为什么说这是“AI 优化 AI”的一步?模型间优化与自我改进区别
2026-11-06 3469103
2026-11-06 0
选择对应支付产品的沙箱验收用例,先查订单状态,再申请退款并确认退款单终态。
用例订单总额 552 分,现金 551 分,券 1 分。以下 Python 3.9.6 脚本只构造和解析本地 XML 样例。三个检查点分别对应查单、退款申请响应和退款查询响应。
from xml.etree import ElementTree as ET
def response(fields):
root = ET.Element("xml")
for key, value in fields.items():
ET.SubElement(root, key).text = value
return ET.tostring(root, encoding="unicode")
def read(xml):
fields = {item.tag: item.text for item in ET.fromstring(xml)}
if fields.get("return_code") != "SUCCESS" or fields.get("result_code") != "SUCCESS":
raise ValueError("request failed")
return fields
common = dict(return_code="SUCCESS", result_code="SUCCESS")
order = "demo-552"
refund = "refund-552"
paid = response(dict(common, out_trade_no=order, total_fee="552",
cash_fee="551", coupon_fee="1", trade_state="SUCCESS"))
accepted = response(dict(common, out_trade_no=order, out_refund_no=refund))
finished = response(dict(common, out_trade_no=order, out_refund_no_0=refund,
refund_fee_0="552", refund_status_0="SUCCESS"))
p = read(paid)
assert p["out_trade_no"] == order and p["trade_state"] == "SUCCESS"
assert int(p["total_fee"]) == 552 == int(p["cash_fee"]) + int(p["coupon_fee"])
print("payment=SUCCESS")
r = read(accepted)
assert r["out_trade_no"] == order and r["out_refund_no"] == refund
print("refund=ACCEPTED")
q = read(finished)
assert q["out_trade_no"] == order and q["out_refund_no_0"] == refund
assert q["refund_fee_0"] == "552" and q["refund_status_0"] == "SUCCESS"
print("refund=SUCCESS")
# 实际输出
# payment=SUCCESS
# refund=ACCEPTED
# refund=SUCCESS旧版免充值验收用例的接口使用 sandboxnew 路径。在同一仿真环境按用例下单并查单。统一下单取得预支付信息后,查单按 out_trade_no 核对 trade_state、交易单号和总金额。查询接口成功与交易成功是两次判断。
POST /sandboxnew/pay/unifiedorder
POST /sandboxnew/pay/orderquery
# 查单目标:trade_state=SUCCESS,total_fee=552现行 V2 付款码验收指引使用 xdc/apiv2sandbox 路径,并明确该入口不支持统一下单。按所用产品核对路径、凭据和验收用例。
沿用同一沙箱的 out_trade_no,指定独立的 out_refund_no。本例做全额退款,total_fee 与 refund_fee 都是 552 分。申请响应先核对编号,再通过退款查询确认终态。
POST /sandboxnew/pay/refund
POST /sandboxnew/pay/refundquery
# 按 out_refund_no_n 找到目标行,再看 refund_status_n一个订单可能有多条退款,不能固定取第一项。本地样例只有一条,用索引 0 演示。真实接口响应还需按该产品约定验签,本文没有调用商户接口。
跑通后保留同一用例的支付查单和退款查询证据。需要验证通知时,再单独测试验签、重放与丢失后的查询补偿。