python - 手动生成的 JSON 有效，但通过 json.dumps 创建的 JSON 不起作用，即使输出看起来完全相同

Question

我Marketo通过 Python 库marketo-rest-python使用 API 。我可以创建潜在客户并通过以下基本代码更新它们：

leads = [{"email":"joe@example.com","firstName":"Joe"},{"email":"jill@example.com","firstName":"Jill"}]
lead = mc.execute(method='create_update_leads', leads=leads, action='createOnly', lookupField='email', 
                asyncProcessing='false', partitionName='Default')

当我以编程方式创建这个“引导”JSON对象时

leads = []

lead = {}
lead['email'] = "joe@example.com"
lead['firstName'] = "Joe"
leads.append(lead)

lead = {}
lead['email'] = "jill@example.com"
lead['firstName'] = "Jill"
leads.append(lead)

json_leads = json.dumps(leads, separators=(',', ':'))

print(json_leads)

然后输出在 Microsoft Azure Databricks 中完全相同，但 Marketo 系统返回609-> Invalid JSON。

我的输出看起来像

[{"email":"joe@example.com","firstName":"Joe"},{"email":"jill@example.com","firstName":"Jill"}]

它与示例中的完全相同。当我使用示例 JSON 代码行时，它可以工作，但我自己生成的 JSOn 不起作用。

有人知道这应该是什么吗？我在 Microsoft Azure Databricks 中使用 Python。

Answer 1

我相信你不需要打电话json.dumps，就这样吧

leads = []

lead = {}
lead['email'] = "joe@example.com"
lead['firstName'] = "Joe"
leads.append(lead)

lead = {}
lead['email'] = "jill@example.com"
lead['firstName'] = "Jill"
leads.append(lead)

lead = mc.execute(method='create_update_leads', leads=leads, action='createOnly', 
lookupField='email', asyncProcessing='false', partitionName='Default')