Question

我将这些数据加载到一个数组中，打印出来确定。然而，我想要做的是逐行读取文件，然后将整行加载到另一个数组，然后打印出第二个数组的第二个字段。

#!/usr/bin/perl

use strict;
use warnings;

my @fields;

while ( my $line = <DATA> )  {

    @fields = split /,/, $line;

    foreach my $field ( @fields ) {   

        push ( my @one_line_array, $field );
        print my "$one_line_array[1]\n" ;
    }
}

__DATA__

BACKUP, CAPSER_JOB_01_PP, usa-penguin.com, /bin/bash -lc '/usr/bin/run.sh'
PROD, CAPSER_PROD_JOB_01_PS, usa-penguin.com, /bin/bash -lc '/usr/bin/run.sh'
BACKUP, CAPSER_JOB_02_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script02'
PROD, CAPSER_PROD_JOB_02_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script02'
BACKUP, CAPSER_JOB_03_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script03'
PROD, CAPSER_PROD_JOB_03_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script03'
BACKUP, CAPSER_JOB_04_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script04'
PROD, CAPSER_PROD_JOB_04_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script04'
PROD, CAPSER_PROD_JOB_05_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script05'
PROD, CAPSER_PROD_JOB_06_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script06'
BACKUP, CAPSER_JOB_07_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script07'
PROD, CAPSER_PROD_JOB_07_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script07'
BACKUP, CAPSER_JOB_08_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script08'
PROD, CAPSER_PROD_JOB_08_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script08'
BACKUP, CAPSER_JOB_09_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script09'
PROD, CAPSER_PROD_JOB_09_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script09'
BACKUP, CAPSER_JOB_10_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script10'
PROD, CAPSER_PROD_JOB_10_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script10'
BACKUP, CAPSER_JOB_11_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script11'
PROD, CAPSER_PROD_JOB_11_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script11'
PROD, CAPSER_PROD_JOB_12_PS, europe-linux.com, /bin/bash -lc '$HOME/run/script12'
PROD, CAPSER_PROD_JOB_13_PS, europe-linux.com, /bin/bash -lc '$HOME/run/script13'
BACKUP, CAPSER_JOB_14_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script14'
BACKUP, CAPSER_JOB_15_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script15'
 BACKUP, CAPSER_JOB_16_PP, usa-penguin.com, /bin/csh -c '$HOME/TRADEDATE.pl'
PROD, CAPSER_PROD_JOB_17_PS, usa-penguin.com, /bin/csh -c '$HOME/TRADEDATE.pl'
BACKUP, CAPSER_JOB_01_17, usa-penguin.com, /bin/bash -lc '$HOME/run/script17'
PROD, CAPSER_PROD_JOB_18_PS, usa-penguin.com, /bin/bash -lc '$HOME/run/script18'
BACKUP, CAPSER_JOB_19_PP, usa-penguin.com, /bin/bash -lc '$HOME/run/script19'

当我打印出第一个数组时，我可以打印出来 - 这样就可以了：

 BACKUP
 CAPSER_JOB_01_PP
 usa-penguin.com
 /bin/bash -lc '/usr/bin/run.sh'

PROD
 CAPSER_PROD_JOB_01_PS
 usa-penguin.com
 /bin/bash -lc '/usr/bin/run.sh'

BACKUP
 CAPSER_JOB_02_PP
 usa-penguin.com
 /bin/bash -lc '$HOME/run/script02'

PROD
 CAPSER_PROD_JOB_02_PS
 usa-penguin.com
 /bin/bash -lc '$HOME/run/script02'

然而，当我创建第二个数组时，我想要打印的只是该数组的第二个字段：

 CAPSER_JOB_01_PP
 CAPSER_PROD_JOB_01_PS
 CAPSER_JOB_02_PP
 CAPSER_PROD_JOB_02_PS

当我尝试获取第二个数组的第二个字段时，我收到了这些错误。当我用＆＃34;我的＆＃34;初始化它时我收到编译错误。我知道如何切出每一秒的场。

 casper@casper.com 1066$ ./casper_test2.pl
Use of uninitialized value $one_line_array[1] in concatenation (.) or    string at ./casper_test2.pl line 13, <DATA> line 1.
Use of uninitialized value $one_line_array[1] in concatenation (.) or string at ./casper_test2.pl line 13, <DATA> line 2.
Use of uninitialized value $one_line_array[1] in concatenation (.) or string at ./casper_test2.pl line 13, <DATA> line 2.
Use of uninitialized value $one_line_array[1] in concatenation (.) or string at ./casper_test2.pl line 13, <DATA> line 2.
Use of uninitialized value $one_line_array[1] in concatenation (.) or string at ./casper_test2.pl line 13, <DATA> line 2.

casper@casper.com 1068$ ./casper_test2.pl
syntax error at ./casper_test2.pl line 13, near "my "$one_line_array[1]\n""
Execution of ./casper_test2.pl aborted due to compilation errors.
casper@casper.com1069$

@Diab Jerius - 这实际上给了我现在所需要的东西：

 while (my $line = <DATA> ) {
     @fields = split /,/, $line ;
     print "$fields[1]\n" ;
 }

Answer 1

我看到的问题在下面一行：

print my "$one_line_array[1]\n" ; #Why you using my here...!

其次，您可以直接在循环内部尝试打印$ field [1]，看看您是否获得了正确的值。

#!/usr/bin/perl

use strict;
use warnings;

my @fields;

while ( my $line = <DATA> )  {

    @fields = split /,/, $line;
    print $fields[1];

}

...更新

Answer 2

如果您想要的只是第二个字段，那么您可以使用map：

来完成您的工作

use strict;
use warnings;

use Data::Dumper;

my @second_col = map { (split /,/) [1] } <DATA>;

print Dumper \@second_col;

如果那是你的用例，你也可以同时对它进行子串/变换：

my @trimmed_second_col = map { (split /,/)[1] =~ s/\s+//gr } <DATA>;
print Dumper \@trimmed_second_col;

这将按<DATA>的一行进行迭代，在,上拆分每一个，然后应用'删除空白'正则表达式，这要归功于r标志返回结果，以及存储到@trimmed_second_col。

但是我认为你遇到问题的原因是因为你误解了一些事情 - 你正在推动循环的每次迭代，这意味着你做了太多次。你误导了my。

my所做的是声明一个对当前范围是私有的变量。因此，如果你在一个foreach循环中使用它，它每次都会创建一个 new 数组，当循环迭代时它会再次消失。特别是print毫无意义。

那怎么样：

#!/usr/bin/env perl use strict; use warnings; use Data::Dumper; my @second_col; while ( <DATA> ) { my @row = split /,/; # if you need to preserve it in an array. push @second_col, $row[1]; #or if you just need to print it: print $row[1],"\n"; } print Dumper \@second_col;

perl切出每个数组行的第二个字段

2 个答案: