Question

我有下面的代码，我试图只打印出这个2d数组的第一行

# how many columns
for (my $c = 0; $c <= $#list[0]; $c++) {
print $list[0][$c]."\n";

数据应该是

[0] => "ID,Cluster,Version"
[1] => "2,32,v44"

错误：

syntax error at ./connect_qb.pl line 107, near "$#list["
syntax error at ./connect_qb.pl line 107, near "++) "
Execution of ./connect_qb.pl aborted due to compilation errors.

Answer 1

$list[0]

是对数组的引用，因此数组是

@{ $list[0] }

所以该数组的最后一个元素是

$#{ $list[0] }

所以你要用

for my $c (0 .. $#{ $list[0] }) {
   print "$list[0][$c]\n";
}

或

for (@{ $list[0] }) {
   print "$_\n";
}

Answer 2

你应该避免使用c-style for循环。这是一种方法。

use strict;
use warnings;
use feature qw(say);

my @a = (["ID","Cluster","Version"], ["2","32","v44"]);
say for (@{$a[0]});

稍微不那么令人困惑的解除引用：

...
my $ref = $a[0];
say for (@$ref);

Answer 3

这是一个简单的衬垫

print join(",",@{$list[0]}),"\n";

Answer 4

试试这个：

for (my $c = 0; $c <=  (scalar @{$list[0]}); $c++)

表示循环条件

在Perl中打印第二行数组

4 个答案: