Question

在我的项目中，当用户点击链接时，AJAX请求会发送创建CSV所需的信息。 CSV需要很长时间才能生成，因此我希望能够在AJAX响应中包含生成的CSV的下载链接。这可能吗？

我见过的大多数答案都会以下列方式返回CSV：

return Response(
        csv,
        mimetype="text/csv",
        headers={"Content-disposition":
                 "attachment; filename=myplot.csv"})

但是，我不认为这与我发送的AJAX响应兼容：

return render_json(200, {'data': params})

理想情况下，我希望能够在params dict中发送下载链接。但我也不确定这是否安全。这个问题通常是如何解决的？

Answer 1

我认为一个解决方案可能是futures库（pip install futures）。第一个端点可以将任务排队，然后将文件名发回，然后可以使用另一个端点来检索文件。我还包括gzip因为如果您要发送更大的文件可能是个好主意。我认为更强大的解决方案使用Celery或Rabbit MQ或类似的东西。但是，这是一个简单的解决方案，可以满足您的要求。

from flask import Flask, jsonify, Response
from uuid import uuid4
from concurrent.futures import ThreadPoolExecutor
import time
import os
import gzip

app = Flask(__name__)

# Global variables used by the thread executor, and the thread executor itself
NUM_THREADS = 5
EXECUTOR = ThreadPoolExecutor(NUM_THREADS)
OUTPUT_DIR = os.path.dirname(os.path.abspath(__file__))

# this is your long running processing function
# takes in your arguments from the /queue-task endpoint
def a_long_running_task(*args):
    time_to_wait, output_file_name = int(args[0][0]), args[0][1]
    output_string = 'sleeping for {0} seconds. File: {1}'.format(time_to_wait, output_file_name)
    print(output_string)
    time.sleep(time_to_wait)
    filename = os.path.join(OUTPUT_DIR, output_file_name)
    # here we are writing to a gzipped file to save space and decrease size of file to be sent on network
    with gzip.open(filename, 'wb') as f:
        f.write(output_string)
    print('finished writing {0} after {1} seconds'.format(output_file_name, time_to_wait))

# This is a route that starts the task and then gives them the file name for reference
@app.route('/queue-task/<wait>')
def queue_task(wait):
    output_file_name = str(uuid4()) + '.csv'
    EXECUTOR.submit(a_long_running_task, [wait, output_file_name])
    return jsonify({'filename': output_file_name})

# this takes the file name and returns if exists, otherwise notifies it is not yet done
@app.route('/getfile/<name>')
def get_output_file(name):
    file_name = os.path.join(OUTPUT_DIR, name)
    if not os.path.isfile(file_name):
        return jsonify({"message": "still processing"})
    # read without gzip.open to keep it compressed
    with open(file_name, 'rb') as f:
        resp = Response(f.read())
    # set headers to tell encoding and to send as an attachment
    resp.headers["Content-Encoding"] = 'gzip'
    resp.headers["Content-Disposition"] = "attachment; filename={0}".format(name)
    resp.headers["Content-type"] = "text/csv"
    return resp


if __name__ == '__main__':
    app.run()

在Flask中为文件创建下载链接的最佳方法？

1 个答案: