我的Perl脚本不工作，我有一种感觉，这是grep命令 [英] My perl script isn't working, I have a feeling it's the grep command

问题描述

我试图在一个文件中搜索
数字的实例，如果其他文件包含这些数字，则发布。

 ＃！/ usr / bin / perl 
 open（file，textIds.txt）; ＃
 @file =< file>; #file寻找
＃关闭文件; ＃
 while（<>）{
 $ temp = $ _; 
 $ temp =〜tr / | / \ t /;名称和ID之间的#puts选项卡
 @arrayTemp = split（\t，$ temp）; 
 @ found = grep {/ $ arrayTemp [1] /}< file>; 
 if（defined $ found [0]）{
 #if（grep {/ $ arrayTemp [1] /}< file>）{
 print $ _; 
} 
 @ found =（）; 
} 
打印\\\
; 
关闭文件; 
 
＃输入文件行的格式为
 #John | 7791 154 
 #Smith | 5432 290 
 #Conor | 6590 897 
 
＃此文件格式为
＃5432 
＃7791 
＃6590 
＃23140 
  

解决方案

脚本中存在一些问题。 c $ c>和使用警告; 。
这将在您的脚本中事先告诉您有关奇怪的事情。

切勿将裸词作为文件句柄使用，因为它们是全局标识符。使用三参数打开
来代替： open（my $ fh，'&'，'testIds.txt'）;

使用autodie; 或检查打开的结果是否有效。您阅读并存储 testIds。 txt 放入数组 @file 但稍后（在 grep 中），您是
再次尝试从该文件（句柄）中读取（使用< file> ）。正如@PaulL所说，由于该文件已被读取，因此总是
给出 undef （false）。

替换 | 带有制表符，然后在选项卡上拆分并不是必需的。您可以同时在
标签和管道上分割（假设John | 7791 154真的是John | 7791 \ t154）。

您在谈论输入文件和文件中时，并未精确地告诉哪个是哪个。
我假设你的textIds.txt是只有数字的文本，而其他的输入文件是从STDIN读取的
（带有

考虑到这一点，您的脚本可以写成： p>

 ＃！/ usr / bin / perl 
 use strict; 
使用警告; 
 
＃打开'textIds.txt'并将其放入数组@file：
 open（my $ fh，'<'，'textIds.txt'）或者死无法打开file：$！\\\
; 
 my @file =< $ fh>; 
 close（$ fh）; 
 
＃迭代STDIN并与'textIds.txt'中的行进行比较：
 while（my $ line =<>）{
＃splitJohn | 7791 （John，7791，154）：
 my（$ name，$ number1，$ number2）= split（/ \ || \ t /，$ line）; 
 
＃将$ number1与@file的每个成员进行比较，如果找到则打印：
 if（grep（/ $ number1 /，@file））{
 print $ line; 
} 
} 
  

I'm trying for search in the one file for instances of the number and post if the other file contains those numbers

#!/usr/bin/perl
open(file, "textIds.txt"); #
        @file = <file>;         #file looking into
#        close file;            #
while(<>){
        $temp = $_;
        $temp =~ tr/|/\t/;      #puts tab between name and id
        @arrayTemp = split("\t", $temp);
        @found=grep{/$arrayTemp[1]/} <file>;
        if (defined $found[0]){
        #if (grep{/$arrayTemp[1]/} <file>){
                print $_;
        }
        @found=();
}
print "\n";
close file;

#the input file lines have the format of 
#John|7791  154
#Smith|5432 290
#Conor|6590 897

#And in the file the format is 
#5432
#7791
#6590
#23140

解决方案

There are some issues in your script.

1. Always include use strict; and use warnings;. This would have told you about odd things in your script in advance.
2. Never use barewords as filehandles as they are global identifiers. Use three-parameter-open instead: open( my $fh, '<', 'testIds.txt');
3. use autodie; or check whether the opening worked.
4. You read and store testIds.txt into the array @file but later on (in your grep) you are again trying to read from that file(handle) (with <file>). As @PaulL said, this will always give undef (false) because the file was already read.
5. Replacing | with tabs and then splitting at tabs is not neccessary. You can split at the tabs and pipes at the same time as well (assuming "John|7791 154" is really "John|7791\t154").
6. Your talking about "input file" and "in file" without exactly telling which is which. I assume your "textIds.txt" is the one with only the numbers and the other input file is the one read from STDIN (with the |'s in it).

With this in mind your script could be written as:

#!/usr/bin/perl
use strict;
use warnings;

# Open 'textIds.txt' and slurp it into the array @file:
open( my $fh, '<', 'textIds.txt') or die "cannot open file: $!\n";
my @file = <$fh>;
close($fh);

# iterate over STDIN and compare with lines from 'textIds.txt':
while( my $line = <>) {
    # split "John|7791\t154" into ("John", "7791", "154"):
    my ($name, $number1, $number2) = split(/\||\t/, $line);

    # compare $number1 to each member of @file and print if found:
    if ( grep( /$number1/, @file) ) {
        print $line;
    }
}

这篇关于我的Perl脚本不工作，我有一种感觉，这是grep命令的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！