Perl脚本:目录的有效性始终返回false

Question

我想使用perl脚本从文件中读取一些参数.我用grep命令查找thoses参数的值.

#!/usr/bin/perl

if( scalar @ARGV ==0)
{
  die "Number of argument is zero. Please use: perl <script_name> <cdc68.ini> \n";
}


if( !-f $ARGV[0])     ## if the 1st arg is not a file
{
   die "$ARGV[0] is not a valid file type \nInput Arguments not correct!!! \n";
}

my $file_cnf=$ARGV[0];

my $DEST_PATH=`grep relogin_logs $file_cnf | cut -d "=" -f2`;

my $SRC_PATH=`grep dump_logs $file_cnf | cut -d "=" -f2`;
my $FINAL_LOG=`grep final_log $file_cnf | cut -d "=" -f2`;

print "\n$DEST_PATH \n $SRC_PATH \n $FINAL_LOG\n";


if ( !-d $DEST_PATH)
{
  die "$DEST_PATH is not a dir";
}

else
{
  print "ok";
}

我想读的文件是

cat cdc68.ini
reconn_interval=15
relogin_logs=/osp/local/home/linus/pravej/perlD/temp/relogin/
dump_logs=/osp/local/home/linus/pravej/perlD/temp/
final_log=/osp/local/home/linus/pravej/perlD/final_log/
Number_days=11

样本输出:

perl readconfig.pl cdc68.ini
/osp/local/home/linus/pravej/perlD/temp/relogin/
/osp/local/home/linus/pravej/perlD/temp/
/osp/local/home/linus/pravej/perlD/temp/relogin/
is not a dir at readconfig.pl line 26.

任何人都可以在这里暗示我做错了吗？请注意,我不想使用任何perl模块,如config或tiny.pm. 这些目录已经存在于我的unix系统中.提前感谢您的帮助

Answer 1

您可以使用perl解析而不是shell utils

my $file_cnf = $ARGV[0];

open my $fh, "<", $file_cnf or die $!;
my %ini = map /(.+?)=(.+)/, <$fh>;
close $fh;

print "\n$ini{relogin_logs} \n $ini{dump_logs} \n $ini{final_log}\n";