PowerShell解压ZIP后提取XML标签属性值无输出问题求助

安全阿岳

2026-9-29

错误原因

路径定义使用全角中文引号，且分号错误写入路径字符串内部，导致XML文件路径无效
Select-Xml 未指定读取的XML文件路径，且存在左括号缺失的语法错误
需求是提取<Report>标签的属性值，原代码错误取了节点InnerText内容
Set-Content 未传入待写入的内容，所以输出文件为空

修复后代码

echo "unzipping file..."
powershell -Command "Expand-Archive C:\path_to_zip_file -Destinationpath C:\path_to_to_destination_folder -Force;
$Path = 'C:\path_to_to_destination_folder\file.xml';
$reportNode = (Select-Xml -Path $Path -XPath '//Report').Node;
$outputStr = 'name {0} result_many fail, summary {1}, yes {2} , no  {3}, finished {4}' -f $reportNode.name, $reportNode.summary, $reportNode.yes, $reportNode.no, $reportNode.finished;
Set-Content -Path 'C:\path_to_to_destination_folder\result.txt' -Value $outputStr -Encoding UTF8"
echo "done"