python - 处理从 Instagram 获取隐私帐户的期望值

Question

我有一些 instagram 帐户的列表，并通过 Instagram url '?__a=1' 我试图获得帐户的隐私。一切都很好，但是在大约 128 个用户名之后，我收到以下错误：

JSONDecodeError: Expecting value: line 1 column 1 (char 0)

这是我的代码

import pandas as pd
import json
import requests
from pandas import read_csv
import sqlite3

#connecting to database
db = sqlite3.connect('followers_check.db')
c = db.cursor()
cnt = 0

#path to followers name
file = '/Users/Desktop/followers_data/follower.csv'
list = pd.read_csv(file)
loop = list.iterrows()

for i in loop:
    cnt = cnt + 1
    if cnt>127:
        usm = (i[1]['names'])
        page_content = requests.get('https://www.instagram.com/'+usm+'/?__a=1')
        tree = page_content.json()
        privacy_account = tree['graphql']['user']['is_private']

        if privacy_account==False:
            c.execute("INSERT INTO followers_checked(usm) VALUES (?)", [usm])
            db.commit()
            print('done...'+str(cnt))

有人有解决方案或任何建议吗？

标签： pythonjsonweb-scrapinginstagraminstagram-api