Question

我在这里有日志文件我想提取以下信息

想要提取十六进制值。
想要提取小数值，如果该行有第二个括号{0-9}首先转换为十六进制然后提取（9 - > 0x09）
想要提取减值（-25）首先转换为十六进制然后提取

例如-25 - ＆GT; FFE7 - ＆GT;想提取 - ＆gt; 0xFF，0XE7
如果值为0x2789，则拆分并加入0x（2789 - > 0x27,0x89）

my_input_data

     my_info    0x2789  Uint16, unsigned short      
     param_id   0x14    Uint8,unsigned char     
     cell_id    0x05    Uint8,unsigned char     
     Indicator  0x0B    Uint8,unsigned char     
     filler1{3} { 0x00, 0x00, 0x00 }    Uint8,unsigned char     
     rscp_tap   -116    Sint8,signed char       
     filler2{3} { 0x01, 0x00, 0x00 }    Uint8,unsigned char     
     dsp    -101    Sint8,signed char       
     filler3{3} { 0x00, 0x00, 0x00 }    Uint8,unsigned char     
     system_fm_number   0x3601  Uint16, unsigned short      
     filler4{2} { 0x00, 0x00 }  Uint8,unsigned char 
     dsp_input      {7}

我的预期输出：

    0x27,0x89,0x14,0x05,0x0B,0x00,0x00,0x00,0xFF,
    0x8C,0x01,0x00,0x00,0xFF,0x9B,0x36,0x01,0x00,0x00,0x07

my_code

#! /usr/bin/env perl

use strict;
use warnings;

use List::MoreUtils 'true';

use feature qw(say);

use Data::Dumper;

# input variable pass as a input argument
my $variable_name = shift @ARGV;

# variable value pass as a input argument
my $variable_value = shift @ARGV;

#variable value need to be replaced with new value
my $Replacement_var = shift @ARGV;

# Name of the file the data is in
my $input_filename = 'input.txt';

# Name of the file you want to dump the output to
my $output_filename = 'output.txt';

# Open the file
open my $input_fh, "<", $input_filename or die $!;

# Open the output file
open my $output_fh, ">", $output_filename or die $!;

# Array to store the hex data
my @hex_array;
my @data_new;

# Loop over each of the lines of the file
while ( <$input_fh> ) {

    # Find all the matches and push them to the array

    if ( /$variable_name/ and /$variable_value/ ) {
        s/$variable_value/$Replacement_var/;
    }

    print $output_fh $_;

    #here  extracting only hex values from each line
    while ( $_ =~ m/(0x(\d+)(?:[0-9]|[A-f])+)/gi ) {
        push @hex_array, ( $1 );
    }
}

# Close the file
close $input_fh;

# Write the data to the file
@data_new = join( ", ", @hex_array );

print {$output_fh} @data_new;

# Close the file
close $output_fh;

# Exit
exit();

上面的代码用于提取十六进制值，但不用于提取十进制{0-9}和减去-25值并转换回十六进制。

我想我需要修改正则表达式。

Answer 1

＆＃34;我在等待解决方案＆＃34;

您迫切需要阅读并吸收How to ask

我不明白为什么你从@ARGV获取的三个变量用于在每行输入中进行替换

此外，您的代码会将每个修改后的行复制到输出文件中，但它不会出现在您预期的输出中＃34;

你需要做的不仅仅是找到一些看起来可能有效的代码，然后将其破解并将其放在Stack Overflow上以便其他人为你完成。你立即失去了许多人的尊重，你可能很难得到更多问题的答案

此程序按照您的要求执行

use strict;
use warnings 'all';

my ( $infile, $outfile ) = qw/ input.txt output.txt /;

open my $fh, '<', $infile  or die $!;

my @data;

while ( <$fh> ) {

    my ($f2) = / \S \s+ ( \{ [^{}]+ \} | \S+ ) /x;

    while ( $f2 =~ / 0x ( \p{hex}+ ) | ( [+-]?\d+ ) /xg ) {
        push @data, $1 // sprintf '%04X', $2 & 0xFFFF;
    }
}

{
    my $data = join ',', map "0x$_", map { unpack '(A2)*' } @data;

    open my $fh, '>', $outfile or die $!;
    print $fh $data, "\n";
    close $fh;
}

输出

0x27,0x89,0x14,0x05,0x0B,0x00,0x00,0x00,0xFF,0x8C,0x01,0x00,0x00,0xFF,0x9B,0x00,0x00,0x00,0x36,0x01,0x00,0x00,0x00,0x07

无法从日志文件中提取特定信息

my_input_data

我的预期输出：

my_code

1 个答案:

输出