I think this should be easy to solve for experienced people, for me though I was trying and searched on google about this but couldnt find anything that fits my situation. I just want to let $amountok and $amountko give out the values they have at index 0 and 1 but it is not working like this (see # marks).
我认为这对于有经验的人来说应该是很容易解决的,虽然我在谷歌上搜索过,但是找不到适合我的东西。我只是想让$amountok和$amountko给出索引0和1的值,但它不是这样工作的(参见# marks)。
print ERGEBNIS "ID;Amount;Amount OK;Amount KO\n";
foreach my $key (keys %elementhash){
my $sum = 0;
foreach $valueofkey(@{$elementhash{$key}}){
$sum += $valueofkey;
}
#my $amountok= @{$elementhash{$key}[0]};
#my $amountko= @{$elementhash{$key}[1]};
print ERGEBNIS $key.";".$sum.";".$amountok.";".$amountko."\n";
}
this would give me out the error: "Cant use string "7" as an array ref.", this must be possible somehow but i just don't know how! thank you guys
这将给出一个错误:“不能使用字符串”7作为数组ref。这一定是可能的,但我就是不知道怎么做!谢谢你们
this is the Dumper output of %elementhash:
这是%elementhash的倾倒输出:
$VAR1 = { '1000' => [7],
'2000' => [5],
'3000' => [56, 12]
};
1 个解决方案
#1
5
Your hash elements contain array references. Even if there is only one value in the array ref, it's still an array ref. You can use the ->
operator for dereferencing.
哈希元素包含数组引用。即使数组ref中只有一个值,它仍然是一个数组ref。您可以使用->操作符来取消引用。
$elementhash{$key}->[0];
Note that you first have a hash, not a hash reference, so there is no arrow before the {$key}
part. Inside there is an array reference. In fact, you don't strictly need to put a ->
in front of the [0]
because Perl knows that nested data structures are built from references. It's a matter of preference that's been well discussed. I personally like the arrow, but it's not needed here1.
注意,您首先有一个散列,而不是一个散列引用,因此在{$key}部分之前没有箭头。里面有一个数组引用。实际上,您并不需要在[0]前面加上->,因为Perl知道嵌套数据结构是由引用构建的。这是一个已经被充分讨论过的偏好问题。我个人喜欢这支箭,但这里不需要。
What you tried to do with @{$elementhash{$key}[0]}
was take the first element inside the $key
, e. g. the 7
for key 1000
, and deref that as an array. Of course you can't do that, because it's a number, and not an array reference.
您尝试使用@{$elementhash{$ $key}[0]}所做的是获取$key中的第一个元素,例如,7表示key 1000, deref表示数组。当然你不能这么做,因为它是一个数字,而不是一个数组引用。
Because not all of the arrays in your data structure have two values, you should check if the second value exists. You can do that with the //
defined-or operator. It allows 0
or other un-true values, but not undef
.
因为不是数据结构中的所有数组都有两个值,所以应该检查第二个值是否存在。您可以使用//定义或操作符来实现这一点。它允许0或其他非真值,但不允许undef。
Finally, you were missing a my
for $valueofkey
.
最后,你错过了$valueofkey的一个my。
use strict;
use warnings;
my %elementhash = (
'1000' => [7],
'2000' => [5],
'3000' => [ 56, 12 ]
);
print "ID;Amount;Amount OK;Amount KO\n";
foreach my $key ( keys %elementhash ) {
my $sum = 0;
foreach my $valueofkey ( @{ $elementhash{$key} } ) {
$sum += $valueofkey;
}
my $amountok = $elementhash{$key}->[0];
my $amountko = $elementhash{$key}->[1] // 0;
print $key. ";" . $sum . ";" . $amountok . ";" . $amountko . "\n";
}
Note that you can rewrite the last line to use join
.
注意,可以重写最后一行以使用join。
print join( ';', $key, $sum, $amountok, $amountko ), "\n";
If your Perl is at least version 5.10, you can also turn on use feature 'say'
so you don't need to print the "\n"
.
如果您的Perl至少是5.10版本,那么您还可以打开use特性‘say’,这样就不需要打印“\n”了。
say join ';', $key, $sum, $amountok, $amountko;
If you have more complicated CSV, consider using Text::CSV or Text::CSV_XS instead.
如果您有更复杂的CSV,可以考虑使用Text::CSV或Text:::CSV_XS。
1) You do need one arrow if the variable that you are starting from is a reference. But you can omit the following ones.
1)如果开始的变量是引用,则需要一个箭头。但是你可以省略下面的。
my $foo = { bar => [1, 2, 3] };
say $foo->{bar}[2]; # works
say $foo{bar}[2]; # complains that %foo needs explicit package name
#1
5
Your hash elements contain array references. Even if there is only one value in the array ref, it's still an array ref. You can use the ->
operator for dereferencing.
哈希元素包含数组引用。即使数组ref中只有一个值,它仍然是一个数组ref。您可以使用->操作符来取消引用。
$elementhash{$key}->[0];
Note that you first have a hash, not a hash reference, so there is no arrow before the {$key}
part. Inside there is an array reference. In fact, you don't strictly need to put a ->
in front of the [0]
because Perl knows that nested data structures are built from references. It's a matter of preference that's been well discussed. I personally like the arrow, but it's not needed here1.
注意,您首先有一个散列,而不是一个散列引用,因此在{$key}部分之前没有箭头。里面有一个数组引用。实际上,您并不需要在[0]前面加上->,因为Perl知道嵌套数据结构是由引用构建的。这是一个已经被充分讨论过的偏好问题。我个人喜欢这支箭,但这里不需要。
What you tried to do with @{$elementhash{$key}[0]}
was take the first element inside the $key
, e. g. the 7
for key 1000
, and deref that as an array. Of course you can't do that, because it's a number, and not an array reference.
您尝试使用@{$elementhash{$ $key}[0]}所做的是获取$key中的第一个元素,例如,7表示key 1000, deref表示数组。当然你不能这么做,因为它是一个数字,而不是一个数组引用。
Because not all of the arrays in your data structure have two values, you should check if the second value exists. You can do that with the //
defined-or operator. It allows 0
or other un-true values, but not undef
.
因为不是数据结构中的所有数组都有两个值,所以应该检查第二个值是否存在。您可以使用//定义或操作符来实现这一点。它允许0或其他非真值,但不允许undef。
Finally, you were missing a my
for $valueofkey
.
最后,你错过了$valueofkey的一个my。
use strict;
use warnings;
my %elementhash = (
'1000' => [7],
'2000' => [5],
'3000' => [ 56, 12 ]
);
print "ID;Amount;Amount OK;Amount KO\n";
foreach my $key ( keys %elementhash ) {
my $sum = 0;
foreach my $valueofkey ( @{ $elementhash{$key} } ) {
$sum += $valueofkey;
}
my $amountok = $elementhash{$key}->[0];
my $amountko = $elementhash{$key}->[1] // 0;
print $key. ";" . $sum . ";" . $amountok . ";" . $amountko . "\n";
}
Note that you can rewrite the last line to use join
.
注意,可以重写最后一行以使用join。
print join( ';', $key, $sum, $amountok, $amountko ), "\n";
If your Perl is at least version 5.10, you can also turn on use feature 'say'
so you don't need to print the "\n"
.
如果您的Perl至少是5.10版本,那么您还可以打开use特性‘say’,这样就不需要打印“\n”了。
say join ';', $key, $sum, $amountok, $amountko;
If you have more complicated CSV, consider using Text::CSV or Text::CSV_XS instead.
如果您有更复杂的CSV,可以考虑使用Text::CSV或Text:::CSV_XS。
1) You do need one arrow if the variable that you are starting from is a reference. But you can omit the following ones.
1)如果开始的变量是引用,则需要一个箭头。但是你可以省略下面的。
my $foo = { bar => [1, 2, 3] };
say $foo->{bar}[2]; # works
say $foo{bar}[2]; # complains that %foo needs explicit package name