如何獲得多個xml文件的相同值，然後使用powershell命令創建一個文件夾？

Question

樣本文件的XML 1個

enter code here 
<?xml version="1.0"?> 
<catalog> 
<book id="bk101"> 
    <author>Gambardella, Matthew</author> 
    <title>XML Developer's Guide</title>` 
    <genre>Test1</genre> 
    <price>44.95</price> 
    <publish_date>2000-10-01</publish_date> 
    <description>An in-depth look at creating applications 
    with XML.</description> 

樣本文件的XML 2

enter code here 
<?xml version="1.0"?> 
<catalog> 
<book id="bk101"> 
    <author>Gambardella, Matthew</author> 
    <title>XML Developer's Guide</title> 
    <genre>Test2</genre> 
    <price>44.95</price> 
    <publish_date>2000-10-01</publish_date> 
    <description>An in-depth look at creating applications 
    with XML.</description> 

需要得到 「流派」，然後創建每個文件夾的價值？

Answer 1

隨着移動文件的一部分：

Get-ChildItem *.xml | % { 
      [xml]$x=get-content $_ ; 
      $newpath=$x.catalog.book.genre 
      new-item "$newpath"-type Directory -ErrorAction SilentlyContinue 
      if (test-path "$newpath") { 
       move-item "$_" -destination "$newpath" 
      } 
      } 

Answer 2

我知道你想用XML發現每一個流派目錄中的文件的話，如果你有每個文件的一本書：

Get-ChildItem *.xml | % { [xml]$x=get-content $_ ;new-item $x.catalog.book.genre -type Directory -ErrorAction SilentlyContinue }