Perl脚本grep [英] Perl script grep

问题描述

脚本正在打印输入行的数量，我希望它能打印出现在另一个文件中的输入行的数量。

 ＃！/ usr / bin / perl -w 
 open（file，text.txt）; 
 @todd =< file>; 
关闭文件; $（<>）{
 if（grep（/ ^ $ _ $ /，@todd））{
 #if（grep @todd，/ ^ $ _ $ /） {
 print $ _; 
} 
打印\\\
; 
} 
  

如果文件包含

  1 
 3 
 4 
 5 
 7 
  

以及将从中读取的输入文件

  1 
 2 
 3 
 4 
 5 
 6 
 7 
 8 
 9 
  

我希望它打印1,3,4,5和7
，但是正在打印1-9

UPDATE ******
这是我的代码现在，我得到这个错误
readline（）在关闭文件句柄todd在./may6test.pl第3行。

 ＃！/ usr / bin / perl -w 
 open（todd，< text。文本）; 
 @files =< todd>; ＃文件正在查看
关闭todd; 
 while（my $ line =<>）{
 chomp $ line; 
 if（grep（/ ^ $ line $ /，@files））{
 print $ _; 
} 
打印\\\
; 
} 
  

这对我来说没有任何意义，因为我有这个基本上正在做的其他脚本同样的事情

$ p $ ＃！/ usr / bin / perl -w
open（file，< text2.txt）; ＃
@file =< file>; ＃文件正在查看
关闭文件; ＃
while（<>）{
$ temp = $ _;
$ temp =〜tr / | / \ t /;名称和ID之间的#puts选项卡
my（$ name，$ number1，$ number2）= split（\t，$ temp）;
if（grep（/ ^ $ number1 $ /，@file））{
print $ _;
}
}
打印\\\
;


解决方案

好的，这里的问题是 - grep 也设置 $ _ 。因此， grep {$ _} @array 将始终为您提供数组中的每个元素。

在基本层面上 - 您需要：

  while（ my $ line =<>）{
 chomp $ line; 
 if（grep {/ ^ $ line $ /} @todd）{
 #do something 
} 
} 
  

但是我会建议您改为考虑构建一行代码行：

  open（my $ input，'<'，text.txt）或者$！ 
 my％in_todd = map {$ _ => 1}< $ input>; 
 close $ input; 
 while（<>）{
 print in if_todd {$ _}; 
} 
  

注意 - 您可能需要注意拖尾换行。

The script is printing the amount of input lines, I want it to print the amount of input lines that are present in another file

#!/usr/bin/perl -w
open("file", "text.txt"); 
        @todd = <file>;         
        close "file";
while(<>){
        if( grep( /^$_$/, @todd)){
        #if( grep @todd, /^$_$/){
                print $_;
        }
        print "\n";
}

if for example file contains

1
3
4
5
7

and the input file that will be read from contains

1
2
3
4
5
6
7
8
9

I would want it to print 1,3,4,5 and 7 but 1-9 are being printed instead

UPDATE****** This is my code now and I am getting this error readline() on closed filehandle todd at ./may6test.pl line 3.

#!/usr/bin/perl -w
open("todd", "<text.txt");
        @files = <todd>;         #file looking into
        close "todd";
while( my $line = <> ){
        chomp $line;
        if ( grep( /^$line$/, @files) ) {
                print $_;
        }
        print "\n";
}

which makes no sense to me because I have this other script that is basically doing the same thing

#!/usr/bin/perl -w
open("file", "<text2.txt");    #
        @file = <file>;         #file looking into
        close "file";           #
while(<>){
        $temp = $_;
        $temp =~ tr/|/\t/;      #puts tab between name and id
        my ($name, $number1, $number2) = split("\t", $temp);
        if ( grep( /^$number1$/, @file) ) {
                print $_;
        }
}
print "\n";

解决方案

OK, the problem here is - grep sets $_ too. So grep { $_ } @array will always give you every element in the array.

At a basic level - you need to:

while ( my $line = <> ) { 
   chomp $line; 
   if ( grep { /^$line$/ } @todd ) { 
      #do something
   }
}

But I'd suggest instead that you might want to consider building a hash of your lines instead:

open( my $input, '<', "text.txt" ) or die $!;
my %in_todd = map { $_ => 1 } <$input>;
close $input;
while (<>) {
   print if $in_todd{$_};
}

Note - you might want to watch for trailing linefeeds.

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