用XML重命名PowerShell文件

Question

目前，我有一个文件夹C:\Call Archive\，它的XML文件名为index.xml，还有大量的.wav文件，所有这些文件的名称在您查看它时都没有任何意义。像AT1_ID1_TT3_ID6-1623089222.5534.WAV这样的名字

我想要做的是采用XML结构，并根据XML中的标记重命名每个.wav文件。

来源数据：

<all>
  <recording_index recapp_id="" recapp_name="" last_generated="0">
    <day_index date="2021-06-07">
      <recording target_name="Unknown recording" filename="warning%3A" filesize="0" date="1969-12-31 19:00" source="" destination=""/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623089222.5534.WAV" filesize="40K" date="2021-06-07 14:07" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623089304.5536.WAV" filesize="42K" date="2021-06-07 14:08" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091000.5538.WAV" filesize="11K" date="2021-06-07 14:36" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091138.5540.WAV" filesize="14K" date="2021-06-07 14:39" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091201.5542.WAV" filesize="14K" date="2021-06-07 14:40" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091227.5544.WAV" filesize="43K" date="2021-06-07 14:41" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091276.5546.WAV" filesize="31K" date="2021-06-07 14:41" source="John Doe" destination="5555512345"/>
    </day_index>
  </recording_index>
</all>

理想的文件名应该是06-07-2021_John Doe_5555512345.wav。注意，如果可能的话，日期将采用MM-DD-YYYY格式。

什么都能帮上忙！

Answer 1

您只需要遍历并执行重命名。如下所示：

# Establishing test data:
$XML = 
@"
<all>
  <recording_index recapp_id="" recapp_name="" last_generated="0">
    <day_index date="2021-06-07">
      <recording target_name="Unknown recording" filename="warning%3A" filesize="0" date="1969-12-31 19:00" source="" destination=""/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623089222.5534.WAV" filesize="40K" date="2021-06-07 14:07" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623089304.5536.WAV" filesize="42K" date="2021-06-07 14:08" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091000.5538.WAV" filesize="11K" date="2021-06-07 14:36" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091138.5540.WAV" filesize="14K" date="2021-06-07 14:39" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091201.5542.WAV" filesize="14K" date="2021-06-07 14:40" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091227.5544.WAV" filesize="43K" date="2021-06-07 14:41" source="John Doe" destination="5555512345"/>
      <recording target_name="10 Digit Dialing" filename="AT1_ID1_TT3_ID6-1623091276.5546.WAV" filesize="31K" date="2021-06-07 14:41" source="John Doe" destination="5555512345"/>
    </day_index>
  </recording_index>
</all>
"@

$XML = [XML]$XML

# This is the real work:
$xml.all.recording_index.day_index.recording |
ForEach-Object{
    $Date = (Get-Date $_.Date).ToString('MM-dd-yyyy')
    $Newname = "{0}_{1}_{2}.{3}" -f $Date, $_.source, $_.destination, $_.filename.Split('.')[-1]
    $Newname
    Rename-Item -Path $_.filename -NewName $Newname
}

对于您感兴趣的每个XML元素，只需将属性连接起来以获得新名称，然后使用Rename-Item执行重命名。我对文件位置有点怀疑，因为原始文件的完整路径不在XML中。您可能需要使用这些文件从目录或CD运行到目录。

我也关心命名碰撞。我不知道这是否只是一个样本数据的函数，如果是一个问题，您可能需要获得更细粒度的时间戳，以确保唯一的文件名。