擅长:python、mysql、java
<p><code>os.startfile</code>的文档明确指出:</p>
<blockquote>
<p>startfile() returns as soon as the associated application is launched.
There is no option to wait for the application to close, and no way to
retrieve the application’s exit status</p>
</blockquote>
<p>因此,我建议使用另一种方法,例如通过<code>subprocess.Popen</code>启动它,这允许您等待子进程完成。在</p>