The xsl file transforms it into HTML. I can't open the file in IE just fine and view the expected results. I would like to via a PowerShell script convert this in plain HTML. How would I go about that?
xsl文件将其转换为HTML。我无法在IE中打开文件,并查看预期的结果。我想通过PowerShell脚本将其转换为纯HTML。我该怎么办呢?
5
This blog entry has a code snippet that should work. It uses the System.Xml.Xsl.XslCompiledTransform
.NET class to do the XSL transformation. The rest is just for getting the input and displaying the output.
此博客条目有一个应该有效的代码片段。它使用System.Xml.Xsl.XslCompiledTransform .NET类来执行XSL转换。其余的只是用于获取输入和显示输出。
This was originally a comment, but I guess I'll make it an answer so it's easier for other people who are searching for a solution.
这本来是一个评论,但我想我会把它作为答案,这样对于寻找解决方案的其他人来说更容易。
2
The PowerShell Community Extensions has a Convert-Xml
that will do an XSL transform on the XML. If the resulting file isn't valid HTML then you need to work on the XSL file.
PowerShell社区扩展有一个Convert-Xml,它将对XML进行XSL转换。如果生成的文件不是HTML,那么您需要处理XSL文件。
0
If you want to return an xml object instead of writing the output to file so you can do more in-line activities. this will work. in addition i split the creation of the processor into a separate function so that you can create it once and re-use which is more memory friendly.
如果要返回xml对象而不是将输出写入文件,以便可以执行更多内联活动。这会奏效。此外,我将处理器的创建拆分为一个单独的功能,以便您可以创建一次并重新使用,这对内存更友好。
function Invoke-TransformXML($path,$styleSheetPath,$output,$parameters, $compiledtransform)
{
if( ! (test-path $path )) { Throw"XML input file not found: $path"}
$path = resolve-path $path
if ( ! (Test-Path $compiledtransform))
{
if( ! ($compiledtransform.GetType() -eq [System.Xml.Xsl.XslCompiledTransform] ))
{
$ctType = $compiledtransform.GetType() ;
Throw "Compiled transform is wrong type: $ctType"
}
else
{
$xslt = $compiledtransform
}
}
if (($compiledtransform -eq $null) )
{
if( ! (test-path $styleSheetPath ) ) { Throw"XSL template file not found: $styleSheetPath"}
$styleSheetPath = Resolve-Path $styleSheetPath
$xslt = Get-CompiledTransform $styleSheetPath
}
$transformed = New-Object System.IO.MemoryStream
try
{
$xslt.Transform([string]$path, [System.Xml.Xsl.XsltArgumentList]$arglist, [System.IO.Stream]$transformed)
$transformed.Position = 0
#$reader = New-Object System.Xml.XmlTextReader($ms)
$outdoc = New-Object System.Xml.XmlDocument
$outdoc.Load($transformed)
# close stream, we are done with it
$transformed.Close()
return $outdoc
} Finally {
$transformed.Close()
}
}
function Get-CompiledTransform($styleSheetPath)
{
if( ! (test-path $styleSheetPath ) ) { Throw"XSL template file not found: $styleSheetPath"}
$styleSheetPath = Resolve-Path $styleSheetPath
if( [System.Diagnostics.Debugger]::IsAttached )
{
$xslt = New-Object System.Xml.Xsl.XslCompiledTransform( $true )
}
else
{
$xslt = New-Object System.Xml.Xsl.XslCompiledTransform( $false )
}
$arglist = new-object System.Xml.Xsl.XsltArgumentList
foreach( $param in $parms )
{
if ($parms.Name)
{
$paramName = $parms.Name
$paramNamespaceUri = $parms.NamespaceUri
$paramValue = $parms.Value
$arglist.AddParam($paramName, $paramNamespaceUri, $paramValue)
}
}
$xsltSettings = New-Object System.Xml.Xsl.XsltSettings($false,$true)
$xslt.Load($styleSheetPath, $xsltSettings, (New-Object System.Xml.XmlUrlResolver))
return $xslt
}
5
This blog entry has a code snippet that should work. It uses the System.Xml.Xsl.XslCompiledTransform
.NET class to do the XSL transformation. The rest is just for getting the input and displaying the output.
此博客条目有一个应该有效的代码片段。它使用System.Xml.Xsl.XslCompiledTransform .NET类来执行XSL转换。其余的只是用于获取输入和显示输出。
This was originally a comment, but I guess I'll make it an answer so it's easier for other people who are searching for a solution.
这本来是一个评论,但我想我会把它作为答案,这样对于寻找解决方案的其他人来说更容易。
2
The PowerShell Community Extensions has a Convert-Xml
that will do an XSL transform on the XML. If the resulting file isn't valid HTML then you need to work on the XSL file.
PowerShell社区扩展有一个Convert-Xml,它将对XML进行XSL转换。如果生成的文件不是HTML,那么您需要处理XSL文件。
0
If you want to return an xml object instead of writing the output to file so you can do more in-line activities. this will work. in addition i split the creation of the processor into a separate function so that you can create it once and re-use which is more memory friendly.
如果要返回xml对象而不是将输出写入文件,以便可以执行更多内联活动。这会奏效。此外,我将处理器的创建拆分为一个单独的功能,以便您可以创建一次并重新使用,这对内存更友好。
function Invoke-TransformXML($path,$styleSheetPath,$output,$parameters, $compiledtransform)
{
if( ! (test-path $path )) { Throw"XML input file not found: $path"}
$path = resolve-path $path
if ( ! (Test-Path $compiledtransform))
{
if( ! ($compiledtransform.GetType() -eq [System.Xml.Xsl.XslCompiledTransform] ))
{
$ctType = $compiledtransform.GetType() ;
Throw "Compiled transform is wrong type: $ctType"
}
else
{
$xslt = $compiledtransform
}
}
if (($compiledtransform -eq $null) )
{
if( ! (test-path $styleSheetPath ) ) { Throw"XSL template file not found: $styleSheetPath"}
$styleSheetPath = Resolve-Path $styleSheetPath
$xslt = Get-CompiledTransform $styleSheetPath
}
$transformed = New-Object System.IO.MemoryStream
try
{
$xslt.Transform([string]$path, [System.Xml.Xsl.XsltArgumentList]$arglist, [System.IO.Stream]$transformed)
$transformed.Position = 0
#$reader = New-Object System.Xml.XmlTextReader($ms)
$outdoc = New-Object System.Xml.XmlDocument
$outdoc.Load($transformed)
# close stream, we are done with it
$transformed.Close()
return $outdoc
} Finally {
$transformed.Close()
}
}
function Get-CompiledTransform($styleSheetPath)
{
if( ! (test-path $styleSheetPath ) ) { Throw"XSL template file not found: $styleSheetPath"}
$styleSheetPath = Resolve-Path $styleSheetPath
if( [System.Diagnostics.Debugger]::IsAttached )
{
$xslt = New-Object System.Xml.Xsl.XslCompiledTransform( $true )
}
else
{
$xslt = New-Object System.Xml.Xsl.XslCompiledTransform( $false )
}
$arglist = new-object System.Xml.Xsl.XsltArgumentList
foreach( $param in $parms )
{
if ($parms.Name)
{
$paramName = $parms.Name
$paramNamespaceUri = $parms.NamespaceUri
$paramValue = $parms.Value
$arglist.AddParam($paramName, $paramNamespaceUri, $paramValue)
}
}
$xsltSettings = New-Object System.Xml.Xsl.XsltSettings($false,$true)
$xslt.Load($styleSheetPath, $xsltSettings, (New-Object System.Xml.XmlUrlResolver))
return $xslt
}