如何删除CSV中以“#"开头的行或为空|电源外壳 [英] How to delete lines in a CSV that start with "#" or are blank | PowerShell

问题描述

我有一个CSV文件，其中包含将近4,000多个行，其中的随机空白行和注释始终以#"开头.我正在尝试导入CSV文件，同时跳过空白行或以#"开头的行.我已经尝试过了:

I have a CSV file that contains nearly 4,000+ lines, with random blank lines and comments starting with "#" throughout. I'm trying to import the CSV file while skipping lines that are blank or start with "#". I've tried this:

$csvFile = Import-Csv .\example.csv | Where-Object {$_ -notlike '#*'} | Where-Object {$_ -ne ''}

，这不起作用.我也尝试过:

and this does not work. I've also tried:

$csvFile = Import-Csv .\example.csv
Get-Content $csvFile | Where-Object {$_ -notlike '#*' | Where-Object {$_ -ne ''} | Set-Content newCSV.csv

这两种情况都会导致文件以所有行导入，并且与原始文件没有任何变化.有一个更好的方法吗?我很困惑.

Both of these result in the file being imported with all lines, and nothing has changed from the original. Is there a better way to do this? I'm stumped.

推荐答案

这应该可以完成:

Get-Content $csvPath | Where-Object { $_ -notmatch '^#|^$' } | Set-Content $strippedCsvPath

^#匹配以#开头的行，^$匹配空行.

^# matches line starting with #, ^$ matches empty line.

您可以使用以下代码段简化测试:

You can simplify testing using following snippet:

$csv = 'line1',
'#line2',
'',
'line4'

$csv | Where-Object { $_ -notmatch '^#|^$' }

如果空格(不重要)，如果空行表示仅包含空格的行，则可以将RegEx略微更改为^\s*#|^\s*$，如下所示:

If whitespaces are (un)important, if empty lines means lines containing only whitespaces, you can slightly change RegEx to ^\s*#|^\s*$ as follows:

$csv = 'line1',
'#line2',
'',
'    ',
'   #comment',
'line4'

$csv | Where-Object { $_ -notmatch '^\s*#|^\s*$' }

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