Attempt to print out the row containing the earliest/latest date - bash

Question

I'm stuck on what to do. My script will output the date/time itself, but not the whole row containing that value.
And the input looks like the following:
1 UID PID PPID C STIME TTY TIME CMD
2 adz110 5344 5334 0 08:47 pts /2 00:00:00 bash
3 dmq292 6908 6854 0 Jun04 pts /1 00:00:00 bash
4 adz110 7227 7150 0 Jul11 pts /9 00:00:00 who
5 erg474 7466 7461 0 08:54 pts /10 00:00:00 ls
6 dmq292 7966 7960 0 Jun04 pts /13 00:00:00 assign1 . sh if of
7 xle135 8983 8636 0 08:59 pts /15 00:00:00 ssh ctf . cs . utsarr . net
8 zeh458 9057 1980 0 08:59 pts /7 00:00:00 vim prog . c
9 rslavin 9150 9139 0 08:59 pts /16 00:00:00 ps - af
10 xle135 8636 8628 0 08:58 pts /15 00:00:00 bash
This seems to work for printing out "Earliest Start Time: Jun04" and "Latest Start Time: 08:59".
min = $5; next} {if ($5 > min) min = $5;}
{max = 0}{if ($5>max) max=$5}
I'm finishing up the entire script, which looks like this:
sort |
gawk '
$1 !~ /[a-z]{3}[0-9]{3}/ {
next
}
NR==0{min = $5; next} {if ($5 > min) min = $5;}
{max = 0}{if ($5>max) max=$5}
prev!=$1 && prev{
print "User: " prev ORS "\t " val
prev=val=""
}
{
prev=$1
val=(val?val ORS "\t ":"")substr($0, index($0,$8))
} ;
END {
if(prev){
print "User: " prev ORS "\t " val
}; { print "\nEarliest Start Time :\n" } { print min };
{ print "\nLatest Start Time :\n" } { print max };
}'
I thought I would want to change it to "min = $0" and "max = $0" to get the whole row containing the min and max values. But that prints out different rows instead.
The sample output should be:
1 User : adz110
2 bash
3 who
4 User : dmq292
5 bash
6 assign1 . sh if of
7 User : erg474
8 ls
9 User : xle135
10 bash
11 ssh ctf . cs . utsarr . net
12 User : zeh458
13 vim prog . c
14
15 Earliest Start Time :
16 dmq292 6908 6854 0 Jun04 pts /1 00:00:00 bash
17
18 Latest Start Time :
19 xle135 8983 8636 0 08:59 pts /15 00:00:00 ssh ctf . cs . utsarr . net
For me, earliest Start time just comes out as "Jun04" and latest start time comes out as "08:59"

Answer 1

#! /bin/bash
file=$1
awk ' BEGIN {
mon["Jan"]=1
mon["Feb"]=2
mon["Mar"]=3
mon["Apr"]=4
mon["May"]=5
mon["Jun"]=6
mon["Jul"]=7
mon["Aug"]=8
mon["Sep"]=9
mon["Oct"]=10
mon["Nov"]=11
mon["Dec"]=12
}
# main logic here
NR != 1{
# storing user info
users[$1]=users[$1]"\n\t"$9
for (i = 10 ; i <= NF ; ++i)
users[$1]=users[$1]" "$i
# min
if (min != "") {
if (compare(min, $5) == 1) {
min=$5
min_line=$0
}
} else {
min=$5
min_line=$0
}
# max
if (max != "") {
if (compare(max, $5) == -1) {
max=$5
max_line=$0
}
} else {
max=$5
max_line=$0
}
}
# printing here
END {
for (user in users) {
print "User:", user, users[user]
}
print "Earliest Start Time :\n\t"min_line
print "Latest Start Time :\n\t"max_line
}
function compare(arg1, arg2) {
# returns
# -1 if arg1 < arg2
# 0 if arg1 == arg2
# 1 if arg1 > arg2
last_two_arg1=substr(arg1, 3)
last_two_arg2=substr(arg2, 3)
first_two_arg1=substr(arg1, 0, 2)
first_two_arg2=substr(arg2, 0, 2)
if (int(first_two_arg1) == first_two_arg1) {
# arg1 is a time
if (int(first_two_arg2) == first_two_arg2) {
# arg2 is a time aswell
if (first_two_arg1 > first_two_arg2) {
return 1
} else if (first_two_arg1 == first_two_arg2) {
if (last_two_arg1 > last_two_arg2)
return 1
else if (last_two_arg1 == last_two_arg2)
return 0
else
return -1
} else {
return -1
}
} else {
# arg2 is a date means is smaller
return 1
}
} else {
# arg1 is date
if (int(first_two_arg2) == first_two_arg2) {
# arg2 is time, means arg1 is smaller
return -1
} else {
# both arg1 and arg2 are date
first_three_arg1=substr(arg1, 0, 3)
first_three_arg2=substr(arg2, 0, 3)
if (mon[first_three_arg1] > mon[first_three_arg2])
return 1
else if (mon[first_three_arg1] == mon[first_three_arg2]) {
if (last_two_arg1 > last_two_arg2)
return 1
else if (last_two_arg1 == last_two_arg2)
return 0
else
return -1
} else {
return -1
}
}
}
}' "${file}"
The main barrier in writing this script is a function to compare two different formats of time.
The above script should produce the desired result, although I haven't taken care of sorting, if that's needed, you can ask.