Python 将文件从表单上载到S3
我是s3新手,正在尝试上载一些文件,但系统无法找到指定的文件:.jpg我理解这个问题。当文件保存在根目录下时,一切正常。但我不想保存文件。我想在Python 将文件从表单上载到S3,python,amazon-s3,flask,Python,Amazon S3,Flask,我是s3新手,正在尝试上载一些文件,但系统无法找到指定的文件:.jpg我理解这个问题。当文件保存在根目录下时,一切正常。但我不想保存文件。我想在表单中的操作之后直接上传它 def upload_to_s3(file_to_upload, s3_upload_folder): s3 = boto3.resource('s3', aws_access_key_id=app.config['ACCESS_KEY_ID'],
表单中的操作之后直接上传它
def upload_to_s3(file_to_upload, s3_upload_folder):
s3 = boto3.resource('s3',
aws_access_key_id=app.config['ACCESS_KEY_ID'],
aws_secret_access_key=app.config['SECRET_ACCESS_KEY'])
s3.meta.client.upload_file(file_to_upload, app.config['BUCKET_NAME'], s3_upload_folder)
def _user_img_folder(form, file_name):
username = session['name']
vacation_name = slugify(form.test_name.data)
directory = os.path.join(username, test_name)
directory = os.path.join(UPLOAD_FOLDER, directory)
return directory + '/' + file_name
@app.route('/post', methods=['GET', 'POST'])
def test():
if _is_image():
uploaded_images = request.files.getlist('photo')
for image in uploaded_images:
processed_image_name = _hash_image_name(image) # Returns hashed filename with extension
directory = _user_img_folder(form, processed_image_name)
upload_to_s3(str(processed_image_name), str(directory))
return render_template('test.html', form=form, error=error)
谢谢你的帮助
编辑1:
{大量编辑}
{%extends'\u base.html%}
{%block content%}
{{form.csrf_token}
{{form.vacation_name(placeholder=“name Your vacation”)}
{{form.location(placeholder=“它在哪里?”)}
{{form.with_who(占位符='who was with you')}
{{form.description(placeholder=“告诉我们您的假期……或者不。”)}
{form.when(class=“datepicker”,placeholder=“when?”)}
{{form.photo(multiple=“multiple”)}
多恩
{%endblock%}
找到了答案:
我改变了主意
s3.meta.client.upload_file(file_to_upload, app.config['BUCKET_NAME'], s3_upload_folder)
致:
因此,诀窍是,您必须将文件放在磁盘上并提供filepath
和file\u-to\u-upload
,或者提供文件本身,正如我在回答中所演示的那样 更改为methods=['POST']
@hjpotter92实际上它在那里,但我猜它忘了键入。你能发布你的
吗?@hjpotter92已发布^^
s3.meta.client.upload_file(file_to_upload, app.config['BUCKET_NAME'], s3_upload_folder)
s3.Object(app.config['BUCKET_NAME'], s3_upload_folder).put(Body=image)