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python 3 json解析返回错误的特定键

maskeda • 4 天前 • 152 次点击  

我正在构建一个脚本来获取存储在json键下的值 ['records']['short_name'] . 这将返回我们的应用程序的短名称。

JSON修订示例:

{
  "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"
}

我的修订代码:

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.')

我得到的结果是:

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
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文章 [ 1 ]  |  最新文章 4 天前
luigibertaco
Reply   •   1 楼
luigibertaco    4 天前

错误的原因是您试图获取密钥 short_name 从返回的列表中 records .

你只需要改变:

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

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

最终准则是:

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,有些事情是可以改进的,例如。

app_json = json.loads(app_data.text)

可替换为:

app_json = app_data.json()

另外,如果记录列表返回一个空的记录列表,它也会中断。

考虑使用 .get() 从“不安全”的指令中收集数据时。

即:

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