Question

更新

我有一个1000行的日志文件，其中包含一些参考文献。

Time Reference Date of start Date of end
12:00 AT001 13 November 2011 15 November 2011
13:00 AT038 15 December 2012 17 December 2012
14:00 AT076 17 January 2013 19 January 2013

$ ref1 = AT038

基本上，我想解析日志文件并为$ ref1输出（逐行），例如：

Time : 13h
Reference : AT038
Date of start : 15 December 2012
Date of end : 17 December 2012

提前致谢

Answer 1

尝试：

$ref1 = "AT038"
$csv = Import-Csv .\myfile.txt -Delimiter ' '#Import file as CSV with space as delimiter
$csv | ? { $_.reference -EQ $ref1 } | FL #Piping each line of CSV to where-object cmdlet,  filtering only line where value of column reference is equal to $ref1 variable value. Piping the result of the filtering to file-list to have output as requested in OP.

在OP中更改了必要条件后添加的代码：

$ref1 = "AT038"
$txt = gc .\myfile.txt
$txt2 = $txt | % { $b = $_ -split ' '; "$($b[0]) $($b[1]) $($b[2])_$($b[3])_$($b[4]) $($b[5])_$($b[6])_$($b[7])" }
$csv = convertfrom-csv -InputObject $txt2 -Delimiter ' ' 
$csv | ? { $_.reference -EQ $ref1 } | FL

Answer 2

这个怎么样：

Get-Content SourceFileName.txt | 
% { ($_ -Replace '(\d{2}):\d{2} (\w{2}\d{3})', 'Time : $1h|Reference : $2').Split('|')} |
Out-File TargetFileName.txt

Answer 3

以下是我的修订版：

$regex = '(\d{2}):\d{2} (\w{2}\d{3}) (\d{2} \b\w+\b \d{4}) (\d{2} \b\w+\b \d{4})'
$replace = 'Time : $1h|Reference : $2|Date of start : $3|Date of end : $4'
Get-Content SourceFileName.txt | 
% { ($_ -Replace $regex, $replace).Split('|')} |
Out-File TargetFileName.txt

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