从FileInfo打开文件

huangapple go评论87阅读模式
英文:

Opening a File from a FileInfo

问题

在golang中,如果我有一个os.FileInfo,是否有办法从中打开一个*os.File,而不需要原始路径?

假设我有这样的代码:

package main

import (
	"os"
	"path/filepath"
	"strings"
)

var files []os.FileInfo

func walker(path string, info os.FileInfo, err error) error {
	if strings.HasSuffix(info.Name(), ".txt") {
		files = append(files, info)
	}
	return nil
}

func main() {
	err := filepath.Walk("/tmp/foo", walker)
	if err != nil {
		println("Error", err)
	} else {
		for _, f := range files {
			println(f.Name())
			// 这里我们想要打开文件
		}
	}
}

是否有办法将FileInfo转换为*File?我实际上使用的代码不是基于filepath.Walk,但我确实得到了一个[]os.FileInfo切片。我仍然有根目录和文件名,但似乎在这个阶段已经丢失了任何进一步的子树信息。

英文:

In golang, if I have an os.FileInfo, is there any way to open an *os.File from that by itself without the original path?

Let's say I had something like this:

package main

import (
	"os"
	"path/filepath"
	"strings"
)

var files []os.FileInfo

func walker(path string, info os.FileInfo, err error) error {
	if strings.HasSuffix(info.Name(), ".txt") {
		files = append(files, info)
	}
	return nil
}

func main() {
	err := filepath.Walk("/tmp/foo", walker)
	if err != nil {
		println("Error", err)
	} else {
		for _, f := range files {
			println(f.Name())
            // This is where we'd like to open the file
		}
	}
}

Is there a way to convert FileInfo to * File? The code I'm actually working with isn't based on filepath.Walk; but I do get an []os.FileInfo slice back. I still have the root directory, and the file name, but it seems like any further sub-tree information has gone by this stage.

答案1

得分: 24

不。FileInfo接口根本不公开路径,而osioutil包中提供的所有方法都接受路径名作为字符串。

英文:

No. The FileInfo interface simply does not expose the path and all provided methods in the os and ioutil packages accept the pathname as a string.

答案2

得分: 10

附加信息,这是构建包括路径的文件名字符串的方法:

filename := filepath.Join(path, info.Name())
英文:

As additional info, this is how to constructor the filename string including path:

filename := filepath.Join(path, info.Name())

答案3

得分: 4

不,只有 FileInfo 是无法打开文件的。os.Open 只接受字符串作为参数。你应该始终提供文件路径或父路径,因为这是获取 FileInfo 的唯一方式。

英文:

No, a file cannot be opened with just the FileInfo. os.Open only takes a string. You should always have the path or the parent path because that is the only way to get a FileInfo.

huangapple
  • 本文由 发表于 2015年5月5日 05:55:21
  • 转载请务必保留本文链接:https://go.coder-hub.com/30040661.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定