Golang http.NewRequest 总是 returns 404 - Postman ok

Question

我正在尝试编写一个 post 请求函数，将文件从 golang 服务器发送到客户端。代码从这里抄袭了一点 (golang POST data using the Content-Type multipart/form-data)。无论出于何种原因，我总是得到“404 页面未找到”，即使通过 postman 完成相同的请求时到达 endpt 并成功返回。这是奇怪的行为，我不太确定如何调试。（出于测试目的，我已经将 URL 硬编码为命中本地运行服务器。）

这是我的代码：

func PostUpload(values map[string]io.Reader, URL string){
    fmt.Println("inside PostUpload in outbound")
    client := &http.Client{}
    var err error

    var b bytes.Buffer
    w := multipart.NewWriter(&b)
    for key, r := range values {
            var fw io.Writer
            if x, ok := r.(io.Closer); ok {
                    defer x.Close()
            }
            // Add an image file
            if x, ok := r.(*os.File); ok {
                    if fw, err = w.CreateFormFile(key, x.Name()); err != nil {
                            return
                    }
            } else {
                    // Add other fields
                    if fw, err = w.CreateFormField(key); err != nil {
                            return
                    }
            }
            if _, err = io.Copy(fw, r); err != nil {
                fmt.Println("there was an error")
                fmt.Println(err)
            }
        }
    w.Close()

    // Now that you have a form, you can submit it to your handler.
        // req, err := http.NewRequest("POST", URL, &b)
        req, err := http.NewRequest("POST", "http://127.0.0.1:5000/upload", &b)
    if err != nil {
                fmt.Println("there was an error on http.NewRequest in outbound post file upload")
                fmt.Println(err)
    }
    // Don't forget to set the content type, this will contain the boundary.
        req.Header.Set("Content-Type", w.FormDataContentType())


        requestDump, err := httputil.DumpRequest(req, true)
        if err != nil {
            fmt.Println(err)
        }
        fmt.Println(string(requestDump))


    // Submit the request
    resp, err := client.Do(req)
    if err != nil {
                fmt.Println("there was an error on client do in outbound post file upload")
                fmt.Println(err)
        }

        resp_body, _ := ioutil.ReadAll(resp.Body)
        var parsed map[string]interface{}
        err = json.Unmarshal(resp_body, &parsed)
        if err!=nil{
            if typeof(resp_body)=="[]uint8"{
                fmt.Println("received []uint8, converting and displaying")
                fmt.Print(string(resp_body))
            }else{
                fmt.Println("inside unmarshal error")
                fmt.Println(err)
            }
        }

    return
}

这是控制台中的输出：

api      | 21:16:09 app         | inside PostUpload in outbound
api      | 21:16:09 app         | POST /upload HTTP/1.1
api      | Host: 127.0.0.1:5000
api      | Content-Type: multipart/form-data; boundary=0ce8aee17e5d00524acd3875d8b4b41dba5f99d8a7796d56289e38f89c64
api      |
api      | --0ce8aee17e5d00524acd3875d8b4b41dba5f99d8a7796d56289e38f89c64
api      | Content-Disposition: form-data; name="file"; filename="/uploads/TIME^2019-01-26 21:16:09.413514444 +0000 UTCFILE^Any_Publishing.txt"
api      | Content-Type: application/octet-stream
api      |
api      | # © 2016 and later: Unicode, Inc. and others.
api      | # License & terms of use: http://www.unicode.org/copyright.html#License
api      | #
api      | # File: Any_Publishing.txt
api      | # Generated from CLDR
api      | #
api      |
api      | # Variables
api      | $single = \' ;
api      | $space = ' ' ;
api      | $double = \";
api      | $back = \` ;
api      |
api      | --0ce8aee17e5d00524acd3875d8b4b41dba5f99d8a7796d56289e38f89c64--
api      |
api      | 21:16:09 app         | received []uint8, converting and displaying
api      | 404 page not found

有人看到哪里出了问题吗？

编辑：

有人建议将 req 上的转储注释掉，因为它可能会消耗 req 而将请求留空。我试过了，结果和上面一样。

有人提到他们想查看上传成功的 curl 请求，所以这里是：

patientplatypus:~/Downloads:15:48:28$curl -X POST -F "file=@./catpic.jpg" http://127.0.0.1:5000/upload
file uploaded successfully!

带有 -v 标志的 curl 的完整转储：

patientplatypus:~/Downloads:15:48:42$curl -v -X POST -F "file=@./catpic.jpg" http://127.0.0.1:5000/upload
Note: Unnecessary use of -X or --request, POST is already inferred.
*   Trying 127.0.0.1...
* TCP_NODELAY set
* Connected to 127.0.0.1 (127.0.0.1) port 5000 (#0)
> POST /upload HTTP/1.1
> Host: 127.0.0.1:5000
> User-Agent: curl/7.54.0
> Accept: */*
> Content-Length: 264915
> Expect: 100-continue
> Content-Type: multipart/form-data; boundary=------------------------48411aa948b523e2
>
< HTTP/1.1 100 Continue
* HTTP 1.0, assume close after body
< HTTP/1.0 200 OK
< Content-Type: text/html; charset=utf-8
< Content-Length: 27
< Server: Werkzeug/0.14.1 Python/3.6.5
< Date: Sat, 26 Jan 2019 22:12:19 GMT
<
* Closing connection 0
file uploaded successfully!

有人提到他们想看看我试图攻击的服务器，所以这里是：https://github.com/patientplatypus/pythonServer

Answer 1

从表面上看，您没有提出相同的请求。 Curl（感谢包括在内）设置一个表单值，而你的 go 设置 2。在你的 curl 中它是 'file' 但在 go 中，'filename'。我在输出的内容配置行中看到了这一点。使用 -v 卷曲还可能包含有关它发送 f 的内容的更多信息。

Answer 2

该 404 错误字符串看起来与 Go 标准库生成的字符串相同：

https://github.com/golang/go/blob/902fc114272978a40d2e65c2510a18e870077559/src/net/http/fs.go#L630

你确定你真的在点击 Flask 应用程序吗？

Golang http.NewRequest 总是 returns 404 - Postman ok

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