json - python 3 json解析特定键返回错误

Question

我正在构建一个脚本来提取存储在 json key 下的值['records']['short_name']。这应该返回我们应用程序的短名称。

JSON 编辑示例：

{
  "totalRecords": 214575,
  "nextPageAvailable": true,
  "records": [
    {
      "_id": "xxxxxxxxxxxxxxxx",
      "sys_updated_on": "2019-07-18 14:30:52",
      "short_name": "Application Test"
    }
  ],
  "lastUpdated": "2019-11-08T18:43:42.000Z"
}

我的编辑代码：

import json
import requests

url = "https://url.com/api/v3/data"

app_query = {"widgetId":"Assets", "asset_type":"Application", "install_status":"Active"}

headers = {
    'authority': "url.com",
    'accept': "application/json, textplain, */*",
    'authorization': "Bearer key_redacted",
    'Host': "url",
    'Accept-Encoding': "gzip, deflate",
    'Connection': "keep-alive",
    'cache-control': "no-cache"
    }

app_data = requests.request("GET", url, headers=headers, params=app_query)

app_json = json.loads(app_data.text)

if app_data.status_code == 200:
    print(app_json['records']['short_name'][0])

elif app_data.status_code == 404:
    print('404 - Not Found.')

我得到的输出是：

Traceback (most recent call last):
  File "query.py", line 23, in <module>
    print(app_json['records']['short_name'][0])
TypeError: list indices must be integers or slices, not str

标签： jsonpython-3.xparsing

Answer 1

错误的原因是您试图short_name从返回的列表中获取键records。

你只需要改变：

print(app_json['records']['short_name'][0])

至

print(app_json['records'][0]['short_name'])

最终代码将是：

import json
import requests

url = "https://url.com/api/v3/data"

app_query = {"widgetId":"Assets", "asset_type":"Application", "install_status":"Active"}

headers = {
    'authority': "url.com",
    'accept': "application/json, textplain, */*",
    'authorization': "Bearer key_redacted",
    'Host': "url",
    'Accept-Encoding': "gzip, deflate",
    'Connection': "keep-alive",
    'cache-control': "no-cache"
    }

app_data = requests.request("GET", url, headers=headers, params=app_query)

app_json = json.loads(app_data.text)

if app_data.status_code == 200:
    print(app_json['records'][0]['short_name'])

elif app_data.status_code == 404:
    print('404 - Not Found.')

请注意，例如，有些事情可以改进。

app_json = json.loads(app_data.text)

可以替换为：

app_json = app_data.json()

此外，如果记录列表返回一个空的记录列表，它也会中断。

.get()从“不安全”的字典中收集数据时考虑使用。

IE：

app_json.get('records')
# you could optionally set a default value
app_json.get('records', [])