问题描述
我应该在文本文件中读入多个值，例如：
maclawty796 pts / 1 75-30- 120-13.lig Wed Oct 12 19:27 - 19:33
maclawty796 pts / 1 75-30-120-13.lig Wed Oct 12 19:35 - 19:38
hturner pts / 1 tom-nilsons-macb Wed Oct 12 13:30 - 13:32
nnt pts / 2 99-59-5-115.ligh 10月11日星期二15:51 - 15:54
并将它们转换为当月的总计，例如：
Ehowe Sep 145分钟
Ehowe 10月38分钟
maclawty796 9月240分钟
maclawty796 10月155分钟
到目前为止，我的代码是：
$ file ='C：\Users \user\Desktop\perlintro\timelog.txt';
打开（TimeLog，$ file）或死掉Could not open timelog.txt;
use strict;
my $ username;
my $ months;
my $ time;
my $ minutes1;
my $ minutes2;
my $ seconds1;
my $ seconds2;
my $ oneLine;
my％hash =（）; $（< TimeLog>）{
$ oneLine = $ _;
; （（$ username）= /（[[a-y] *（\d | \w）*）/）{
printf％-12s，$ username;
。 （（月））= /（（1月）|（2月）|（3月）|（4月）|（5月）|（6月）|（7月）|
}
。 （8月）|（9月）|（10月）|（11月）|（12月））/）{
printf（％-7s，$ months）; （（$分钟1，$秒1，$分钟2，$秒2）= /（\d\d）:( \d\d）\s（
）
（\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\'） ;
printf（％-3s minutes \\\
，$ time）;
$ b $为我的$用户名（键（％哈希））{
为我的$月（键（％{$ hash {$ username}}））{
print（$ username $ months $ hash {$ username} {$ months} \\\
）;
$ p
$ b $ p $因此，单日，但我需要帮助，使它只能按月打印。
解决方案试试这个：
＃！/ usr / bin / perl
使用警告;
使用strict;
my％time_log; $（< DATA>）{
chomp;
my（$ user，undef，undef，undef，$ mon，undef，$ time1，undef，$ time2）= split; （$（$ user），$ mons）
$ time_log {$ user} - > {$ mon} + = time_diff（$ time1，$ time2）
}
=每个％time_log）{
while（my（$ mon，$ period）= each％$ mons）{
print$ user $ mon $ period\\\
;
}
}
sub time_diff {
my（$ time1，$ time2）= @_;
my（$ hh1，$ mm1，$ hh2，$ mm2）= split /：/，$ time1：$ time2;
return（$ hh2 - $ hh1）* 60 +（$ mm2 - $ mm1）;
}
__DATA__
maclawty796 pts / 1 75-30-120-13.lig Wed Oct 12 19:27 - 19:33
maclawty796 pts / 1 75 -30-120-13.lig Wed Oct 12 19:35 - 19:38
hturner pts / 1 tom-nilsons-macb Wed Oct 12 13:30 - 13:32
nnt pts / 2 99 -59-5-115.ligh 10月11日星期二15:51 - 15:54
I am supposed to read in a text file with multiple values such as:
maclawty796 pts/1 75-30-120-13.lig Wed Oct 12 19:27 - 19:33
maclawty796 pts/1 75-30-120-13.lig Wed Oct 12 19:35 - 19:38
hturner pts/1 tom-nilsons-macb Wed Oct 12 13:30 - 13:32
nnt pts/2 99-59-5-115.ligh Tue Oct 11 15:51 - 15:54
and turn them into totals for the month, such as:
Ehowe Sep 145 minutes
Ehowe Oct 38 minutes
maclawty796 Sep 240 minutes
maclawty796 Oct 155 minutes
So far my code is:
$file = 'C:\Users\user\Desktop\perlintro\timelog.txt';
open(TimeLog, $file) or die "Couldn't open timelog.txt";
use strict;
my $username;
my $months;
my $time;
my $minutes1;
my $minutes2;
my $seconds1;
my $seconds2;
my $oneLine;
my %hash = ();
while (<TimeLog>) {
$oneLine=$_;
if ( ($username) = /([[a-y]*(\d|\w)*)/) {
printf "%-12s", $username;
}
if ( ($months) = /((Jan) | (Feb) | (Mar) | (Apr) | (May) | (Jun) | (Jul) | (Aug) | (Sep) | (Oct) | (Nov) | (Dec) )/ ) {
printf ("%-7s", $months);
}
if ( ($minutes1, $seconds1, $minutes2, $seconds2) = /(\d\d):(\d\d)\s-\s(\d\d):(\d\d)/ ) {
$time = ($minutes2 * 60 + $seconds2) - ($minutes1 * 60 + $seconds1);
printf ("%-3s minutes\n", $time);
}
for my $username (keys(%hash)) {
for my $months (keys(%{ $hash{$username} })) {
print("$username $months $hash{$username}{$months}\n");
}
}
}
So, currently it prints every single day, but I need help to make it only print by month.
解决方案 Try this:
#!/usr/bin/perl
use warnings;
use strict;
my %time_log;
while (<DATA>) {
chomp;
my ($user, undef, undef, undef, $mon, undef, $time1, undef, $time2) = split;
$time_log{$user}->{$mon} += time_diff($time1, $time2)
}
while (my ($user, $mons) = each %time_log) {
while (my ($mon, $period) = each %$mons) {
print "$user $mon $period\n";
}
}
sub time_diff {
my ($time1, $time2) = @_;
my ($hh1, $mm1, $hh2, $mm2) = split /:/, "$time1:$time2";
return ($hh2 - $hh1) * 60 + ($mm2 - $mm1);
}
__DATA__
maclawty796 pts/1 75-30-120-13.lig Wed Oct 12 19:27 - 19:33
maclawty796 pts/1 75-30-120-13.lig Wed Oct 12 19:35 - 19:38
hturner pts/1 tom-nilsons-macb Wed Oct 12 13:30 - 13:32
nnt pts/2 99-59-5-115.ligh Tue Oct 11 15:51 - 15:54
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