条带API过滤争议请求

Question

我正在向Python Stripe API请求所有争议。我目前检索所有争议的代码如下。你知道吗

import stripe

stripe.api_key = "12345"

disputes = stripe.Dispute.list(limit=100)

但是，我只想请求包含status的needs_response的争议。在stripeapi中是否有具体的方法来请求这些争议？你知道吗

Answer 1

import stripe
stripe.api_key = get_stripe()
from datetime import datetime, timedelta
import delorean
dt = datetime.utcnow()
lte = delorean.Delorean(dt, timezone="UTC").epoch
a =  datetime.now() + timedelta(days=-2)
gte = delorean.Delorean(a, timezone="UTC").epoch
chuck = stripe.Dispute.list(created={'gte':int(gte),'lte':int(lte)})
print "NUMBER OF DISPUTES:", len(chuck)

ch_lst = []
id_lst = []
name_lst = []
am_lst = []
f_lst = []


for d in chuck.auto_paging_iter():
    for d in chuck.auto_paging_iter():
        if (d['status'] == 'needs_response' or d['status'] == 'warning_needs_response'):
        c = stripe.Charge.retrieve(str(d['charge']))
        ch_lst.append(str(d['charge']))
        id_lst.append(str(d['id']))
        am_lst.append(str(c['amount']))
        name_lst.append(str(c['source']['name']))
        f_lst.append(str(c['source']['fingerprint']))

Answer 2

在这里，最好的办法是在从API中检索disputes的完整列表之后在本地进行过滤。使用auto-pagination，这可以简单到：

disputes = stripe.Dispute.list(limit=3)

for dispute in disputes.auto_paging_iter():
    if (dispute.status == 'needs_response'):
        # Do something with the dispute

作为一个推广，如果它不在arguments list for that resource中，那么您只需要在本地进行进一步的过滤。你知道吗