[英]How to give exactly path in send_from_directory Flask
As we know, the" send_from_directory" need two required parameters 1- directory 2- filename我们知道,“send_from_directory”需要两个必填参数1-目录2-文件名
example:例子:
app.config['UPLOAD_FOLDER'] = r"C:\Users\user\Desktop\upload_foder"
filename =" report.docx"
send_from_directory(app.config['UPLOAD_FOLDER'], filename)
My question is there any way to give the send_from_directory the exact path, for example我的问题是有什么办法可以给 send_from_directory 确切的路径,例如
exact_path = r"C:\Users\user\Desktop\upload_foder\report.docx"
send_from_directory(exact_path)
If you want to pass the exact file path (aka return report.docx)如果你想传递确切的文件路径(又名 return report.docx)
Use send_file(exactpath) instead of send_from_directory使用 send_file(exactpath) 而不是 send_from_directory
Check the documentation here https://flask.palletsprojects.com/en/2.1.x/api/在此处查看文档https://flask.palletsprojects.com/en/2.1.x/api/
The answer is not possible because of the method flask.send_from_directory(directory, path, filename=None, **kwargs)
is required two parameters one is a directory and another one is the path.答案是不可能的,因为方法flask.send_from_directory(directory, path, filename=None, **kwargs)
需要两个参数,一个是目录,另一个是路径。 In official docs from Flask suggest using send_file
to Send a file from within a directory.在 Flask 的官方文档中,建议使用send_file
从目录中发送文件。 ( see more ) ( 查看更多)
Now you can use the method send_file
from a flask that provides the exact path with the file name.现在您可以使用 flask 中的send_file
方法,该方法提供带有文件名的确切路径。 Below this the example code:下面是示例代码:
from flask import send_file
@app.route('/download')
def download():
try:
exact_path = r"C:\Users\user\Desktop\upload_foder\report.docx"
return send_file(exact_path, as_attachment=True)
except Exception as e:
return str(e)
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