如何从/etc/passwd中检索给定用户名的条目? [英] How can I retrieve an entry from /etc/passwd for a given username?

问题描述

我有一个文件:/etc/passwd，我必须从该文件中选择有关作为参数传递的用户的信息(该文件包含用户和有关他们的一些信息，我必须仅选择信息和打印).

I have the file : /etc/passwd and I have to select from this file the informations about an user that is passed as an argument ( the file contains users and some informations about them and i have to select only the information and print it).

推荐答案

该问题被标记为 bash ，我可以这样做:

As this question is tagged bash, I could purpose:

getUserDetails() {
    local dir gid name pass shell uid user
    while IFS=':' read user pass uid gid name dir shell ;do
        [ "$user" = "$1" ] &&
            printf "    %-14s %s\n" User "$user" UID "$uid" GID "$gid" \
                "Full name" "$name" Directory "$dir" "Default shell" "$shell"
    done </etc/passwd
}

getUserDetails user
    User           user
    UID            1000
    GID            1000
    Full name      Linux User,,,
    Directory      /home/user
    Default shell  /bin/sh

或更多 bash 面向工具的问题

or more bash tool oriented:

declare -A UserDetail
getUserDetails() {
    local dir gid name pass shell uid user
    while IFS=':' read user pass uid gid name dir shell ;do
        [ "$user" = "$1" ] && UserDetail=( [user]=$user [name]=$name
                                           [dir]=$dir   [shell]=$shell
                                           [UID]=$uid   [GID]=$gid )
    done </etc/passwd
}

getUserDetail user
printf "The full name is: %s.\n" "${UserDetail[name]}" 
Linux User

declare -p UserDetail
declare -A UserDetail='([name]="Linux User,,," [user]="user" [GID]="1000" [shell]="/bin/sh" [dir]="/home/user" [UID]="1000" )'

paste <(printf "%s\n" "${!UserDetail[@]}") <(printf "%q\n" "${UserDetail[@]}")
name    Linux User\,\,\,
user    user
GID     1000
shell   /bin/sh
dir     /home/user
UID     1000

这种方法非常有效，它设置了一个 global 变量，而没有分叉.

This way is very efficient, it set a global variable without forks.

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