python3的dict

 dict1 = {getlistUrl:getlistData,getskuUrl:getskuData, approveUrl:approveData, approvedlistUrl:approvedlistData, searchpresellUrl:searchpresellData, deletepresellUrl:deletepresellData} 

 

然后

import collections
info = dict(name='cold', blog='linuxzen.com')
for key, value in info.items():
    print （key, ':',  value）

还学会了，妈蛋，调用函数，蠢哭了 嘤嘤嘤：

def kolApprove(url1, value):
    kolRequest = requests.post(url=url1,json=value,cookies=userlogin.cookies)
    print(kolRequest.url)
    pprint(kolRequest.json())
    if kolRequest.status_code == 200:
        print (kolRequest.status_code,kolRequest.reason,'\n')
    else:
        print (kolRequest.read())

def main():
    print ('let\'s try it'+'\n' )
    for ur, va in dict1.items():
        kolApprove(ur,va)
# 用main来开始调用
main()

后来发现，用dict ，不是顺序请求接口的，于是改成list：

# 搞定了，原来要单独一对的里面才可以用.items()
def kolApprove(url1, value):
    kolRequest = requests.post(url=url1,json=value,cookies=userlogin.cookies)
    print(kolRequest.url)
    pprint(kolRequest.json())
    if kolRequest.status_code == 200:
        print (kolRequest.status_code,kolRequest.reason,'\n')
    else:
        print (kolRequest.read())

def main():
    print ('let\'s try it'+'\n' )
    for number in list1:
        for ur, va in number.items():
            kolApprove(ur,va)

 

 接着，听说了一个OrderedDict，有序字典，先装utils库，然后：

from collections import OrderedDict
'''
这样的形式也行，但是一定要ordereddict来格式化一下numbers列表，不可以先写成dict，再用ordereddict 

'''
numbers = ((getlistUrl,getlistData),(getskuUrl,getskuData),(approveUrl,approveData),(approvedlistUrl,approvedlistData),(searchpresellUrl,searchpresellData),(deletepresellUrl,deletepresellData))
ordered_dict = OrderedDict(numbers)
for k, v in ordered_dict.items():
    print (k,v)

不可以直接用numbers={a:b,c:d}这样的字典形式写，

应先写成numbers =((a,b),(c,d))

然后用ordered_numbers = OrderedDict(numbers) 来生成有序化字典，

然后就可以按顺序输出啦，就可以按顺序请求接口啦，啦啦啦啦🌶