关于python中urllib.parse中的bug

需求：将字典类型的URL参数，进行URL编码

将

{"query_hash":"69cba40317214236af40e7efa697781d","variables":{"id":"15756602186","first":12}}

转化为：

query_hash=69cba40317214236af40e7efa697781d&variables=%7B%22id%22%3A%2215756602186%22%2C%22first%22%3A12%7D

如果直接调用urlencode编码出来的URL参数，和浏览器看到的不一样，多了+号

from urllib.parse import urlencode
params ={"query_hash":"69cba40317214236af40e7efa697781d","variables":{"id":"15756602186","first":12}}
print(urlencode(params))

输出 query_hash=69cba40317214236af40e7efa697781d&variables=%7B%27id%27%3A+%2715756602186%27%2C+%27first%27%3A+12%7D

问了一圈有大佬说这是urllib.parse的bug

urlencode编码嵌套的字典就会出现这样的问题

转了一圈，解决方案有两种：

• 方案一、

params = quote(json.dumps({"id":"15756602186","first":12},separators=(',', ':')))
payload = f'query_hash=69cba40317214236af40e7efa697781d&variables={params}'
print(payload)

• 方案二、

params = {"query_hash":"69cba40317214236af40e7efa697781d","variables":'{"id":"15756602186","first":12}'}
print(urlencode(params))

这种方案有个风险，比如通过传参的方式传入variables，就无法成功编码

        dynamic_params = {"id":"15756602186","first":50}
        params = {"query_hash": "69cba40317214236af40e7efa697781d", "variables": f'{dynamic_params}'}
        print(params)