perl序列號發生器

Question

在bash中，序列號例如222R5555

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

該行可以在perl中更短（少代碼）嗎？有沒有辦法在perl中使用重複運算符？

謝謝

Answer 1

使用較少的代碼？不。Perl的字符串增量不允許將數字放在字母之前，因此您必須將其分成兩個範圍：'000' .. '999'和'A0000' .. 'Z9999'並連接值。這肯定會佔用超過68個字符的代碼。

Answer 2

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

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; 
    ... 
}