api.add_resource(Todo,'/v1_todo/<id>')，比方说这样只能获取单个参数，但要怎么样才能获取从前端传过来的列表参数呢？最近刚接触 flask，还有很多不熟悉的地方，请多多包涵，先谢谢各位大大的解答了。</id>
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This topic created in 2895 days ago, the information mentioned may be changed or developed.
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1
mrchi Oct 27, 2018
在 URL 里传列表？
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2
ltoddy Oct 27, 2018
from flask import request
在 request 里面. |
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3
gnozix Oct 27, 2018
parser = RequestParser()
parser.add_argument("id", location="json", action="append") req = parser.parse_args(strict=True) id_ = req.get("id") |
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4
Mrkon Oct 29, 2018
楼主的意思是传表单, 3L 正解, 以下是官方文档的示例,楼主要多看官方文档哦
from flask_restful import reqparse parser = reqparse.RequestParser() parser.add_argument('rate', type=int, help='Rate to charge for this resource') args = parser.parse_args() rate = args['rate'] |
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5
IceCola1 Oct 31, 2018
from flask import request
requst.json.get() 别用 flask_restful 接收 List,有坑 |
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