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How to Find: last login information in terms of year-month-date

Moin,

I'm not sure, if I understood your question. Do you mean this?

Code:

last | cut -c1-9,40-56 | while read u d; do date -d "$d" +"$u %Y-%m-%d"; done

Jan

thanks for answer.

can you tell me script to find the diff between 2 dates.

like
d1=2009-23-03
d2=2008-22-78

thanks
avklinux

Quote:

Originally Posted by avklinux (Post 3459969)

can you tell me script to find the diff between 2 dates.

Script it such that the date is presented in UNIX time format, then do subtraction.

Example of UNIX time output:

Code:

date -u +%s

BTW, I'm moving this to Linux - General. Good luck.

can u explain in terms of d1 and d2.

means d1=1236430420 and d2=1236430460

then how do i script it ...???

Code:

d1=1236430420

d2=1236430460

echo "scale=5; ${d1}-${d2}"|bc

Moin,

Quote:

Originally Posted by avklinux (Post 3459969)

can you tell me script to find the diff between 2 dates.

like
d1=2009-23-03
d2=2008-22-78

d2 is a strange date, isn't it? ;-) I'll use the standard format yyyy-mm-dd for an example:

Code:

jan@jack:~/tmp> d1=2009-03-23

jan@jack:~/tmp> d2=2008-08-22

jan@jack:~/tmp> expr `date -d "$d1" +%s` - `date -d "$d2" +%s`

18406800 # seconds

jan@jack:~/tmp> expr \( `date -d "$d1" +%s` - `date -d "$d2" +%s` \) / 60

306780 # minutes

jan@jack:~/tmp> expr \( `date -d "$d1" +%s` - `date -d "$d2" +%s` \) / 3600

5113 # hours

jan@jack:~/tmp> expr \( `date -d "$d1" +%s` - `date -d "$d2" +%s` \) / 86400

213 # days

Jan

The Bash script below does basically what jan61's post above does. Input dates in the same format to get total days, hours, minutes between two dates.

Code:

#!/bin/bash

read -p "Date1: " d1 

read -p "Date2: " d2

d3=$(date --date "$d1" +%s)

d4=$(date --date "$d2" +%s)

d5=$(($d4 - $d3))

printf "Days: "

echo "scale=0; (${d5} / 86400) " | bc 

printf "Hours: " 

echo "scale=0; ((${d5} / 86400) * 24) " | bc 

printf "Minutes: " 

echo "scale=0; ((${d5} /86400) * (24) * 60)" | bc

The bash script below here takes same input format to seconds and produces the difference between two dates not just total days, hours, minutes. Example enter date: 2009-03-03 02:30:15, and second date:
2009-03-05 10:45:30, get output:

Days: 2
Hours: 8
Minutes: 15
Seconds: 15

Code:

#!/bin/bash

read -p "Date1: " d1 

read -p "Date2: " d2

d3=$(date --date "$d1" +%s)

d4=$(date --date "$d2" +%s)

d5=$(($d4 - $d3))

printf "Days: "

echo "scale=0; (${d5} / 86400) " | bc 

printf "Hours: " 

echo "scale=0; ((${d5} % 86400) / 3600) " | bc

printf "Minutes: "

echo "scale=0; ((${d5} % 86400) % 3600) / 60 " | bc

printf "Seconds: "

echo "scale=0; ((${d5} % 86400) % 3600) % 60 % 60" | bc

First actual Bash scripts I've done so I imagine there are easier ways to do it. They work! Been playing around trying to get what I wanted to work here for so long I forgot why I was doing it.