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[英]I need a grep command that will pull the usernames only from the /etc/passwd file in linux
[英]Pull out corresponding info from /etc/passwd
我有一個名為names.txt的文件,其中包含名稱列表。 其中一些名稱與/ etc / passwd(第5個字段)中的名稱不對應,而有些則與。 對於文件中具有用戶名的名稱,我要打印其用戶名。 例如,如果名稱Bill Gates在names.txt文件中,而該行在/etc/passwd bgates:x:23246:879:Bill Gates:/co/bgates:/bin/bash
我將打印出“ Bill Gates存在,並且用戶名“ bgates””
這是我一直在嘗試的方法,但是它只是打印出整個/ 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
謝謝
也許是這樣的嗎?
#!/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
您接受使用awk解決問題嗎?
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
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