python - Flask 显示 TypeError : send_from_directory() missing 1 required positional argument: 'path'

Question

当我在 Azure 上部署 Flask 应用程序时， View 引发 TypeError: send_from_directory() missing 1 required positional argument: 'path' .当我在本地运行时，这不会发生。

from flask import send_from_directory

@app.route('/download/<path:filename>', methods=['GET', 'POST'])
def download(filename):
    uploads = os.path.join(app.root_path, app.config['UPLOAD_FOLDER'])
    return send_from_directory(directory=uploads, filename=filename)

Answer 1

将最后一行更改为 return send_from_directory(uploads, filename) .
请参阅有关 send_from_directory 的 Flask 文档.底部的更改日志显示“在 2.0 版中更改:path 替换了 filename 参数。”
如果仍要使用命名参数，请更改 filename=至 path= . send_from_directory(directory=uploads, path=filename)