Golang替代python / flask send_from_directory（）

Question

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!

Answer 1

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.