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Objective: Output from Linux username, passwordlastchanged, passwordexpires in human readable format.

I pulled user info from /etc/passwd and /etc/shadow for users with shell /bin/bash

join -1 1 -2 1 -t : /etc/passwd /etc/shadow |grep bin/bash|awk -F\: '{print $1";"$9";$14}' where $1 = username $9 = number of days since last password change, calculated from 1/1_1970 $14 = number of days to password expiry date, calculated from 1/1_1970

I can get a similar number as $9 and $14 for the current date

expr $(date +%S) /86400

Question: How do I output the value of $9-$(date +%s)/86400 and $14-$(date +%s)/86400 in my current one liner?

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2 Answers 2

Reset to default
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Please be aware that field 8 in /etc/shadow does NOT contain "password expiry date" but according to man shadow (on linux):

account expiration date

The date of expiration of the account, expressed as the number of days since Jan 1, 1970. Note that an account expiration differs from a password expiration. . . An empty field means that the account will never expire.

Do you want "password expiration" or "account expiration"?

With the join in your question, all "shadow" fields will be shifted by 6 for further evaluation. The grep command is unnecessary; awk can do that. Try this adaption of your one-liner: 

join -1 1 -2 1 -t : /etc/passwd /etc/shadow | awk -F: -v"TD=$(date +%s)" 'BEGIN {TD = TD / 86400} $7 ~ /bash$/ {print $1, int($9-TD), $14?int($14-TD):"never"}' OFS=";" user1;-1429;never user2;-1429;never
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  • Ah yes: I've used the account expiration date on field 8 instead of maximum days valid in field 5. That changes the task slightly. Commented Oct 16, 2018 at 21:08
  • I'm after the number of days until password expiration or like in your example: the word 'never' if that is the case. Commented Oct 17, 2018 at 8:18
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    There's no "password expiration" field in shadow. You'd need to calculate it from "date of last password change" plus "maximum password age". Why don't you adapt the proposal and report back? Commented Oct 17, 2018 at 9:26
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Use echo as follows:

 echo "$(($9-$(date +%s)/86400 ))"

$((...)) is an arithmetic expansion.

The usage of ... in $((...)) will be interpreted as an arithmetic expression. For example, a hexadecimal string will be interpreted as a number and converted to decimal. The whole expression will then be replaced by the numeric value that the expression evaluates to.

$((...)) should be quoted as to not be affected by the shell's word splitting and filename globbing.

or

echo "$9-$(date +%s)/86400" | bc

bc stands for basic calculator and it can handle mathematical operations.

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