好的,我放弃了.我试图发布包含JSON的文件的内容.该文件的内容如下所示:
{
"id”:99999999,
"orders":[
{
"ID”:8383838383,
"amount":0,
"slotID":36972026
},
{
"ID”:2929292929,
"amount":0,
"slotID":36972026
},
{
"ID”:4747474747,
"amount":0,
"slotID":36972026
}]
}
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这里的代码可能与标记不符:
#!/usr/bin/env python3
import requests
import json
files = {'file': open(‘example.json’, 'rb')}
headers = {'Authorization' : ‘(some auth code)’, 'Accept' : 'application/json', 'Content-Type' : 'application/json'}
r = requests.post('https://api.example.com/api/dir/v1/accounts/9999999/orders', files=files, headers=headers)
Run Code Online (Sandbox Code Playgroud) 看起来这应该是如此简单,但对于我的生活,我找不到答案.我从数据库中提取两个日期时间/时间戳:
2015-08-10 19:33:27.653
2015-08-10 19:31:28.209
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如何从第二个中减去第一个,最好是以毫秒为单位的结果?是的,我也有约会,因为我需要它在午夜左右工作.
使用 python 3,我想下载 API 数据,该数据以 JSON 形式返回,然后我想仅将特定的(列或字段或其他?)插入到 sqlite 数据库中。所以,这就是我所得到的和我遇到的问题:
使用python的请求模块:
##### import modules
import sqlite3
import requests
import json
headers = {
'Authorization' : 'ujbOsdlknfsodiflksdonosB4aA=',
'Accept' : 'application/json'
}
r = requests.get(
'https://api.lendingclub.com/api/investor/v1/accounts/94837758/detailednotes',
headers=headers
)
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好的,第一个问题是如何将请求的 JSON 数据放入 python 可以使用的东西(字典?)中。就是它...
jason.loads(r.text)
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然后我创建要插入特定数据的表:
curs.execute('''CREATE TABLE data(
loanId INTEGER NOT NULL,
noteAmount REAL NOT NULL,
)''')
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没问题...但是现在,即使 JSON 数据看起来像这样(尽管有数百条记录)...
{
"myNotes": [
{
"loanId":11111,
"noteId":22222,
"orderId":33333,
"purpose":"Debt consolidation",
"canBeTraded":true,
"creditTrend":"DOWN",
"loanAmount":10800,
"noteAmount":25,
"paymentsReceived":5.88,
"accruedInterest":12.1,
"principalPending":20.94,
},
{
"loanId":11111,
"noteId":22222,
"orderId":33333,
"purpose":"Credit card …
Run Code Online (Sandbox Code Playgroud) 这里的问题很奇怪。我有一个运行24/7的Python 3脚本,并使用Selenium和Firefox进入网页,每5分钟从一个下载链接下载一个文件(我不能只使用urllib或其他方式下载文件,因为即使下载文件的链接地址保持不变,文件中的数据不断变化,并且每次重新加载页面时都不同,这也取决于指定的条件。该脚本几乎一直都能正常运行,但是我无法摆脱这种错误,该错误有时会突然弹出,从而终止脚本。这是错误:
Traceback (most recent call last):
File "/Users/Shared/ROTH_1/Folio/get_F_notes.py", line 248, in <module>
driver.get(search_url)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages/selenium/webdriver/remote/webdriver.py", line 187, in get
self.execute(Command.GET, {'url': url})
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages/selenium/webdriver/remote/webdriver.py", line 173, in execute
response = self.command_executor.execute(driver_command, params)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages/selenium/webdriver/remote/remote_connection.py", line 349, in execute
return self._request(command_info[0], url, body=data)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages/selenium/webdriver/remote/remote_connection.py", line 379, in _request
self._conn.request(method, parsed_url.path, body, headers)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/http/client.py", line 1090, in request
self._send_request(method, url, body, headers)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/http/client.py", line 1118, in _send_request
self.putrequest(method, url, **skips)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/http/client.py", line 966, in putrequest
raise CannotSendRequest(self.__state) …
Run Code Online (Sandbox Code Playgroud) 我在运行优胜美地的Mac上在rails环境中设置了ruby和ruby,我正在尝试通过教程.当我在bash终端中执行以下操作时,我立即停止这样做:
$ rvm use ruby-2.2.1@ learn-rails --create
Unknown ruby interpreter version (do not know how to handle): learn-rails.
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我相信一切都设置正确.这是我在执行以下操作时得到的结果:
$ rvm list
rvm rubies
ruby-2.2.0 [ x86_64 ]
=* ruby-2.2.1 [ x86_64 ]
# => - current
# =* - current && default
# * - default
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和:
$ ruby -v
ruby 2.2.1p85 (2015-02-26 revision 49769) [x86_64-darwin14]
$ rails -v
Rails 4.2.1
$ rvm -v
rvm 1.26.10 (latest) by Wayne E. Seguin <wayneeseguin@gmail.com>, Michal Papis <mpapis@gmail.com> [https://rvm.io/]
$ echo $PATH …
Run Code Online (Sandbox Code Playgroud) 我有一个元组列表(列表):
('2015-06-19', 3453455, 5, 'Scheduled')
('2015-05-19', 6786788, 6, 'Overdue')
('2015-04-19', 2342344, 2, 'Not Received')
('2015-03-19', 9438549, 0, 'Not Received')
('2015-02-19', 6348759, 7, 'Not Received')
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当我运行这个时,我得到了这个:
>>> print(max(list))
('2015-06-19', 3453455, 5, 'Scheduled')
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显然,max(list)根据元组列表中的第一个值确定最大值.我只是想知道这是否是关于元组列表的max(list)的默认行为; 只检查元组中的第一项?
而且,如果我想根据第二个项目/列返回具有最大值的元组,该怎么办?我该怎么办?