dongxi3859
dongxi3859
2017-06-03 18:09

Golang替代python / flask send_from_directory()

已采纳

I have this image urls:

/book/cover/Computer_Science.png

but the location of the image actually exists under

/uploads/img/Computer_Science.png

I'm using Gin framework. Is there any command like Flask's send_from_directory() in Gin or in in-built Golang functions?

If not could you share a snippet of how to do it?

Thanks!

  • 点赞
  • 写回答
  • 关注问题
  • 收藏
  • 复制链接分享
  • 邀请回答

1条回答

  • doubi7739 doubi7739 4年前

    Use gin's Context.File to serve file content. This method internally calls http.ServeFile builtin function. The code snippets will be:

    import "path/filepath"
    
    
    // ...
    router := gin.Default()
    // ... 
    
    router.GET("/book/cover/:filename", func(c *gin.Context) {
        rootDir := "/uploads/img/"
        name := c.Param("filename")
        filePath, err :=  filepath.Abs(rootDir + name)
        if err != nil {
            c.AbortWithStatus(404)
        }
    
        //Only allow access to file/directory under rootDir
        //The following code is for ilustration since HasPrefix is deprecated.
        //Replace with correct one when https://github.com/golang/dep/issues/296 fixed
        if !filepath.HasPrefix(filePath, rootDir) {
            c.AbortWithStatus(404)
        }
    
        c.File(filePath)
    })
    

    Update

    As pointed by zerkms, the path name must be sanitized before passing it Context.File. Simple sanitizer is added in the snippet. Please adapt to your needs.

    点赞 评论 复制链接分享