英文:
direct file upload to Google bucket with REST API
问题
根据你提供的代码，你想要将其包装成一个用于文件上传的 REST API。你希望文件直接上传到存储桶，而不是先上传到服务器。要实现这一点，你可以修改代码，使其接受文件的字节流而不是文件路径作为输入。以下是修改后的代码示例：
func uploadFile(bucket string, fileBytes []byte, destFilePath string) error {
os.Setenv("GOOGLE_APPLICATION_CREDENTIALS", "./credential.json")
ctx := context.Background()
client, err := storage.NewClient(ctx)
if err != nil {
return fmt.Errorf("storage.NewClient: %v", err)
}
defer client.Close()
ctx, cancel := context.WithTimeout(ctx, time.Second*50)
defer cancel()
o := client.Bucket(bucket).Object(destFilePath)
o = o.If(storage.Conditions{DoesNotExist: true})
// Upload an object with storage.Writer.
wc := o.NewWriter(ctx)
if _, err = wc.Write(fileBytes); err != nil {
return fmt.Errorf("wc.Write: %v", err)
}
if err := wc.Close(); err != nil {
return fmt.Errorf("Writer.Close: %v", err)
}
return nil
}
func main() {
bucket := "g1-mybucket-001"
fileBytes := []byte("your file content here")
destFilePath := "it_poc_test_folder/somefile.txt"
err := uploadFile(bucket, fileBytes, destFilePath)
if err != nil {
fmt.Println(fmt.Errorf("Error uploading file: %v", err))
} else {
fmt.Printf("File uploaded to %s.\n", destFilePath)
}
}
在修改后的代码中，uploadFile 函数接受文件的字节流 fileBytes 作为输入，而不是文件路径。你可以将这个函数用于处理 REST API 请求中的文件上传。
英文:
Following the Go example, I have this code to upload a file to a Google bucket:
func uploadFile(bucket string, uploadFilePath string, destFilePath string) error {
os.Setenv("GOOGLE_APPLICATION_CREDENTIALS", "./credential.json")
ctx := context.Background()
client, err := storage.NewClient(ctx)
if err != nil {
return fmt.Errorf("storage.NewClient: %v", err)
}
defer client.Close()
// Open local file.
f, err := os.Open(uploadFilePath)
if err != nil {
return fmt.Errorf("os.Open: %v", err)
}
defer f.Close()
ctx, cancel := context.WithTimeout(ctx, time.Second*50)
defer cancel()
o := client.Bucket(bucket).Object(destFilePath)
o = o.If(storage.Conditions{DoesNotExist: true})
// Upload an object with storage.Writer.
wc := o.NewWriter(ctx)
if _, err = io.Copy(wc, f); err != nil {
return fmt.Errorf("io.Copy: %v", err)
}
if err := wc.Close(); err != nil {
return fmt.Errorf("Writer.Close: %v", err)
}
return nil
}
func main() {
bucket := "g1-mybucket-001"
targetFilePath := "./somefile.txt"
destFilePath := "it_poc_test_folder/somefile.txt"
err := uploadFile(bucket, targetFilePath, destFilePath)
if err != nil {
fmt.Println(fmt.Errorf("Error uplolading file: %v", err))
} else {
fmt.Printf("%s uploaded to %s.\n", targetFilePath, destFilePath)
}
}
Now I plan to wrap it to create a REST API for file upload. I then realize that the code is for upload a local file only. How can I wrap it so that the file would go straight to the bucket, without first uploading it to the server?
答案1
得分: 1
有很多文件上传的示例，主要思路如下：
func uploadFile(w http.ResponseWriter, r *http.Request) {
// ...
r.ParseMultipartForm(10 << 20)
f, handler, err := r.FormFile("myFile")
if err != nil {
// 处理错误
}
defer f.Close()
// ...
}
所以，只需将文件上传的 `f` 主体简单地传递给云上传逻辑，就像它是一个本地文件一样：
// 相同的代码 - 不同的 `f` 来源
// 使用 storage.Writer 上传对象
wc := o.NewWriter(ctx)
if _, err = io.Copy(wc, f); err != nil {
return fmt.Errorf("io.Copy: %v", err)
}
以上是要翻译的内容。
英文:
There are plenty of file upload examples out there, the gist being:
func uploadFile(w http.ResponseWriter, r *http.Request) {
// ...
r.ParseMultipartForm(10 << 20)
f, handler, err := r.FormFile("myFile")
if err != nil {
// handle err
}
defer f.Close()
// ...
}
So just simply pass the f body of the file upload on to your cloud upload logic as if it were a local file:
// same code - different `f` source
// Upload an object with storage.Writer.
wc := o.NewWriter(ctx)
if _, err = io.Copy(wc, f); err != nil {
return fmt.Errorf("io.Copy: %v", err)
}
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