如何从 Powershell 中 7Zip 的错误结果中仅获取错误文件名？

Question

我浏览了很多帖子，但找不到对这个问题的引用。我有一个代码可以测试一组名称中具有特定模式的 zip 文件。测试后，我看到以下错误格式和 none 错误文件。我可以使用 Archives with Error 解析它，但我无法获取错误文件的名称。我的最终目标是从下面的 7zip 错误结果中获取所有错误文件的文件名。

你能帮我解决这个问题吗？谢谢


7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30

Scanning the drive for archives:
1 file, 40021368 bytes (39 MiB)

Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#10-AT_2020-08-06_13268.txt.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#10-AT_2020-08-06_13268.txt.gz
Type = gzip
Headers Size = 10


Sub items Errors: 1

Archives with Errors: 1

Sub items Errors: 1

7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30

Scanning the drive for archives:
1 file, 40021368 bytes (39 MiB)

Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#210-AT_2020-08-06_13268.txt - Copy.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#210-AT_2020-08-06_13268.txt - Copy.gz
Type = gzip
Headers Size = 10


Sub items Errors: 1

Archives with Errors: 1

Sub items Errors: 1

7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30

Scanning the drive for archives:
1 file, 56581 bytes (56 KiB)

Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#DT-F@_2020-08-06_13268.txt.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#DT-F@_2020-08-06_13268.txt.gz
Type = gzip
Headers Size = 10

Everything is Ok

Size:       504716
Compressed: 56581

Answer 1

然后使用正则表达式获取 Path = 文件名

对于演示，我使用的是 Here-String。在现实生活中，您可能需要使用

从文件中加载它

$zipLog = Get-Content -Path 'The7ZipErrorLog.txt' -Raw

（此处需要 -Raw 才能在一个多行字符串中获取所有内容）

$zipLog = @"
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30

Scanning the drive for archives:
1 file, 40021368 bytes (39 MiB)

Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#10-AT_2020-08-06_13268.txt.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#10-AT_2020-08-06_13268.txt.gz
Type = gzip
Headers Size = 10


Sub items Errors: 1

Archives with Errors: 1

Sub items Errors: 1

7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30

Scanning the drive for archives:
1 file, 40021368 bytes (39 MiB)

Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#210-AT_2020-08-06_13268.txt - Copy.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#210-AT_2020-08-06_13268.txt - Copy.gz
Type = gzip
Headers Size = 10


Sub items Errors: 1

Archives with Errors: 1

Sub items Errors: 1

7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30

Scanning the drive for archives:
1 file, 56581 bytes (56 KiB)

Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#DT-F@_2020-08-06_13268.txt.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#DT-F@_2020-08-06_13268.txt.gz
Type = gzip
Headers Size = 10

Everything is Ok

Size:       504716
Compressed: 56581
"@

$zipLog -split '7-Zip.+Igor Pavlov.+\d{4}-\d{2}-\d{2}' | Where-Object { $_ -match 'Archives with Errors' } | ForEach-Object {
    ([regex] '(?im)^Path = (.+)').Match($_).Groups[1].Value
}

输出：

C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#10-AT_2020-08-06_13268.txt.gz
C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#210-AT_2020-08-06_13268.txt - Copy.gz

^{在正则表达式中，(?im) 意味着使匹配工作 case-insensitive 并让 ^ 或 $ 锚匹配换行符，因为每个text bloxk 是多行字符串。}

如何从 Powershell 中 7Zip 的错误结果中仅获取错误文件名？

How can I get only Error file names from error result of 7Zip in Powershell?

powershell

parsing

error-handling

7zip