使用powershell打开并保存xml配置文件

Question

我正在尝试创建一个脚本来修改配置文件的内容，将其保存并启动相关程序。该文件显然是一个XML文件，并与我创建的脚本我把它保存在一个正常的文本文件。这可能是我的程序无法启动的原因。那么我怎样才能保存在XML？

下面是脚本：

$content = [XML](Get-Content("path\file.config")) 
$content = $content.replace("IP address","Other IP address") 
$content | out-file "path\file.config" 

在此先感谢

本

Answer 1

$content = Get-Content "path\file.config" | Out-String 
$content = $content.replace("IP address","Other IP address") 
$content = ([xml]$content).save("path\file.config") 

Answer 2

这个代码是没有意义的，因为它是部分处理xml和文字部分。我试图用下面的评论来说明这一点，以及如何解决这个问题。

实施例输入

<?xml version="1.0"?> 
<mytag>IP address</mytag> 

击穿

# this line imports the file, and turns it into an XML object. so far so good. 
$content = [XML](Get-Content("path\file.config")) 

# this line leads to an error: 
# Method invocation failed because [System.Xml.XmlDocument] does not contain a method named 'replace' 
$content = $content.replace("IP address","Other IP address") 

# this line will try to export the xml object (unsuccessfully) 
$content | out-file "path\file.config" 

方法1 - 当作文本

$content = (Get-Content("path\file.config")) 
$content = $content.replace("IP address","Other IP address") 
$content | out-file "path\file.config" 

方法2 - 当作XML

$content = [XML](Get-Content("path\file.config")) 
$content = $content.mytag = "Other IP Address" 
$content.OuterXml | out-file "path\file.config"