Question

我想从xml文件中提取所有窗口和所有视点节点中name-attributes的值。

<?xml version='1.0' encoding='utf-8' ?>
        <workbook>
            <windows>
                <window class='dashboard' name='D1'>
                  <viewpoints>
                    <viewpoint name='V1'> </viewpoint>
                    <viewpoint name='V2'> </viewpoint>
                    <viewpoint name='V3'> </viewpoint>
                </viewpoints>
                </window>
                <window class='dashboard' name='D2'>
                  <viewpoints>
                    <viewpoint name='V10'> </viewpoint>
                    <viewpoint name='V11'> </viewpoint>
                  </viewpoints>
                 </window>
            </windows>
        </workbook>

这是我写的代码：

[XML]$doc = get-content -path 'W:\Demo1.xml'


$objs = @()
$dashboards = $doc.SelectNodes("//window[@class = 'dashboard']")

foreach ($dashboard in $dashboards)
{
   $worksheets = $dashboard.SelectNodes("//viewpoint[@name]")

    foreach ($worksheet in $worksheets)
    {
        $obj = new-object psobject -prop @{Dashboard=$dashboard.name; Worksheet = $worksheet.name}; 
        $objs += $obj;

    }  
 } 

$objs

我的期望：

dashboard   worksheet   
D1          V1
D1          V2
D1          V3
D2          V10
D2          V11

我得到了什么：

dashboard   worksheet   
D1          V1
D1          V2
D1          V3
D1          V10
D1          V11
D1          V14
D2          V1
D2          V2
D2          V3
D2          V10
D2          V11
D2          V14

这里有什么问题？结果与我对嵌套循环如何工作的理解完全矛盾。

Answer 1

以/字符开头的XPath意味着它将从文档根节点开始。要从上下文节点创建相对XPath，您需要在.之前放置/。

所以你的代码应该是：

[XML]$doc = get-content -path 'W:\Demo1.xml'


$objs = @()
$dashboards = $doc.SelectNodes("//window[@class = 'dashboard']")

foreach ($dashboard in $dashboards)
{
   $worksheets = $dashboard.SelectNodes(".//viewpoint[@name]")

    foreach ($worksheet in $worksheets)
    {
        $obj = new-object psobject -prop @{Dashboard=$dashboard.name; Worksheet = $worksheet.name}; 
        $objs += $obj;

    }  
 } 

$objs

Answer 2

您还可以简化并采用xpath从信息中心窗口中选择所有viewpoint个节点：

[XML]$doc = get-content -path 'W:\Demo1.xml'

$objs = @()
$worksheets = $doc.SelectNodes("//window[@class = 'dashboard']//viewpoint[@name]")

foreach ($worksheet in $worksheets)
{
    $obj = new-object psobject -prop @{Dashboard=$worksheet.ParentNode.ParentNode.name; Worksheet = $worksheet.name}; 
    $objs += $obj;
}  

$objs

这是一个没有xpath的解决方案：

[XML]$doc = get-content -path 'W:\Demo1.xml'

$objs = $doc.DocumentElement.windows.window | Where class -eq 'dashboard' | foreach {
    $dashboardName = $_.name
    $_.viewpoints.viewpoint | foreach {
        new-object psobject -prop @{Dashboard=$dashboardName; Worksheet = $_.name}
    }
}

使用powershell通过xml节点嵌套循环

2 个答案: