Question

对于一个数组我检查在foreach循环中设置一些条件。只有当它为真时，我才想将$_分配给另一个数组..但是简单的@stale = $_;没有附加到它......也不是push或任何其他函数。。

知道如何解决这个问题吗？

编辑：OP评论的代码

# To give a generic ORACLE_HOME and HOSTNAME path for all the instances.
#
#using shell ENV variable in perl
my $home  = $ENV{'ORACLE_HOME'};
#
#concatenating the paths
my $subdir = "/ccr/hosts/";
my $subdir1 = $home.$subdir;
my $host = "`hostname`/state/review";
my $subdir2 = $subdir1.$host;
#
#assigning each row of command output to an array
@infile = `ls -ltr $subdir2`;
#
#printing each value of the array i.e. file details
foreach (@infile) {
        chop($_);
        $m_age = -M $_;
        if (($stale = `date` - $m_age) < 1.0) {
#
#
#change/correct this line :(
#
#
                printf "\nFile %s was last modified 24hrs back. stale collections. \n" , $_ , $m_age ,$stale;
                push @stale_files , $_;
        }
        else {
        @not_stale_files = $_[$q++];
        }
        $count++;
}
print "\n for stale:\n";
foreach $i (@stale_files) {
print "$stale_files[$i]\n";
}

此外，每次循环执行时，$_值都应附加到@stale。但只有$_的最后一个值（在上次循环运行期间获得）才会被分配到@stale。

Answer 1

好的，让我们回顾一下编辑过的代码

请使用

use warnings;
use strict;

代码中的

：它将帮助您检测错误

# To give a generic ORACLE_HOME and HOSTNAME path for all the instances.
#
#using shell ENV variable in perl
my $home  = $ENV{'ORACLE_HOME'};
#
#concatenating the paths
my $subdir = "/ccr/hosts/";
my $subdir1 = $home.$subdir;

您不需要复杂的构造

my $subdir1 = $ENV{'ORACLE_HOME} . '/ccr/hosts';

就够了

下一行不起作用：后退标记不会在"引用的字符串中展开

my $host = "`hostname`/state/review";
my $host = `hostname` . '/state/review';

my $subdir2 = $subdir1.$host;
#
#assigning each row of command output to an array
@infile = `ls -ltr $subdir2`;

此处@infile将包含多个文件。每个元素的输出都为ls -ltr。像

这样的东西

total 16
-rw-r--r--  1 corti  corti   94 Sep 14 11:40 test.pl~
-rw-r--r--  1 corti  corti  979 Sep 14 11:49 test.pl

#
#printing each value of the array i.e. file details
foreach (@infile) {
    chop($_);

这里你正在做-M -rw-r--r-- 1 corti corti 94 Sep 14 11:40 test.pl~。

    $m_age = -M $_;
    if (($stale = `date` - $m_age) < 1.0) {
       #
       #
       #change/correct this line :(
       #
       #
       printf "\nFile %s was last modified 24hrs back. stale collections. \n" , $_ , $m_age ,$stale;

这是对的。问题出在$_

的内容中

            push @stale_files , $_;
    }
    else {

$_是@infile的元素。你想要实现$_[]的目标是什么？什么是$q？

    @not_stale_files = $_[$q++];
    }

$count永远不会被初始化

    $count++;
}
print "\n for stale:\n";
foreach $i (@stale_files) {
  print "$stale_files[$i]\n"; 
}

Answer 2

使用“push @stale, $_;”（添加到数组的末尾）而不是“@stale = $_;”（完全替换数组的现有内容）应该适用于这种情况。

修改：证明push做了所要求的事。

代码：

#!/usr/bin/env perl

use strict;
use warnings;

my @numbers = (1, 8, 3, 5, 2, 8, 9, 0);
my @stale;

for (@numbers) {
    if ($_ < 5) {
        printf "\nNumber %s is less than five.\n", $_;
        push @stale, $_;
    }   
}

print join ", ", @stale;
print "\n";

输出：

$ ./add_to_array.pl 

Number 1 is less than five.

Number 3 is less than five.

Number 2 is less than five.

Number 0 is less than five.
1, 3, 2, 0
$

无法弄清楚这个简单的perl错误？

2 个答案: