[已解决]求助批处理删除文本中 第六列与第七列

KLL

文本中共8列 , 以逗号分隔 , 批处理删除 第六列与第七列
如下:
2023/08/01 23:50,15721,15722.25,15716.75,15717.75,3.25,-0.0207,997,
2023/08/01 23:55,15718,15719.75,15711,15717.5,-0.25,0.0016,1851,
2023/08/02 00:00,15717.25,15719,15714,15716.75,-0.75,1.048,534,

批处理删除后成为如下

2023/08/01 23:50,15721,15722.25,15716.75,15717.75,997,
2023/08/01 23:55,15718,15719.75,15711,15717.5,1851,
2023/08/02 00:00,15717.25,15719,15714,15716.75,534,

KLL

原文件a.txt  处理后 生成 b.txt 文件

谢谢

Five66

不考虑其他细节,应该是这样吧

1. @for /f "tokens=1-5,6,7,* delims=," %%i in (a.txt) do @echo %%i,%%j,%%k,%%l,%%m,%%p >b.txt

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hfxiang

回复 1# KLL

用gawk（http://bcn.bathome.net/tool/5.1.0/gawk.exe）也可实现

1. 
   
2. gawk -v"FS=," -v"OFS=," "{print $1,$2,$3,$4,$5,$8 FS}" a.txt>b.txt
   

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77七

1. @echo off
   
2. (for /f "useback tokens=1-7* delims=," %%a in ("a.txt") do (
   
3.         echo %%a,%%b,%%c,%%d,%%e,%%h
   
4. ))>"b.txt"
   
5. pause

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Batcher

回复 1# KLL


请参考Q-04和Q-05把bat文件和txt文件都保存为ANSI编码：
https://mp.weixin.qq.com/s/Koje4FufWxWBs7ioDy_LJA

1. @echo off
   
2. cd /d "%~dp0"
   
3. (for /f "tokens=1-8 delims=," %%a in (a.txt) do (
   
4.     echo %%a,%%b,%%c,%%d,%%e,%%h,
   
5. ))>b.txt

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KLL

运行都OK , 谢谢先进们的热心指导,Thanks

Nsqs

1. @echo off
   
2. powershell -noprofile -executionpolicy bypass "(gc 1.txt) -replace '((?:\d+\.?\d*,){5}).+,(.+)','$1$2'" >new.txt
   
3. pause

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