Question

我正在尝试从计算机中获取程序列表，并将该程序列表上载到共享服务器上的文本文件中。代码和输出如下。当我运行代码行：wmic /output:C:\Users\%username%\Desktop\temp.txt product get name,version pause时，出现Invalid file name.错误。如果有人可以告诉我如何摆脱这个错误，我真的很感激。

代码：

echo. >> "N:\Individual Files\Jerry\Projects\TroubleshootingPCIssues\audits\Audit of Software Programs on Each Computer.txt"
echo. >> "N:\Individual Files\Jerry\Projects\TroubleshootingPCIssues\audits\Audit of Software Programs on Each Computer.txt"
echo %computername% >> "N:\Individual Files\Jerry\Projects\TroubleshootingPCIssues\audits\Audit of Software Programs on Each Computer.txt"
echo -------------------------------------------------------------------------------------- >> "N:\Individual Files\Jerry\Projects\TroubleshootingPCIssues\audits\Audit of Software Programs on Each Computer.txt"
wmic /output:C:\Users\%username%\Desktop\temp.txt  product get name,version
pause
type "C:\Users\%username%\Desktop\temp.txt">>"N:\Individual Files\Jerry\Projects\TroubleshootingPCIssues\audits\Audit of Software Programs on Each Computer.txt"
pause

输出：（2个空白行，计算机名称和虚线）（尚未列出程序名称）

`


ANTHEMDESKTOP23 
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Answer 1

这可以解决您的问题吗？ %username%可以包含空格，需要照顾它。

wmic /output:"C:\Users\%username%\Desktop\temp.txt"  product get name,version

正如Andrew所说，这比上述更可靠：

wmic /output:"%userprofile%\Desktop\temp.txt"  product get name,version

WMIC重定向：文件名错误无效

1 个答案: