duanluo5096
duanluo5096
2015-03-03 22:17

如何解析Content-Disposition标头以检索filename属性?

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Using go, how can I parse the Content-Disposition header retrieved from an http HEAD request to obtain the filename of the file?

Additionally, how do I retrieve the header itself from the http HEAD response? Is something like this correct?

resp, err := http.Head("http://example.com/")
//handle error
contentDisposition := resp.Header.Get("Content-Disposition")

The mime/multipart package specifies a method on the Part type that returns the filename (called FileName), but it's not clear to me how I should construct a Part, or from what.

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1条回答

  • doumou3883 doumou3883 6年前

    You can parse the Content-Disposition header using the mime.ParseMediaType function.

    disposition, params, err := mime.ParseMediaType(`attachment;filename="foo.png"`)
    filename := params["filename"] // set to "foo.png"
    

    This will also work for Unicode file names in the header (e.g. Content-Disposition: attachment;filename*="UTF-8''fo%c3%b6.png").

    You can experiment with this here: http://play.golang.org/p/AjWbJB8vUk

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