在批处理或 PoSh 的引号之间删除逗号 [英] Remove comma when between quotes from batch or PoSh

问题描述

我有一个包含以下内容的 CSV 文件:

A,B,CD,"E,F",GH,I,"J,K,L"

我需要删除引号之间的逗号(也删除引号，但这并不重要):

A,B,CD,EF,GH,I,JKL

我查看了 PoSh

正则表达式是 "([^"]*)" 匹配 "，然后捕获到组 1(即 $match.Groups[1].Value) 除 " 之外的任何零个或多个字符，然后匹配 ".如果您对引号进行了转义，则需要增强，但方法是相同的.

I have a CSV file with a content like:

A,B,C
D,"E,F",G
H,I,"J,K,L"

I need to remove the commas when between quotes (also remove the quotes, but that is not so important):

A,B,C
D,EF,G
H,I,JKL

I looked at PoSh -replace operator but I can't get it to capture multiple group values:

 PS >"D,`"E,F`",G" -replace "`"((?:[^,`"]+)\,?)+`"", '$1'
 D,F,G

as you can see when the group is repeated, only the last value captured is preserved. Is there a way to do the transformation I want?

https://regex101.com/r/ON1rgp/1/

解决方案

You may define a callback to pass to the Regex::Replace method where you may just grab the part between quotes and remove all , there:

$callback = {  param($match) $match.Groups[1].Value.Replace(',','') }
$s =  "D,`"E,F`",G"
$rex = [regex]'"([^"]*)"'
$rex.Replace($s, $callback)

The regex is "([^"]*)" that matches ", then captures into Group 1 (i.e. the $match.Groups[1].Value) any zero or more chars other than " and then matches ". It will need enhancing in case you have escaped quotes, but the approach will be the same.

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