使用REST API将文件直接上传到Google存储桶。

huangapple
go评论126阅读模式
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：

  1. func uploadFile(bucket string, fileBytes []byte, destFilePath string) error {
  2.     os.Setenv("GOOGLE_APPLICATION_CREDENTIALS", "./credential.json")
  3.     ctx := context.Background()
  4.     client, err := storage.NewClient(ctx)
  5.     if err != nil {
  6.         return fmt.Errorf("storage.NewClient: %v", err)
  7.     }
  8.     defer client.Close()
  10.     ctx, cancel := context.WithTimeout(ctx, time.Second*50)
  11.     defer cancel()
  12.     o := client.Bucket(bucket).Object(destFilePath)
  13.     o = o.If(storage.Conditions{DoesNotExist: true})
  15.     // Upload an object with storage.Writer.
  16.     wc := o.NewWriter(ctx)
  17.     if _, err = wc.Write(fileBytes); err != nil {
  18.         return fmt.Errorf("wc.Write: %v", err)
  19.     }
  20.     if err := wc.Close(); err != nil {
  21.         return fmt.Errorf("Writer.Close: %v", err)
  22.     }
  23.     return nil
  24. }
  26. func main() {
  27.     bucket := "g1-mybucket-001"
  28.     fileBytes := []byte("your file content here")
  29.     destFilePath := "it_poc_test_folder/somefile.txt"
  30.     err := uploadFile(bucket, fileBytes, destFilePath)
  31.     if err != nil {
  32.         fmt.Println(fmt.Errorf("Error uploading file: %v", err))
  33.     } else {
  34.         fmt.Printf("File uploaded to %s.\n", destFilePath)
  35.     }
  36. }

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:

  1. func uploadFile(bucket string, uploadFilePath string, destFilePath string) error {
  2.     os.Setenv("GOOGLE_APPLICATION_CREDENTIALS", "./credential.json")
  3.     ctx := context.Background()
  4.     client, err := storage.NewClient(ctx)
  5.     if err != nil {
  6.         return fmt.Errorf("storage.NewClient: %v", err)
  7.     }
  8.     defer client.Close()
  10.     // Open local file.
  11.     f, err := os.Open(uploadFilePath)
  12.     if err != nil {
  13.         return fmt.Errorf("os.Open: %v", err)
  14.     }
  15.     defer f.Close()
  17.     ctx, cancel := context.WithTimeout(ctx, time.Second*50)
  18.     defer cancel()
  19.     o := client.Bucket(bucket).Object(destFilePath)
  20.     o = o.If(storage.Conditions{DoesNotExist: true})
  21.     
  22.     // Upload an object with storage.Writer.
  23.     wc := o.NewWriter(ctx)
  24.     if _, err = io.Copy(wc, f); err != nil {
  25.         return fmt.Errorf("io.Copy: %v", err)
  26.     }
  27.     if err := wc.Close(); err != nil {
  28.         return fmt.Errorf("Writer.Close: %v", err)
  29.     }
  30.     return nil
  31. }
  33. func main() {
  34.     bucket := "g1-mybucket-001"
  35.     targetFilePath := "./somefile.txt"
  36.     destFilePath := "it_poc_test_folder/somefile.txt"
  37.     err := uploadFile(bucket, targetFilePath, destFilePath)
  38.     if err != nil {
  39.         fmt.Println(fmt.Errorf("Error uplolading file: %v", err))
  40.     } else {
  41.         fmt.Printf("%s uploaded to %s.\n", targetFilePath, destFilePath)
  42.     }
  43. }

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：

  1. func uploadFile(w http.ResponseWriter, r *http.Request) {
  3.     // ... 
  4.     r.ParseMultipartForm(10 << 20)
  5.     f, handler, err := r.FormFile("myFile")
  6.     if err != nil {
  7.         // 处理错误
  8.     }
  9.     defer f.Close()
  10.     // ...
  11. }
  13. 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：
  15. // 相同的代码 - 不同的 `f` 来源
  17. // 使用 storage.Writer 上传对象
  18. wc := o.NewWriter(ctx)
  19. if _, err = io.Copy(wc, f); err != nil {
  20.     return fmt.Errorf("io.Copy: %v", err)
  21. }

以上是要翻译的内容。

英文:

There are plenty of file upload examples out there, the gist being:

  1. func uploadFile(w http.ResponseWriter, r *http.Request) {
  3.     // ... 
  4.     r.ParseMultipartForm(10 &lt;&lt; 20)
  5.     f, handler, err := r.FormFile(&quot;myFile&quot;)
  6.     if err != nil {
  7.         // handle err
  8.     }
  9.     defer f.Close()
  10.     // ...
  11. }

So just simply pass the f body of the file upload on to your cloud upload logic as if it were a local file:

  1. // same code - different `f` source
  3. // Upload an object with storage.Writer.
  4. wc := o.NewWriter(ctx)
  5. if _, err = io.Copy(wc, f); err != nil {
  6.     return fmt.Errorf(&quot;io.Copy: %v&quot;, err)
  7. }

huangapple
  • 本文由 发表于 2023年3月24日 17:49:30
  • 转载请务必保留本文链接：https://go.coder-hub.com/75832132.html
  • go
  • google-cloud-platform
  • rest
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