xml - 比较Powershell中同一父节点的两个xml元素

Question

<department>
    <employee name="ABC">
    <salary value="basic">5000</salary>
    <salary value="allowance">8000</salary>
    </employee>
    <employee name="XYZ">
    <salary value="basic">6000</salary>
    <salary value="allowance">3000</salary>
    </employee>
    </department>

如何在Powershell中获得津贴额大于基本津贴的雇员姓名。

我在下面尝试过，但没有给我结果

$result= $xml |Select-Xml '//department'| 
ForEach{
$basic=$_.Node.employee.salary.value | where {$_.Node.employee.salary.value -eq 'basic'}
$allowance=$_.Node.employee.salary.value | where {$_.Node.employee.salary.value -eq 'allowance'}
if ($allowance -ge $basic)
{
$name=$_.Node.employee.name
}
}
echo $result

Answer 1

如果要使用Select-Xml，可以执行以下操作:

$node = $xml | Select-Xml -XPath "//department/employee[./salary[@value='basic'] < ./salary[@value='allowance']]"
$node.node.name # Output the employee name

有关xpath表达式示例，请参见XPath Examples。有关XPath运算符的更多信息，请参见Boolean, Comparison, and Set Expressions，包括在必要时转义<和>。