以编程方式正从用户名UID和GID在Unix的?(Programmatically getting

2019-07-28 21:33发布

我想使用的setuid()和setgid()设置程序的各个ID的从根放弃特权下来,而是利用他们,我需要知道我要改变用户的UID和GID。

是否有一个系统调用来做到这一点? 我不想硬编码或从/ etc / passwd文件解析。

此外,我想这是通过编程而不是使用:

ID -u USERNAME

任何帮助将不胜感激

Answer 1:

看一看在getpwnam()和getgrnam()函数。



Answer 2:

你想用的系统调用的getpw *系列,一般在pwd.h 。 它本质上是一个C级接口在/ etc / passwd文件中的信息。



Answer 3:

#include <sys/types.h>
#include <pwd.h>
#include <stdlib.h>
#include <unistd.h>
#include <stdio.h>

int main()
{
    char *username = ...

    struct passwd *pwd = calloc(1, sizeof(struct passwd));
    if(pwd == NULL)
    {
        fprintf(stderr, "Failed to allocate struct passwd for getpwnam_r.\n");
        exit(1);
    }
    size_t buffer_len = sysconf(_SC_GETPW_R_SIZE_MAX) * sizeof(char);
    char *buffer = malloc(buffer_len);
    if(buffer == NULL)
    {
        fprintf(stderr, "Failed to allocate buffer for getpwnam_r.\n");
        exit(2);
    }
    getpwnam_r(username, pwd, buffer, buffer_len, &pwd);
    if(pwd == NULL)
    {
        fprintf(stderr, "getpwnam_r failed to find requested entry.\n");
        exit(3);
    }
    printf("uid: %d\n", pwd->pw_uid);
    printf("gid: %d\n", pwd->pw_gid);

    free(pwd);
    free(buffer);

    return 0;
}


Answer 4:

看看getpwnam和struct passwd文件。



Answer 5:

您可以使用下面的代码片段:

#include <pwd.h>
#include <grp.h>

gid_t Sandbox::getGroupIdByName(const char *name)
{
    struct group *grp = getgrnam(name); /* don't free, see getgrnam() for details */
    if(grp == NULL) {
        throw runtime_error(string("Failed to get groupId from groupname : ") + name);
    } 
    return grp->gr_gid;
}

uid_t Sandbox::getUserIdByName(const char *name)
{
    struct passwd *pwd = getpwnam(name); /* don't free, see getpwnam() for details */
    if(pwd == NULL) {
        throw runtime_error(string("Failed to get userId from username : ") + name);
    } 
    return pwd->pw_uid;
}

编号: getpwnam() getgrnam()



文章来源: Programmatically getting UID and GID from username in Unix?