Powershell - 將.txt數據輸出為CSV

Question

我希望您可以幫助PowerShell初學者將.txt文件中的數據輸出到CSV。

要記住的事情：

• 每個文本文件都包含多個需要提取的條目
• 終端將始終為A，后跟3位數
• 詢問者總是從C開始，但可以是不同的長度
• 文件中有兩個出生日期模式
• 這些空格不是在txt中以制表符分隔，也不是所有相同的間距量。

這是我想要的CSV輸出（道歉我沒有足夠的代表來上傳圖片！）

+------------+------+----------+----------+-------------------+----------+
| Date       | Time | Terminal | Enquirer | Enquiry           | DOB      |
+------------+------+----------+----------+-------------------+----------+
| 29/05/2017 | 1227 | A999     | CA75849  | DOE/JOHN          |  / /1988 |
+------------+------+----------+----------+-------------------+----------+
| 29/05/2017 | 1424 | A999     | CA75849  | SMITH/SIMON/PETER |  / /1967 |
+------------+------+----------+----------+-------------------+----------+
|            |      |          |          |                   |          |
+------------+------+----------+----------+-------------------+----------+

數據.txt文件的示例

START                 TERMINAL    USER        ENQUIRER                  TERMINAL IP

========================================================================================================================

29/05/17 1227       A999        CA75849       8875849 OCBA NCPS RBC/12/1960



        SEARCH REF                      NAME : DOE/JOHN PAGE CODE =  

                                    DATE OF BIRTH :   /  /1988




========================================================================================================================
29/05/17 1424       A999        CA75849       8875849 OCBA NCPS RBC/60/2111                                

        SEARCH REF                        NAME : SMITH/SIMON/PETER CHAPTER CODE =  

                                    DATE OF BIRTH :   /  /1967


========================================================================================================================

請不要因為我的廢話Powershell腳本而暈倒。

我設法獲得的最好的是擁有正確的信息，但是單元格中的'System.Object []'而不是實際的結果字符串。

$XmlDocument = Get-Content -Path "C:\Script Projects\Log\880006.txt"
$TodaysCSV = "C:\Script Projects\Log\880006.csv"

$TopLine = $XmlDocument | Select-String "A1" 
$Enquiry = $XmlDocument | Select-String "Search"
$DOB = $XmlDocument | Select-String "BIRTH"

$toptop = $topline -split " "

$Date = $toptop | Select-String -Pattern "^\d{2}/\d{2}/\d{2}$"
$Time = $toptop | Select-String -Pattern "^\d{4}$"
$Terminal = $toptop | Select-String -Pattern "^A\d{3}$"
$Enquirer = $toptop | Select-String -Pattern "C\w\d{5}" 

$csv = New-Object -TypeName PSObject 

Add-Member -InputObject $csv -MemberType NoteProperty -Name "Date"  -Value $Date
Add-Member -InputObject $csv -MemberType NoteProperty -Name "Time"  -Value $Time
Add-Member -InputObject $csv -MemberType NoteProperty -Name "Terminal"  -value $Terminal
Add-Member -InputObject $csv -MemberType NoteProperty -Name "Enquirer"  -value $Enquirer
Add-Member -InputObject $csv -MemberType NoteProperty -Name "Enquiry"  -value $Enquiry
Add-Member -InputObject $csv -MemberType NoteProperty -Name "DOB"  -value $DOB


$csv | Export-Csv $TodaysCSV -NoTypeInformation

我得到了這個輸出

    +-----------------+-----------------+-----------------+-----------------+-----------------+-----------------+
| Date            | Time            | Terminal        | Enquirer        | Enquiry         | DOB             |
+-----------------+-----------------+-----------------+-----------------+-----------------+-----------------+
| System.Object[] | System.Object[] | System.Object[] | System.Object[] | System.Object[] | System.Object[] |
+-----------------+-----------------+-----------------+-----------------+-----------------+-----------------+
|                 |                 |                 |                 |                 |                 |
+-----------------+-----------------+-----------------+-----------------+-----------------+-----------------+
|                 |                 |                 |                 |                 |                 |
+-----------------+-----------------+-----------------+-----------------+-----------------+-----------------+

我也嘗試過這種方式，但我只得到日期或玩耍我可以獲得所有數據，但只能在單列中

$Total2 = "$date2", "$time2", "$Terminal2", "$Enquirer", "$Enquiry" 

$Columns = "Date", "Time", "Terminal", "Enquirer", "Enquiry"

$total2 | ConvertFrom-Csv -Header $Columns| export-csv $TodaysCSV -NoTypeInformation

但是不記得我是怎么做到的，如果我從該文件導入PowerShell並導出為CSV，我仍然不會讓我水平顯示它。

我設法得到的最好的是這個

Date    
----    
29/05/17
29/05/17




Time
----
1227
1424




Terminal
--------
A999    
A999    




Enquirer
--------
CA75849 
CA75849 




Enquiry                                                                         
-------                                                                         
SEARCH REF                      NAME : DOE/JOHN CHAPTER CODE =        
SEARCH REF                      NAME : SMITH/SIMON/PETER CHAPTER CODE = 

預先感謝您的幫助！

Answer 1

我用了不同的方法，

• 將文件拆分為用虛線和。分隔的部分
• 3個不同的RegExes使用命名（捕獲組）來grep每個行類型的屬性。
  RegEx101.com上的第一個
• 所有找到的屬性都存儲在一個中間$ Row對象中，如果所有RegEx匹配都收集在$Csv

---

## Q:\Test\2018\07\06\SO_51209341.ps1
$FileIn   = '.\SO_51209341_data.txt'
$TodayCsv = '.\SO_51209341_data.csv'

$RE1 = [RegEx]'(?m)(?<Date>\d{2}\/\d{2}\/\d{2}) (?<Time>\d{4}) +(?<Terminal>A\d{3}) +(?<User>C[A-Z0-9]+) +(?<Enquirer>.*)$'
$RE2 = [RegEx]'\s+SEARCH REF\s+NAME : (?<Enquiry>.+?) (PAGE|CHAPTER) CODE ='
$RE3 = [RegEx]'\s+DATE OF BIRTH : (?<DOB>[0-9 /]+?/\d{4})'

$Sections = (Get-Content $FileIn -Raw) -split "={30,}`r?`n" -ne ''

$Csv = ForEach($Section in $Sections){
    $Row= @{} | Select-Object Date,Time,Terminal,User,Enquirer,Enquiry,DOB
    $Cnt = 0
    If ($Section -match $RE1){++$Cnt
        $Row.Date     = $Matches.Date
        $Row.Time     = $Matches.Time
        $Row.Terminal = $Matches.Terminal
        $Row.User     = $Matches.User
        $Row.Enquirer = $Matches.Enquirer.Trim()
    }
    If ($Section -match $RE2){++$Cnt
        $Row.Enquiry  = $Matches.Enquiry
    }
    If ($Section -match $RE3){++$Cnt
        $Row.DOB      = $Matches.DOB
    }
    if ($Cnt -eq 3){$Row}
}

$csv | Format-Table
$csv | Export-Csv $Todaycsv -NoTypeInformation

示例輸出修訂版

> . Q:\Test\2018\07\06\SO_51209341.ps1

Date     Time Terminal User    Enquirer                      Enquiry           DOB
----     ---- -------- ----    --------                      -------           ---
29/05/17 1227 A999     CA75849 8875849 OCBA NCPS RBC/12/1960 DOE/JOHN            /  /1988
29/05/17 1424 A999     CA75849 8875849 OCBA NCPS RBC/60/2111 SMITH/SIMON/PETER   /  /1967

Answer 2

這里的問題是您需要在導出之前將對象轉換為字符串。

要使代碼正常工作，您可以稍微修改對象創建：

$csv = @()
for ($i = 0;$i -lt $Date.Length; $i++) {
$obj = New-Object -TypeName PSObject 
  Add-Member -InputObject $obj -MemberType NoteProperty -Name "Date"  -Value $Date[$i]
  Add-Member -InputObject $obj -MemberType NoteProperty -Name "Time"  -Value $Time[$i]
  Add-Member -InputObject $obj -MemberType NoteProperty -Name "Terminal"  -value $Terminal[$i]
  Add-Member -InputObject $obj -MemberType NoteProperty -Name "Enquirer"  -value $Enquirer[$i]
  Add-Member -InputObject $obj -MemberType NoteProperty -Name "Enquiry"  -value $Enquiry[$i]
  Add-Member -InputObject $obj -MemberType NoteProperty -Name "DOB"  -value $DOB[$i]
  $csv += $obj
}

說明：

這里的問題是你應該創建一個數組對象，同時你應該創建對象數組。 這就是為什么在導出時你得到System.Object[]而不是期望值。

注意 ：看看你粘貼的代碼和文件格式。 這一行：

$TopLine = $XmlDocument | Select-String "A1" 

應該

$TopLine = $XmlDocument | Select-String "A999"