如何以编程方式读取linux文件权限,而不是使用“ls”然后解析结果。
答案 0 :(得分:23)
stat(2)
系统调用返回包含struct stat
成员的st_mode
。这是ls -l
显示的模式位。
在我的系统上,man 2 stat
页面显示:
The following flags are defined for the st_mode field:
S_IFMT 0170000 bitmask for the file type bitfields
S_IFSOCK 0140000 socket
S_IFLNK 0120000 symbolic link
S_IFREG 0100000 regular file
S_IFBLK 0060000 block device
S_IFDIR 0040000 directory
S_IFCHR 0020000 character device
S_IFIFO 0010000 FIFO
S_ISUID 0004000 set UID bit
S_ISGID 0002000 set-group-ID bit (see below)
S_ISVTX 0001000 sticky bit (see below)
S_IRWXU 00700 mask for file owner permissions
S_IRUSR 00400 owner has read permission
S_IWUSR 00200 owner has write permission
S_IXUSR 00100 owner has execute permission
S_IRWXG 00070 mask for group permissions
S_IRGRP 00040 group has read permission
S_IWGRP 00020 group has write permission
S_IXGRP 00010 group has execute permission
S_IRWXO 00007 mask for permissions for others (not in group)
S_IROTH 00004 others have read permission
S_IWOTH 00002 others have write permission
S_IXOTH 00001 others have execute permission
答案 1 :(得分:6)
这是C中的一个函数,它以字符串格式“rwxr-r - ”
返回文件权限char* permissions(char *file){
struct stat st;
char *modeval = malloc(sizeof(char) * 9 + 1);
if(stat(file, &st) == 0){
mode_t perm = st.st_mode;
modeval[0] = (perm & S_IRUSR) ? 'r' : '-';
modeval[1] = (perm & S_IWUSR) ? 'w' : '-';
modeval[2] = (perm & S_IXUSR) ? 'x' : '-';
modeval[3] = (perm & S_IRGRP) ? 'r' : '-';
modeval[4] = (perm & S_IWGRP) ? 'w' : '-';
modeval[5] = (perm & S_IXGRP) ? 'x' : '-';
modeval[6] = (perm & S_IROTH) ? 'r' : '-';
modeval[7] = (perm & S_IWOTH) ? 'w' : '-';
modeval[8] = (perm & S_IXOTH) ? 'x' : '-';
modeval[9] = '\0';
return modeval;
}
else{
return strerror(errno);
}
}