如何在 powershell 中过滤列中的空值

Question

我有输入文件text.csv

"#P","A","E","S"
"1","2","3","4"
"3","5","6",""
"A","B","C",""
"D","F","G","H"

我想要输出 csv 文件 export.csv as

"P","A","E","S"
"1","2","3","4"
"D","F","G","H"




 $File1 = Import-Csv -Path "test.csv" -Header "P","A","E","S" | Where { $_.S -ne $null } | Export-Csv "export.csv" -Force -NoTypeInformation

Answer 1

如评论中所述，csv 文件通常以纯文本格式存储表格数据（数字和文本），在这种情况下，每一行将具有相同数量的字段。

基于此，您可能会认为您永远不应该有一个

$Null

，而只是一个空字符串：

$Csv = @'
1,2,3,4
3,5,6,
A,B,C,
D,F,G,H
'@
$Csv |ConvertFrom-Csv -Header "P","A","E","S" |
    ForEach-Object { if ($Null -eq $_.S) { 'Null' } else { 'String' } }

String
String
String
String

但要小心这个假设，不幸的是，情况并非总是如此。如果不仅省略了最后一个值，还省略了前面的逗号（请注意，在以下示例中，

A,B,C

后面没有逗号）：

$Csv = @'
1,2,3,4
3,5,6,
A,B,C
D,F,G,H
'@
$Csv |ConvertFrom-Csv -Header "P","A","E","S" |
    ForEach-Object { if ($Null -eq $_.S) { 'Null' } else { 'String' } }

String
String
Null
String

除此之外，

Import-Csv

和

ConvertFrom-Csv

命令中存在一个错误，其中 ConvertFrom-Csv 读取的空列的最后一个单元格不一致

$Null

#17702

：

$Csv = @'
1,2,3,
3,5,6,
A,B,C,
D,F,G,
'@
$Csv |ConvertFrom-Csv -Header "P","A","E","S" |
    ForEach-Object { if ($Null -eq $_.S) { 'Null' } else { 'String' } }

String
String
String
Null

因此，我会比抱歉更安全，并测试

$Null

(

$Null -ne $_.S

)¹⁾ and 一个空字符串 (

$_.S -ne ''

) 可以简化为：

... |Where-Object { $_.S } |...

这个结果在所谓的“Truthy”条件下，如果相关字段不为空（或

$Null

），另请参见Conversion to Bool：

$Csv = @'     # Or as in your revised question:
1,2,3,4       # "1","2","3","4"
3,5,6,        # "3","5","6",""
A,B,C,        # "A","B","C",""
D,F,G,H       # "D","F","G","H"
'@
# Import-Csv -Path "test.csv" -Header "P","A","E","S" |
#     Where-Object { $_.S } |Export-Csv "export.csv" -Force -NoTypeInformation
$Csv |ConvertFrom-Csv -Header "P","A","E","S" |
    Where-Object { $_.S } |ConvertTo-Csv

"P","A","E","S"
"1","2","3","4"
"D","F","G","H"

1)如果你检查

$Null

;

$Null

应该在比较运算符的左边