Python Can'；正确格式化API json输出

输入：

输出：

import requests
import json
apikey2 = '1234'
headers = {'API-Key':apikey2,'Content-Type':'application/json'}
data = {"url":urlz2, "visibility": "private"}
r2 = requests.post('https://urlscan.io/api/v1/scan/',headers=headers, data=json.dumps(data))
print(r2)
print(r2.json())

我尝试了一些类似于

message: Submission successful
uuid: xxxxxxxx
result: https://urlscan.io/result/xxxxxxxxxx/
visibility: private

但是get有一个错误，它是一个字符串，

response.json（）

方法返回一个字典。您可以迭代此字典的键和值并打印

for match in r2.json().get('message', []):
    print(f'uuid: {message.get("uuid", {}).get("uuid", "Unknown uuid")}')

如果要打印特定键的值，如果找不到该键，则会显示一条消息，可以使用下面的代码。这段代码迭代感兴趣的键，如果该键不在响应字典中，则可以打印消息。否则，打印键和值

import requests
import json
apikey2 = '1234'
headers = {'API-Key':apikey2,'Content-Type':'application/json'}
data = {"url":urlz2, "visibility": "private"}
r2 = requests.post('https://urlscan.io/api/v1/scan/',headers=headers, data=json.dumps(data))
print(r2)
dictionary = r2.json()

for key, value in dictionary.items():
    print(f"{key} = {value}")

---

非常感谢！出于某种原因，这项任务比我在工具im building lol上的第一个api的输出更重要

import requests
import json
apikey2 = '1234'
headers = {'API-Key':apikey2,'Content-Type':'application/json'}
data = {"url":urlz2, "visibility": "private"}
r2 = requests.post('https://urlscan.io/api/v1/scan/',headers=headers, data=json.dumps(data))
print(r2)
dictionary = r2.json()

for key, value in dictionary.items():
    print(f"{key} = {value}")

keys = ["message", "uuid", "result", "visibility"]
response_dict = r2.json()
for key in keys:
    value = response_dict.get(key, None)
    if value is None:
        print(f"{key} = Unknown {key}")
    else:
        print(f"{key} = {value}")