Zapier出错是因为我没有定义输出。以下是他们在常见问题解答中提供的示例

return {'name': input_data['name']}

在我从Postman复制的Python中，JSON负载看起来不一样，所以我没有相同的地方来定义输出。这是我的代码

import requests
import json

url = "https://stackoverflowpost.com"

payload = json.dumps({
  "LeadInfo": "Lead Info",
  "CustomerFID": None,
  "RegCommunityFID": "None",
  "NameLast": "input_data['lastName']",
  "NameMiddle": "None",
  "NameFirst": "input_data['firstName']",
  "Title": "Mr",
  "NameLast2": "TEST",
  "NameMiddle2": "Zap",
  "NameFirst2": "Zap",
  "Title2": "",
  "Suffix2": "",
  "StreetAddress": "Zap St",
  "City": "Townsville",
  "ZipCode": "12345",
  "StateCode": "DE",
  "PhoneHome": "input_data['phone']",
  "EmailHome": "input_data['email']",
  "Rating": "Rating",
  "LeadSource": "LeadSource",
  "RegistrationNotes": "string",
  "ExternalID": "EXID",
  "Status": "Status",
  "ProspectNumber": "2",
  "WCCustBuyerInfo": {
    "ContractName": "ThisContract",
    "WCBuyers": [
      {}
    ]
  }
})
headers = {
  'Content-Type': 'application/json'
}

response = requests.request("POST", url, headers=headers, data=payload)

print(response.text)

我可以在哪里定义输出，以便它只输出JSON主体，而不会在代码中触发不同的错误？
每当我在代码中添加'return'时，我都会得到一个新的错误。