无法在golang中保存多个文件

时间:2019-03-04 18:59:05

标签: go file-upload

我知道如何上传单个文件,但是现在我试图在同一个golang函数中上传多个文件。

这是我的代码:

func PhotoCreatePOST(w http.ResponseWriter, r  *http.Request) {
    var err error

r.ParseMultipartForm(32 << 20) // 32MB is the default used by FormFile
fhs := r.MultipartForm.File["files"]

var fileNames []string
var filename string
var ext string
for _, file := range fhs { //Iterate over multiple uploaded files
    if err != nil {
        log.Fatal(err)
    } else {
        dir, err := os.Getwd()
        if err != nil {
            log.Fatal(err)
        }
        ext = strings.ToLower(path.Ext(file.Filename))

        filename = path.Join(random.RandString(10) + ext)

        destFolder := "/media/photos"
        if _, err := os.Stat(destFolder); os.IsNotExist(err) {
            os.Mkdir(destFolder, 0755)
        }

        //destination Path. The string which is Saving in DB
        savePath := destFolder + "/" + filename

        err = ioutil.WriteFile(savePath, file, 0777) //<--Here is the problem
        if err != nil {
            log.Println(err)
            io.WriteString(w, err.Error())
            return
        }

    }
}

//Add file url to the slice
fileNames = append(fileNames, filename)

}

但是我得到这个错误:

cannot use file (type *multipart.FileHeader) as type []byte in argument to ioutil.WriteFile

我试图将file读成这样的字节:

b, err := ioutil.ReadFile(file) 
    if err != nil {
        fmt.Print(err)
    }

并保存b而不是file

但随后我又收到一个错误

can not read *multipart.Fileheader as string 

我该如何解决?

2 个答案:

答案 0 :(得分:0)

这大致就是您的操作方式。

fileHeaders := r.MultipartForm.File["files"]
var fileNames []string

for _, fileHeader := range fileHeaders {
    file, err := fileHeader.Open()
    if err != nil {
        return err
    }
    defer file.Close()

    // Generate the destination filename randomly. Using even the passed
    // extension is a security vulnerability unless you have a whitelist.
    dest, err := ioutil.TempFile("/media/photos", "")
    if err != nil {
        return err
    }
    defer dest.Close()

    if _, err := io.Copy(dest, file); err != nil {
        return err
    }

    fileNames = append(fileNames, dest.Name())
}

答案 1 :(得分:0)

好的,感谢this的精彩博文,我找到了答案,在这里为记录提供了完整的解决方案:

// PhotoCreatePOST saves multiple photo uploads
func PhotoCreatePOST(w http.ResponseWriter, r *http.Request) {


    err := r.ParseMultipartForm(100000)
    if err != nil {
        fmt.Println("error parsing multiplepart form", err)
        return
    }

    files := r.MultipartForm.File["files"]

    for i, _ := range files { //Iterate over multiple uploaded files

        file, err := files[i].Open()
        defer file.Close()
        if err != nil {
            fmt.Println("error opening file ", err)
            return
        }

        ext := path.Ext(files[i].Filename)

        //TODO: Verify extension is valid

        filename := GetRandomString(10) + ext

        //create destination file making sure the path is writeable.
        dst, err := os.Create("media/photos/" + filename)
        defer dst.Close()
        if err != nil {
            fmt.Println("error creating destination ", err)
            return
        }

        //copy the uploaded file to the destination file
        if _, err := io.Copy(dst, file); err != nil {
            fmt.Println("error copying file", err)
            return
        }

        fmt.Println("Image upload success: ", files[i].Filename)
    }

    fmt.Println("all are uploaded")

    PhotoCreateGET(w, r)
    return 

}

//Generate random filename (never trust user input!)
var letterRunes = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ123456789")

// GetString returns a random string
func GetRandomString(n int) string {
    rand.Seed(time.Now().UnixNano())
    b := make([]rune, n)
    for i := range b {
        b[i] = letterRunes[rand.Intn(len(letterRunes))]
    }
    return string(b)
}

上传表格:

<form method="post" action="/create/photo" enctype="multipart/form-data">
        <input type="file"  name="files"  multiple>
        <input type="hidden" name="token" value="{{.token}}">
        <button title="submit" type="submit" >Save </button>
</form>

Voila。希望它可以为其他人节省一些时间。