byte(*v)
}
if input.ContentType != nil {
entry.Attributes.Mime = *input.ContentType
}
}); err != nil {
glog.Errorf("NewMultipartUpload error: %v", err)
return nil, s3err.ErrInternalError
@ -121,6 +124,9 @@ func (s3a *S3ApiServer) completeMultipartUpload(input *s3.CompleteMultipartUploa
entry.Extended[k] = v
}
}
if pentry.Attributes.Mime != "" {
entry.Attributes.Mime = pentry.Attributes.Mime
}
})
if err != nil {

View file

@ -36,6 +36,10 @@ func (s3a *S3ApiServer) NewMultipartUploadHandler(w http.ResponseWriter, r *http
createMultipartUploadInput.Metadata[k] = aws.String(string(v))
}
contentType := r.Header.Get("Content-Type")
if contentType != "" {
createMultipartUploadInput.ContentType = &contentType
}
response, errCode := s3a.createMultipartUpload(createMultipartUploadInput)
glog.V(2).Info("NewMultipartUploadHandler", string(s3err.EncodeXMLResponse(response)), errCode)