如何在Go中获取文件的组ID(GID)? [英] How can I get a file's group ID (GID) in Go?

问题描述

os.Stat()返回 FileInfo 对象，该对象具有Sys()返回没有任何方法的Interface{}的方法.

os.Stat() returns a FileInfo object, which has a Sys() method that returns an Interface{} with no methods.

尽管我可以fmt.Printf()来看到""Gid"，但无法以编程方式访问"Gid".

Though I am able to fmt.Printf() it to "see" the "Gid", I am unable to access "Gid" programmatically.

如何在此处检索文件的"Gid"?

How do I retrieve the "Gid" of a file here?

file_info, _ := os.Stat(abspath)
file_sys := file_info.Sys()
fmt.Printf("File Sys() is: %+v", file_sys)

打印:

File Sys() is: &{Dev:31 Ino:5031364 Nlink:1 Mode:33060 Uid:1616 Gid:31 X__pad0:0 Rdev:0 Size:32 Blksize:32768 Blocks:0 Atim:{Sec:1564005258 Nsec:862700000} Mtim:{Sec:1563993023 Nsec:892256000} Ctim:{Sec:1563993023 Nsec:893251000} X__unused:[0 0 0]}

注意:我不需要便携式解决方案，它只需要在Linux上工作即可(值得注意的是， Sys() 被认为是片状).

Note: I do not need a portable solution, it just has to work on Linux (notable because Sys() is known to be flaky).

可能相关:转换界面{}以在Golang中进行映射

Possibly related: Convert interface{} to map in Golang

推荐答案

reflect 模块显示Sys()返回的数据类型为*syscall.Stat_t，因此这似乎可以将文件的Gid作为字符串获取:

The reflect module showed that the data type for Sys()'s return is *syscall.Stat_t, so this seems to work to get the Gid of a file as a string:

file_info, _ := os.Stat(abspath)
file_sys := file_info.Sys()
file_gid := fmt.Sprint(file_sys.(*syscall.Stat_t).Gid)

如果有更好的方法，请告诉我.

Please let me know if there is a better way to do this.

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