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http - 我可以发布 Content-Type : multipart/form-data 吗

转载 作者：IT王子 更新时间：2023-10-29 01:38:39 24 4
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如何使用 Content-Type: multipart/form-data、[]byte 参数和字符串参数 POST 到 API？我试过了，但是失败了。

错误信息:

details: "[301 301 Moved Permanently]<!DOCTYPE HTML PUBLIC \"-//IETF//DTD HTML 2.0//EN\">\r\n<html>\r\n301 Moved Permanently\r\n<body bgcolor=\"white\">\r\n301 Moved Permanently\r\n<p>The requested resource has been assigned a new permanent URI.</p >\r\n<hr/>Powered by Tengine/2.1.0</body>\r\n</html>\r\n"

去代码:

func NewPost2(url string) ([]byte, error) {
    m := make(map[string]interface{}, 0)
    m["fileName"] ="good"
    m["name"] = Base64ToByte("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")
b, _ := json.Marshal(m)

    httpReq, err := http.NewRequest("POST", url, bytes.NewBuffer(b))
    httpReq.Header.Set("Content-Type", "multipart/form-data;charset=UTF-8")

    client := &http.Client{}
    resp, err := client.Do(httpReq)
    if err != nil {
        return nil, err
    }

    defer resp.Body.Close()

    if resp.StatusCode < 200 || resp.StatusCode >= 300 {
        b, _ := ioutil.ReadAll(resp.Body)
        return nil, fmt.Errorf("[%d %s]%s", resp.StatusCode, resp.Status, string(b))
    }

    respData, err := ioutil.ReadAll(resp.Body)
    if err != nil {
        return nil, err
    }

    return respData, nil
}

最佳答案

现在，我很高兴有心情分享我的解决方案

func NewPostFile(url string, paramTexts map[string]interface{}, paramFile FileItem) ([]byte, error) {
// if paramFiles ==nil {
//  return NewPost(url,paramTexts,header,transport)
// }

bodyBuf := &bytes.Buffer{}
bodyWriter := multipart.NewWriter(bodyBuf)

for k, v := range paramTexts {
    bodyWriter.WriteField(k, v.(string))
}
fileWriter, err := bodyWriter.CreateFormFile(paramFile.Key, paramFile.FileName)
if err != nil {
    fmt.Println(err)
    //fmt.Println("Create form file error: ", error)
    return nil, err
}
fileWriter.Write(paramFile.Content)
contentType := bodyWriter.FormDataContentType()
bodyWriter.Close()
fmt.Println(bodyBuf.String())

resp, err := http.Post(url, contentType, bodyBuf)
if err != nil {
    return nil, err
}
defer resp.Body.Close()
fmt.Println(resp)

if resp.StatusCode < 200 || resp.StatusCode >= 300 {
    b, _ := ioutil.ReadAll(resp.Body)
    return nil, fmt.Errorf("[%d %s]%s", resp.StatusCode, resp.Status, string(b))
}
respData, err := ioutil.ReadAll(resp.Body)
if err != nil {
    return nil, err
}
fmt.Println(string(respData))
return respData, nil

type FileItem struct {
Key      string //image_content
FileName string //test.jpg
Content  []byte //[]byte

关于http - 我可以发布 Content-Type : multipart/form-data 吗，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44302374/

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