如何有选择地替换多个空白字符？

Question

我有一个文本文件“ c：\ zz.txt”，其摘录如下：

#N ABSA ALL ROUNDER FoF
#D UT-ABSAAG
#P 20190215    393.83    393.83   0.00

#N ABSA BALANCED FUND
#D UT-ABSABA
#P 20190215    432.28    432.28   0.00

#N COMMUNITY GILT
#D UT-COM-G
#P 20190215    151.21    151.21   8.59

我想将其导出为以下格式的.TXT文件：

UT_Name,UT_Code,Date,Value1,Value2,Vol
ABSA ALL ROUNDER FoF,UT-ABSAAG,20190215,393.83,393.83,0
ABSA BALANCED FUND,UT-ABSABA,20190215,432.28,432.28,0
COMMUNITY GILT,UT-COM-G,20190215,151.21,151.21,8.59

我的下面的代码

clear-Host
get-content -raw "c:\zz.txt" | % { $_ -replace '(#N)',"`r`n"} |% { $_ -replace '(#D|#P)',','}|Set-Content ZZ1.txt

输出：

ABSA ALL ROUNDER FoF, UT-ABSAAG, 20190215    393.83    393.83   0.00
ABSA BALANCED FUND, UT-ABSABA, 20190215    432.28    432.28   0.00
COMMUNITY GILT, UT-COM-G, 20190215    151.21    151.21   8.59

问题：问题原来是原始数据集中的数据集“ #P”是固定长度的，这意味着我不能简单地用逗号替换所有空格，因为这也会影响字段名“ #N”和“ #D”，在此之上我不想影响。

如何有选择地替换空格？

Answer 1

这使用命名的捕获组来获取项目，然后将其导出到CSV文件。

# fake reading in a raw text file
#    in real life, use Get-Content -Raw
$InStuff = @'
#N ABSA ALL ROUNDER FoF
#D UT-ABSAAG
#P 20190215    393.83    393.83   0.00

#N ABSA BALANCED FUND
#D UT-ABSABA
#P 20190215    432.28    432.28   0.00

#N COMMUNITY GILT
#D UT-COM-G
#P 20190215    151.21    151.21   8.59
'@

# split into blocks, trim unwanted whitespace, filter out the blank block
$SplitInStuff = ($InStuff -split '#N').Trim().Where({$_})

$Results = foreach ($SIS_Item in $SplitInStuff)
    {
    $Null = $SIS_Item -match '(?sm)(?<UT_Name>^.+$).*#D (?<UT_Code>.+).*#P (?<Date>\d+)\s+(?<Value1>[0-9.]+)\s+(?<Value2>[0-9.]+)\s+(?<Vol>[0-9.]+)'
    [PSCustomObject]@{
        # the ".Trim()" was needed to remove leftover EOL/NewLine/space chars
        UT_Name = $Matches.UT_Name.Trim()
        UT_Code = $Matches.UT_Code.Trim()
        Date = $Matches.Date.Trim()
        Value1 = $Matches.Value1.Trim()
        Value2 = $Matches.Value2.Trim()
        Vol = $Matches.Vol.Trim()
        }
    }

$Results |
    Export-Csv -LiteralPath "$env:TEMP\milkywaypizza_ProductInfo.csv" -NoTypeInformation

CSV文件的内容...

"UT_Name","UT_Code","Date","Value1","Value2","Vol"
"ABSA ALL ROUNDER FoF","UT-ABSAAG","20190215","393.83","393.83","0.00"
"ABSA BALANCED FUND","UT-ABSABA","20190215","432.28","432.28","0.00"
"COMMUNITY GILT","UT-COM-G","20190215","151.21","151.21","8.59"

Answer 2

其他方法：

$Data = @'
#N ABSA ALL ROUNDER FoF
#D UT-ABSAAG
#P 20190215    393.83    393.83   0.00

#N ABSA BALANCED FUND
#D UT-ABSABA
#P 20190215    432.28    432.28   0.00

#N COMMUNITY GILT
#D UT-COM-G
#P 2019021    151.21    151.21   8.59
'@

#template for learn schema
$template=@'
#N {Vol_ABSA_ALL_ROUNDER_FoF*:ABSA ALL ROUNDER FoF}
#D UT-{UT_Name:ABSAAG}
#P {Date:20190215}    {Value1:393.83}    {Value2:393.83}   {Vol:0.00}

#N {Vol_ABSA_ALL_ROUNDER_FoF*:ABSA ALL ROUNDER FoF 2}
#D UT-{UT_Name:ABSAAG2}
#P {Date:20190216}    {Value1:393.83}    {Value2:393.83}   {Vol:0.00}
'@

$Data | ConvertFrom-String -TemplateContent $template | export-csv "c:\temp\result.csv" -NoType

Answer 3

如何有选择地替换多个空白字符？

例如，通过仅使用一个量词-replace " {2,}",','

选择多个空格

以下脚本基于您的一个班轮：

## Q:\Test\2019\03\28\SO_55387785.ps1
$FileIn = '.\zz.txt'
$FileOut= '.\zz1.txt'

 Set-Content $FileOut -Value "UT_Name,UT_Code,Date,Value1,Value2,Vol"
(Get-Content $FileIn -raw) -replace "(`r?`n)?#N " -replace "`r?`n(#D|#P) | {2,}",',' |
 Add-Content $FileOut

产生以下输出：

> Get-Content .\ZZ1.txt
UT_Name,UT_Code,Date,Value1,Value2,Vol
ABSA ALL ROUNDER FoF,UT-ABSAAG,20190215,393.83,393.83,0.00
ABSA BALANCED FUND,UT-ABSABA,20190215,432.28,432.28,0.00
COMMUNITY GILT,UT-COM-G,20190215,151.21,151.21,8.59

如何将多行字符串导出为单行文本

3 个答案:

如何有选择地替换多个空白字符？