Question

从下面的代码中，我收到错误

Use of uninitialized value $fileName in concatenation <.> or string at ... line 8.

我也得到错误

Can't open from: No such file or directory at ... line 8.

这是家庭作业参考的示例代码。我想要理解它。

use strict;
use warnings;

my $dayLength = $ARGV[0];
my $byteLength = $ARGV[1];
my $fileName = $ARGV[2];

open INPUT, "< $fileName" or die "Can't open from: $!";
my $takeIn = <INPUT>;

print "$takeIn\n";

my @files = split / /, $takeIn;


print "@files\n";



foreach my $file (@files){
if(-M $file > $dayLength && -s $file > $byteLength){
print "\n::::::::$file::::::::\n";
open INPUT, "< $file" or die "Can't open from: $!";
my $i = <INPUT>;
while($i = <INPUT>){
print $i;}
    }}

Answer 1

正如Hunter所提到的那样，似乎没有第三个参数，并且没有设置$ fileName。

尝试在调用open（）之前添加它，以确保用户提供文件名。

if (!defined($filename) or length($filename) == 0))
{
    die "filename argument is required";
}

在连接中使用未初始化的值错误

1 个答案: