如何更改日期格式列文本文件中的Windows

Question

我有在多行下列数据的txt文件：

10/01/2015,125.99,129.00,125.00,128.97,1615600 
10/02/2015,129.92,129.92,127.01,128.81,509800 
10/05/2015,129.50,130.00,125.50,126.37,876500 
10/06/2015,128.00,131.70,128.00,130.67,1043000 
10/07/2015,134.00,135.45,132.60,134.49,1181500 
10/08/2015,133.90,133.90,131.60,132.25,736700 
10/09/2015,135.01,138.50,135.00,137.80,1493200 
10/12/2015,139.90,142.80,138.12,141.85,1879800 
10/13/2015,138.80,139.50,137.75,137.99,1103600 
10/14/2015,138.34,138.79,137.11,137.92,505200 
10/15/2015,138.00,138.80,134.50,135.21,481600 
10/16/2015,135.10,140.31,133.51,138.62,2431700 
10/19/2015,138.51,139.89,137.50,137.94,521700 

我想要的日期格式在使用.bat script所有行改为YYYY/MM/DD。

我曾尝试以下：

@echo off&setlocal 
for /f "tokens=1,2,3,* delims=,/ " %%i in ('type "kam3.txt"') do (
echo %%k/%%i/%%j,%%l 
)>"newfile.txt" 

不过，我只看到了最新的界线正在发生变化，数据的其余部分没有显示。

Answer 1

以防万一，如果你是开了PowerShell脚本，你可以试试这个：

Get-Content data.txt | Foreach { $_ -replace "^(\d\d)/(\d\d)/(\d\d\d\d),(.*)",'$3/$1/$2,$4' } 

结果：

2015/10/01,125.99,129.00,125.00,128.97,1615600 
2015/10/02,129.92,129.92,127.01,128.81,509800 
2015/10/05,129.50,130.00,125.50,126.37,876500 
2015/10/06,128.00,131.70,128.00,130.67,1043000 
2015/10/07,134.00,135.45,132.60,134.49,1181500 
2015/10/08,133.90,133.90,131.60,132.25,736700 
2015/10/09,135.01,138.50,135.00,137.80,1493200 
2015/10/12,139.90,142.80,138.12,141.85,1879800 
2015/10/13,138.80,139.50,137.75,137.99,1103600 
2015/10/14,138.34,138.79,137.11,137.92,505200 
2015/10/15,138.00,138.80,134.50,135.21,481600 
2015/10/16,135.10,140.31,133.51,138.62,2431700 
2015/10/19,138.51,139.89,137.50,137.94,521700 

您需要在PowerShell中运行这个（在命令键入powershell提示，例如）。 Get-Conetent获取输入文件的内容。 Foreach命令遍历每一行。命令replace与正则表达式匹配，并将行替换为匹配的变量（但顺序不同）。如果不清楚，请随时要求澄清。

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如果您想要不同的分隔符，请使用此命令而不是上述命令。它不查找特定的分隔符（如逗号或分号）。

Get-Content data.txt | Foreach { $_ -replace "^(\d\d)/(\d\d)/(\d\d\d\d)(.*)",'$3/$1/$2$4' }