从/ etc / passwd中提取相应的信息

时间:2014-01-20 00:21:28

标签: linux bash

我有一个名为names.txt的文件,其中包含一个名称列表。其中一些名称与/ etc / passwd(第5个字段)中的名称不对应,而某些名称则与。对于文件中具有名称的用户的名称,我要打印其用户名。例如,如果名称比尔盖茨在names.txt文件中并且此行在/etc/passwd bgates:x:23246:879:Bill Gates:/co/bgates:/bin/bash中,我会打印出“比尔盖茨存在且用户名为'bgates'”

这是我一直在尝试的,但它只打印出整个/ etc / passwd文件。

while read name; do

if cut -d: -f5 '/etc/passwd' | grep -q "$name"; then
        userName=$(cat /etc/passwd | cut -d: -f6)
        echo "$name exists and has the username $userName"
else
        echo "no such person '$line'"
fi
done < names.txt

谢谢

2 个答案:

答案 0 :(得分:3)

也许是这样的?

#!/bin/bash

#set -x
set -eu
set -o pipefail

function get_pwent_by_name
{
    full_name="$1"
    while read pwent
    do
        pw_full_name=$(echo "$pwent" | awk -F':' ' { print $5 }')
        if echo "$pw_full_name" | egrep -iq "$full_name"
        then
            echo "$pwent"
            break
        fi
    done < /etc/passwd
}

while read name
do
    pwent=$(get_pwent_by_name "$name")
    if [ "$pwent" != "" ]
    then
        userName=$(echo "$pwent" | awk -F':' ' { print $1 }')
        echo "$name exists and has the username $userName"
    else
        echo "No such person as $name"
    fi
done < names.txt

答案 1 :(得分:2)

你接受使用awk来解决你的问题吗?

awk -F: 'NR==FNR{a[$5]=$1;next} 
         {print ($0 in a)?$0 " exists and has the username " a[$0]:"no such person " $0}' /etc/passwd names.txt