python gspread从响应体更新多个单元格

Question

我正在使用这个 python 脚本从 Progresso API 获取响应： http://docs.progresso.apiary.io/#reference/behaviour/behaviour-events-collection/get-behaviour-events

from urllib2 import Request, urlopen
import smtplib import gspread 
from oauth2client.service_account import ServiceAccountCredentialseaders = {
'Authorization': 'Bearer [CURRENT_TOKEN]'
}
request = Request('https://private-anon-ae5edf57e7-progresso.apiary-
mock.com/BMEvents/?Behaviour=new', headers=headers)
response_body = urlopen(request).read()
scope = ['https://spreadsheets.google.com/feeds']
credentials = ServiceAccountCredentials.from_json_keyfile_name('ProgressoAPI-
2f6ecaa6635c.json', scope)
gc = gspread.authorize(credentials)
wks  = gc.open("Progresso Test").sheet1
wks.clear()
cell_list = wks.range('A1:H20')
for cell in cell_list:
cell.value = response_body
wks.update_cells(cell_list)

我知道 cell.value = 响应正文是错误的，我不知道如何才能正确 - 我被卡住了。

它出现在每个单元格中，如下所示： "{ ""BehaviourEntryId"": 13798177, ""LearnerId"": 245277, ""LearnerCode"": ""2009-0080"", ""RegGroup"": ""U6-RWE"", ""行为"": ""负面"", ""事件日期"": ""2017-02-07"", ""主题"": ""BE"", ""位置"": ""CLS"", ""已发布"": ""是"", ""创造者"": ""DhDr"", ""编辑器"": null, ""受让人"": ""DiRo"", ""状态"": ""已完成"", ““细节””： [ { ""类别"": ""CL"", ""类型"": ""CLatt"", ""严重性"": ""S2"", “点”：0 }, { ""类别"": ""CL"", ""类型"": ""CLBEH"", ""严重性"": ""S2"", “点”：2 } ], ““评论””： [ { ""BehaviourEntryCommentId"": 5648278, ""机密""：真实， ""评论"": ""被要求上厕所，去了最远的那个，只是为了浪费时间。"" }, { ""BehaviourEntryCommentId"": 5648279, ""机密"": false, ""评论"": ""把口香糖吐在地板上"" }, { ""BehaviourEntryCommentId"": 5648280, ""机密"": false, ""评论"": ""对员工很粗鲁"" } ], ""行动"": [ ""HTO"", ""ISO"" ] }"

如何在单元格范围内按照我想要的方式分隔文本并批量更新？

Answer 1

如果您的意思是两列，其中一行是“BehaviourEntryId”，另一行是 13798177，您可以尝试这样的操作：

import json
response = json.loads(response_body) #decode the json response string, returns a dict
response_pairs = list(response.items)

for i in range(1, len(response_body)+1):
    current_pair = response_pairs[i-1]
    current_key =  current_pair[0]
    current_value = current_pair[1]
    wks.update_acell('A{}'.format(i), current_key)
    wks.update_acell('B{}'.format(i), current_value)