如何提供烧瓶中路线的链接？

Question

我正在通过Flask向用户提供报告。但是，要获取报告，我需要提供下载报告的链接。在localhost中，我传递消息

href =“ http://127.0.0.1:8080/apps/bas/download_full_report”>本地下载文件

即使应用程序部署在像openshift这样的容器平台上，我也传递相同的链接吗？

@application.route(app_route+'download_full_report')
def generate_report():
    reportType = session['context_variables']['report_type']
    df =pickle.loads(zlib.decompress(r.get(reportType+broker_tax_id+"full_report")))
    execel_file = StringIO()
    filename = "%s.csv" % ('output file')
    df.to_csv(execel_file, encoding='utf-8')
    csv_output = execel_file.getvalue()
    execel_file.close()
    resp = make_response(csv_output)
    resp.headers["Content-Disposition"] = ("attachment; filename=%s" % filename)
    resp.headers["Content-Type"] = "text/csv"
    return resp