此代码当前适用于从 Flask 应用程序上传文件。用户给我一个整数和一个文件。然后我将该文件存储在 GridFS 中。现在，如果有人尝试访问该文件，则 gridfs 需要将其返回给他们(如果他们知道特定的整数)。当他们访问 URL/uploads/
时我们会得到这个/uploads/<spacenum>内部我们得到号码并调用 readfile()这会将所需的文件写入 uploads/文件夹。但是我们怎样才能获取该文件并发送它呢？我这样做正确吗？想必我之后需要从磁盘中删除该文件。但我没有file反对像我一样在 upload() 中发送或取消链接功能。
我还有一个问题。我希望 mongodb 在存储这些条目后大约 10 分钟删除它们，我该怎么做。我知道我应该为 gridfs 提供一个特殊的字段，但我只是不知 Prop 体如何做。
app.config['UPLOAD_FOLDER'] = 'uploads/'
db = "spaceshare"
def get_db(): # get a connection to the db above
conn = None
try:
conn = pymongo.MongoClient()
except pymongo.errors.ConnectionFailure, e:
print "Could not connect to MongoDB: %s" % e
sys.exit(1)
return conn[db]
# put files in mongodb
def put_file(file_name, room_number):
db_conn = get_db()
gfs = gridfs.GridFS(db_conn)
with open('uploads/' + file_name, "r") as f:
gfs.put(f, room=room_number)
# read files from mongodb
def read_file(output_location, room_number):
db_conn = get_db()
gfs = gridfs.GridFS(db_conn)
_id = db_conn.fs.files.find_one(dict(room=room_number))['_id']
#return gfs.get(_id).read()
with open(output_location, 'w') as f:
f.write(gfs.get(_id).read())
@app.route('/')
def home():
return render_template('index.html')
@app.route('/upload',methods=['POST'])
def upload():
#get the name of the uploaded file
file=request.files['file']
#print "requested files"
space=request.form['space']
# if the file exists make it secure
if file and space: #if the file exists
#make the file same, remove unssopurted chars
filename=secure_filename(file.filename)
#move the file to our uploads folder
file.save(os.path.join(app.config['UPLOAD_FOLDER'],filename))
put_file(filename,space)
# remove the file from disk as we don't need it anymore after database insert.
os.unlink(os.path.join( app.config['UPLOAD_FOLDER'] , filename))
# debugging line to write a file
f = open('debug.txt', 'w')
f.write('File name is '+filename+' or ' +file.name+' the space is :'+ str(space) )
return render_template('index.html', filename = filename ,space = space) ##take the file name
else:
return render_template('invalid.html')
@app.route('/uploads/<spacenum>', methods=['GET'])
def return_file(spacenum):
read_file(app.config['UPLOAD_FOLDER'] ,spacenum)
send_from_directory(app.config['UPLOAD_FOLDER'], filename)
return render_template('thanks.html' , spacenum = spacenum)
以下是我使用过的源代码、我构建此代码的示例以及完整的源代码。预先感谢您的所有帮助!
最佳答案
显然我遇到的问题与文件路径有关。无论如何，这是该问题的解决方案。使用字符串表示形式作为文件名。
# read files from mongodb
def read_file(output_location, room_number):
db_conn = get_db()
gfs = gridfs.GridFS(db_conn)
_id = db_conn.fs.files.find_one(dict(room=room_number))['_id']
with open(output_location + str(room_number) , 'w') as f:
f.write(gfs.get(_id).read())
return gfs.get(_id).read()
@app.route('/upload/<spacenum>', methods=['GET'])
def download(spacenum):
unSecurefilename = read_file(app.config['UPLOAD_FOLDER'] ,spacenum)
return send_from_directory(app.config['UPLOAD_FOLDER'], str(spacenum) )
#return render_template('index.html' , spacenum = spacenum)
关于python - 检索 GridFS 中要从 Flask 发送的文件？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27761674/