Perl 序列号生成器

Question

在 bash 中，序列号例如 222R5555

echo {0..9}{0..9}{0..9}{A..Z}{0..9}{0..9}{0..9}{0..9}  > seqList.txt

在 perl 中该行可以做得更短（更少的代码）吗？有没有办法在 perl 中的范围内使用重复运算符？

谢谢

Answer 1

my $g0to9 = '{'.join(',', '0'..'9').'}';
my $gAtoZ = '{'.join(',', 'A'..'Z').'}';
my %glob = join('', $g0to9 x 3, $gAtoZ, $g0to9 x 4);
while (my $_ = glob($glob)) {
   ...
}

或

[ Deleted ]

或

for my $p1 ('000'..'999') {
   for my $p2 ('A0000'..'Z9999') {
      my $_ = "$p1$p2";
      ...
   }
}

或或

for my $ch0 ('0'..'9') {
for my $ch1 ('0'..'9') {
for my $ch2 ('0'..'9') {
for my $ch3 ('A'..'Z') {
for my $ch4 ('0'..'9') {
for my $ch5 ('0'..'9') {
for my $ch6 ('0'..'9') {
for my $ch7 ('0'..'9') {
   my $_ = join '', $ch0, $ch1, $ch2, $ch3, $ch4, $ch5, $ch6, $ch7;
   ...
}}}}}}}}

或

use Algorithm::Loops qw( NestedLoops );
my $i = NestedLoops([
   (['0'..'9'])x3,
   (['A'..'Z']),
   (['0'..'9'])x4,
]);
while (my @chs = $i->()) {
   my $_ = join '', @chs;
   ...
}

Answer 2

用更少的代码？不可以。Perl 的字符串增量不允许数字位于字母之前，因此您必须将其分成两个范围： '000' .. '999' 和 'A0000' 。 . 'Z9999' 并连接这些值。这肯定需要超过 68 个字符的代码。