如何使用perl迭代對象內部的哈希數組

Question 1

我試圖遍歷存儲為“Person”對象的'_skills屬性的哈希數組，但遇到了問題。

test1.pl

my $person = new Person( "Hubert", "Blu",
    [
       {"name" => "Python", "Weight" => 80, "group","IT SKILLS", "id"=>0002},
       {"name" => "Magic", "Weight" => 100, "group", "MISC SKILLS", "id"=>0001},
       {"name" => "JavaScript","Weight" => 70, "group"=> "IT SKILLS", "id"=>0004},
       {"name" => "Mage","Weight" => 100, "group"=>"JOB TITLE", "id" =>0003},
       {"name" => "Assassin", "Weight" => 85, "group"=>"JOB TITLE"}
    ]
);

and Person.pl

sub new {
    my ($class,$first_name,$last_name,@skills) = @_ ;
    my $self = {};
    $self->{"_first_name"} = $first_name;
    $self->{"_last_name"} = $last_name;
    @{$self->{"_skills"}} = @skills;
    # Print all the values just for clarification.
    print "First Name is $self->{_first_name}\n";
    print "Last Name is $self->{_last_name}\n";
    print "Skills : " . Data::Dumper::Dumper($self->{_skills}) . "\n";


    print"---SKILLS---\n";
    my $i = 1;
    foreach my $skill_set (@{$self->{_skills}}) {
        print "skill ".$i .": \n";
        $i++;
        #print Data::Dumper::Dumper($skill_set) . "\n";
        print "$skill_set->{'_first_name'}";
    };
    bless $self, $class;
    return $self;
}

第一次轉儲顯示了陣列，看起來很好。循環(huán)中的第二個轉儲再次顯示整個數組，它只循環(huán)一次，然后返回以下錯誤

Not a HASH reference at /home/izoom/PerlTrainingGround/LeeR/lib/Person/Person.pm line 26.

被困了好幾個小時，感謝您一如既往的幫助！

Question 2

您正在將三個參數傳遞給new和invocant。最后一個是對數組的引用。這意味著@skills以一個元素結束，該元素的值就是該引用。

my $person = new Person( "Hubert", "Blu",
    [
       {"name" => "Python", "Weight" => 80, "group","IT SKILLS", "id"=>0002},
       {"name" => "Magic", "Weight" => 100, "group", "MISC SKILLS", "id"=>0001},
       {"name" => "JavaScript","Weight" => 70, "group"=> "IT SKILLS", "id"=>0004},
       {"name" => "Mage","Weight" => 100, "group"=>"JOB TITLE", "id" =>0003},
       {"name" => "Assassin", "Weight" => 85, "group"=>"JOB TITLE"}
    ]
);

should be

my $person = new Person( "Hubert", "Blu",
   {"name" => "Python", "Weight" => 80, "group","IT SKILLS", "id"=>0002},
   {"name" => "Magic", "Weight" => 100, "group", "MISC SKILLS", "id"=>0001},
   {"name" => "JavaScript","Weight" => 70, "group"=> "IT SKILLS", "id"=>0004},
   {"name" => "Mage","Weight" => 100, "group"=>"JOB TITLE", "id" =>0003},
   {"name" => "Assassin", "Weight" => 85, "group"=>"JOB TITLE"}
);

or

    my ($class,$first_name,$last_name,@skills) = @_ ;

    @{$self->{"_skills"}} = @skills;     # Useless shallow array copy
       # -or-
    $self->{"_skills"} = [ @skills ];    # Same[1] but clearer
       # -or-
    $self->{"_skills"} = \@skills;       # No copy of elements

should be

    my ($class,$first_name,$last_name,$skills) = @_ ;

    @{$self->{"_skills"}} = @$skills;    # Shallow array copy
       # -or-
    $self->{"_skills"} = [ @$skills ];   # Same[1] but clearer
       # -or-
    $self->{"_skills"} = $skills;        # No copy of elements

因為此時$self->{"_skills"}是undef。

Answer 1

您正在將三個參數傳遞給new和invocant。最后一個是對數組的引用。這意味著@skills以一個元素結束，該元素的值就是該引用。

my $person = new Person( "Hubert", "Blu",
    [
       {"name" => "Python", "Weight" => 80, "group","IT SKILLS", "id"=>0002},
       {"name" => "Magic", "Weight" => 100, "group", "MISC SKILLS", "id"=>0001},
       {"name" => "JavaScript","Weight" => 70, "group"=> "IT SKILLS", "id"=>0004},
       {"name" => "Mage","Weight" => 100, "group"=>"JOB TITLE", "id" =>0003},
       {"name" => "Assassin", "Weight" => 85, "group"=>"JOB TITLE"}
    ]
);

should be

my $person = new Person( "Hubert", "Blu",
   {"name" => "Python", "Weight" => 80, "group","IT SKILLS", "id"=>0002},
   {"name" => "Magic", "Weight" => 100, "group", "MISC SKILLS", "id"=>0001},
   {"name" => "JavaScript","Weight" => 70, "group"=> "IT SKILLS", "id"=>0004},
   {"name" => "Mage","Weight" => 100, "group"=>"JOB TITLE", "id" =>0003},
   {"name" => "Assassin", "Weight" => 85, "group"=>"JOB TITLE"}
);

or

    my ($class,$first_name,$last_name,@skills) = @_ ;

    @{$self->{"_skills"}} = @skills;     # Useless shallow array copy
       # -or-
    $self->{"_skills"} = [ @skills ];    # Same[1] but clearer
       # -or-
    $self->{"_skills"} = \@skills;       # No copy of elements

should be

    my ($class,$first_name,$last_name,$skills) = @_ ;

    @{$self->{"_skills"}} = @$skills;    # Shallow array copy
       # -or-
    $self->{"_skills"} = [ @$skills ];   # Same[1] but clearer
       # -or-
    $self->{"_skills"} = $skills;        # No copy of elements

因為此時$self->{"_skills"}是undef。

如何使用perl迭代對象內部的哈希數組

熱門問答

在Shining應用程序中使用group by進行描述性統計

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第一次匹配后停止regex

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我需要用numpy來加速python代碼

class Job: # 初始化 def init(self, idx, max_ol): self.idx = idx self.start = 0 # 加工開始時間 self.end = 0 # 加工結束時間 self.op = 0 self.max_ol = max_ol self.Gap = 0 self.l = 0 self.T = []

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如何使用perl迭代對象內部的哈希數組

熱門問答

在Shining應用程序中使用group by進行描述性統計

ForkJoinPool執(zhí)行任務兩次

背景位置不居中/剪輯路徑不工作

第一次匹配后停止regex

在Python中的一行中指定并檢查變量？

我需要用numpy來加速python代碼

class Job: # 初始化 def __init__(self, idx, max_ol): self.idx = idx self.start = 0 # 加工開始時間 self.end = 0 # 加工結束時間 self.op = 0 self.max_ol = max_ol self.Gap = 0 self.l = 0 self.T = []

如何在Python中使用相對路徑和絕對路徑

在Android中，DownloadManager和HttpURLConnection下載有何不同

TensorFlow對于初學者來說學習曲線是怎樣的

保存html文件后目錄自動生成一個".filename.html"

antd layout刷新頁面當前選中會回到默認值怎么解決？

class Job: # 初始化 def init(self, idx, max_ol): self.idx = idx self.start = 0 # 加工開始時間 self.end = 0 # 加工結束時間 self.op = 0 self.max_ol = max_ol self.Gap = 0 self.l = 0 self.T = []

在Android中，DownloadManager和HttpURLConnection下載有何不同

antd layout刷新頁面當前選中會回到默認值怎么解決？