无法使用python flask将文件上载到s3 amazon。FileNotFoundError
我正在尝试将文件上载到AmazonS3。然而,我似乎无法成功上传,尽管我在网上学习了很多教程。我一直遇到FileNotFoundError:[Errno 2]没有这样的文件或目录:-filename-。以下是我的代码: init.py无法使用python flask将文件上载到s3 amazon。FileNotFoundError,flask,amazon-s3,file-not-found,Flask,Amazon S3,File Not Found,我正在尝试将文件上载到AmazonS3。然而,我似乎无法成功上传,尽管我在网上学习了很多教程。我一直遇到FileNotFoundError:[Errno 2]没有这样的文件或目录:-filename-。以下是我的代码: init.py from flask import Flask, render_template, request, redirect from werkzeug.utils import secure_filename app = Flask(__name__) impor
from flask import Flask, render_template, request, redirect
from werkzeug.utils import secure_filename
app = Flask(__name__)
import boto3, botocore
from .config import S3_KEY, S3_SECRET, S3_BUCKET
s3 = boto3.client(
"s3",
aws_access_key_id=S3_KEY,
aws_secret_access_key=S3_SECRET
)
@app.route('/', methods=['GET', 'POST'])
def index():
if request.method == 'POST':
file = request.files['file']
filename = ""
if file:
filename = secure_filename(file.filename)
s3.upload_file(Bucket=S3_BUCKET, Filename=filename, Key=filename)
return redirect(url_for('home.html'))
return render_template('home.html')
if __name__ == '__main__':
app.run(debug=True)
config.py
import os
S3_BUCKET = os.environ.get("S3_BUCKET")
S3_KEY = os.environ.get("S3_KEY")
S3_SECRET = os.environ.get("S3_SECRET_ACCESS_KEY")
SECRET_KEY = os.environ.get('SECRET_KEY')
S3_LOCATION = 'http://{}.s3-ap-southeast-1.amazonaws.com/'.format(S3_BUCKET)
home.html
{% extends "layout.html" %}
{% block content %}
<div>
<form method='post' enctype='multipart/form-data'>
<div class='form-group'>
<label for='file'> Upload </label>
<input type='file' id='file' name='file'>
</div>
<div class='form-group'>
<button type='submit' class='btn btn-primary'> Submit </button>
</div>
</form>
</div>
{% endblock %}
{%extends“layout.html”%}
{%block content%}
上传
提交
{%endblock%}
我可以得到如何解决这个问题的建议吗?我还安装了aws cli,并在开始使用这些代码之前完成了配置部分。FileNotFoundError是因为文件不存在。upload()中的代码实际上并没有保存上传的文件,它只是为它获取了一个安全名称,并立即尝试打开它,但失败了 获取安全文件名后,尝试使用save()将文件保存到文件系统:
upload_file = request.files['file']
filename = secure_filename(upload_file.filename)
upload_file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
然后将其上载(假设您已配置上载文件夹):
与
open(os.path.join(app.config['UPLOAD_FOLDER'], filename), 'rb') as f