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os和ioutil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.
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