Question

我正在尝试列出路径的第一级和第二级文件夹。该脚本可以正常工作，但是我遇到此错误“您无法在空值表达式上调用方法”。知道为什么吗？

$folderPath = '\\FILSERVER\DATA$'
$PathScript = "C:\Users\adm\Desktop\Script_V.2"
$sites = "Madrid"
foreach ($site in $Sites){

#Get_Level_1_Folders
$PathShare = "\\FILSERVER\DATA$\Data_$site"
Get-ChildItem -Path $PathShare -Directory -Force -ErrorAction SilentlyContinue | Select-Object FullName | out-file "${PathScript}\level_1_${site}.txt"
(get-content "${PathScript}\level_1_${site}.txt") -notmatch "--------" | out-file "${PathScript}\level_1_${site}.txt"
(get-content "${PathScript}\level_1_${site}.txt").replace("\\FILSERVER\DATA$\Data_$site\","" ) | out-file "${PathScript}\level_1_${site}.txt"
(get-content "${PathScript}\level_1_${site}.txt") -notmatch "FullName" | out-file "${PathScript}\level_1_${site}.txt"
(get-content "${PathScript}\level_1_${site}.txt") | Foreach {$_.TrimEnd()} | Set-Content "${PathScript}\level_1_${site}.txt"
(get-content "${PathScript}\level_1_${site}.txt") | ? {$_.trim() -ne "" } | set-content "${PathScript}\level_1_${site}.txt"

#Get_Level_2_Folders
$Level_Folders = get-content "${PathScript}\level_1_${site}.txt"
foreach($lv1 in $Leve1_Folders){
Get-ChildItem -Path $PathShare\$lv1 -Directory -Force -ErrorAction SilentlyContinue | Select-Object FullName | out-file "${PathScript}\level_2_${site}_${lv1}.txt"
(get-content "${PathScript}\level_2_${site}_${lv1}.txt") -notmatch "--------" | out-file "${PathScript}\level_2_${site}_${lv1}.txt"
(get-content "${PathScript}\level_2_${site}_${lv1}.txt").replace("\\FILSERVER\DATA$\Data_$site\","") |  out-file "${PathScript}\level_2_${site}_${lv1}.txt"
(get-content "${PathScript}\level_2_${site}_${lv1}.txt") -notmatch "FullName" | out-file "${PathScript}\level_2_${site}_${lv1}.txt"
(get-content "${PathScript}\level_2_${site}_${lv1}.txt") | Foreach {$_.TrimEnd()} | Set-Content "${PathScript}\level_2_${site}_${lv1}.txt"
(get-content "${PathScript}\level_2_${site}_${lv1}.txt") | ? {$_.trim() -ne "" } | set-content "${PathScript}\level_2_${site}_${lv1}.txt"
}

Answer 1

如评论中所述，原因可能是此可扩展字符串：

"${PathScript}\level_2_${site}_${lv1}.txt"

...解析为 empty 的文件的路径。

Get-Content将打开文件-这就是为什么您不会出现任何“找不到文件”错误的原因-然后立即返回而不输出任何内容，因为在空文件中没有有意义的“行”要消耗

因此，(Get-Content ...)表达式的结果为$null，您收到了相关的错误。

您可以使用-replace运算符，该运算符将使用任意数量的字符串（包括无字符串）作为输入-只需确保您对参数进行转义即可：

(Get-Content "${PathScript}\level_2_${site}_${lv1}.txt") -replace [regex]::Escape("\\FILSERVER\DATA$\Data_$site\") |Out-File ...

或者让管道负责枚举输出，而不是依赖隐式属性枚举：

Get-Content "${PathScript}\level_2_${site}_${lv1}.txt" |ForEach-Object {$_.Replace("\\FILSERVER\DATA$\Data_$site\","")} |Out-File ...

为什么在使用replace方法时出现此错误

1 个答案: