How to set Content-Type for a form filed using 'multipart' in Go

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英文:

How to set Content-Type for a form filed using 'multipart' in Go

问题

我正在尝试上传一个文件,需要为API设置特定的Content-Type。当我这样做时:

file, err := os.Open("helloWorld.wav")
buf := new(bytes.Buffer)
writer := multipart.NewWriter(buf)
audioFile, _ := writer.CreateFormFile("file", "helloWorld.wav")
_, err = io.Copy(audioFile, file)
if err != nil {
     return nil, 0, err
}
writer.Close()

它正确地创建了multipart表单,但是假设了这个内容类型:

Content-Type: application/octet-stream

我需要能够将其设置为:

Content-Type: audio/wav;rate=8000

当然,我可以为net/http设置头部,但是我不知道如何为multipart表单中的单个字段设置头部。

英文:

I am attempting to upload a file that requires me to set a specific Content-Type for the API. When I do this:

file, err := os.Open("helloWorld.wav")
buf := new(bytes.Buffer)
writer := multipart.NewWriter(buf)
audioFile, _ := writer.CreateFormFile("file", "helloWorld.wav")
_, err = io.Copy(audioFile, file)
if err != nil {
     return nil, 0, err
}
writer.Close()

It creates the multipart form properly, but assumes this content type:

Content-Type: application/octet-stream

I need to be able to set it to:

Content-Type: audio/wav;rate=8000

While of course I may set the headers for net/http, I am not seeing how to do this for individual fields in a multipart form.

答案1

得分: 18

查看源代码mime/multipart,似乎不可能实现,但你可以实现类似的功能(注意:它没有正确处理文件名的转义)。

func CreateAudioFormFile(w *multipart.Writer, filename string) (io.Writer, error) {
    h := make(textproto.MIMEHeader)
    h.Set("Content-Disposition", fmt.Sprintf(`form-data; name="%s"; filename="%s"`, "file", filename))
    h.Set("Content-Type", "audio/wav;rate=8000")
    return w.CreatePart(h)
}

输出结果:

--0c4c6b408a5a8bf7a37060e54f4febd6083fd6758fd4b3975c4e2ea93732
Content-Disposition: form-data; name="file"; filename="helloWorld.wav"
Content-Type: audio/wav;rate=8000
--0c4c6b408a5a8bf7a37060e54f4febd6083fd6758fd4b3975c4e2ea93732--

完整示例请参见playground

编辑:要写入文件数据,也将其复制到写入器中,就像原始示例中一样。

audioFile, _ := CreateAudioFormFile(writer2, "helloWorld.wav")
io.Copy(audioFile, file)

包含文件数据的完整示例请参见更新的playground

英文:

Looking at the source code mime/multipart it's not possible, but you could implement something like this (note: it's not handling the escaping of filename correctly)

    func CreateAudioFormFile(w *multipart.Writer, filename string) (io.Writer, error) {
	  h := make(textproto.MIMEHeader)
	  h.Set("Content-Disposition", fmt.Sprintf(`form-data; name="%s"; filename="%s"`, "file", filename))
	  h.Set("Content-Type", "audio/wav;rate=8000")
	  return w.CreatePart(h)
}

Outputs

>
--0c4c6b408a5a8bf7a37060e54f4febd6083fd6758fd4b3975c4e2ea93732
Content-Disposition: form-data; name="file"; filename="helloWorld.wav"
Content-Type: audio/wav;rate=8000
--0c4c6b408a5a8bf7a37060e54f4febd6083fd6758fd4b3975c4e2ea93732--

See playground for full example.

Edit: To write the file data, also copy it to the writer as in the original example.

audioFile, _ := CreateAudioFormFile(writer2, "helloWorld.wav")
io.Copy(audioFile, file)

See updated playground for the full example that includes the file data.

huangapple
  • 本文由 发表于 2014年1月15日 14:33:42
  • 转载请务必保留本文链接:https://go.coder-hub.com/21130566.html
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