在Python中打开文件名+日期为csv [英] Open a file name +date as csv in Python

问题描述

我想要能够自动打开文件名并将其保存为.csv文件，我生成的文件总是被称为同一事物+今天的日期。例如，今天的电子表格可以调用：

 TODAYS SHEET+今天date.xls 
  
 
 
 
存储在位置
  C： \\ A \B \C\D 
  
如何让代码在今天打开。 xls文件并将其另存为.csv位置
  C：\A\B\C\D\\ \\ e 
  
我最终希望直接从此.csv文件加载数据，以便与webscraper进行比较，因此可能有一种方法可以将.xls文件打开为.csv格式，而不将其另存为.csv格式。
解决方案
它应该看起来像这样：
  import datetime 
 today_string = datetime.datetime.today ）.strftime（'％x'）
 
 with open（'C：/ A / B / C / D / TODAYS SHEET'+ today_string +'.csv'，'w'）as my_file： 
 my_file.write（'a，a，a，a，a，a'）
  
您可以查看字符串格式为strftime函数。另请查看打开功能以及您可以执行的操作使用文件 
 
I want to be able to open a file name automatically and save it as a .csv, the files I produce are always called the same thing + todays date. For example todays spreadsheet could be called:
"TODAYS SHEET" + Todays date.xls
Stored in location
C:\A\B\C\D
How would I get the code to open todays .xls file and save it as a .csv in location 
C:\A\B\C\D\E
I ultimately want to load data directly from this .csv file for comparison with a webscraper, so there may well be a method to open a .xls file as a .csv without saving it as a .csv in a second location.
解决方案
It should look like something close to that:
import datetime
today_string = datetime.datetime.today().strftime('%x')

with open('C:/A/B/C/D/TODAYS SHEET' + today_string + '.csv', 'w') as my_file:
    my_file.write('a,a,a,a,a,a')
You can have a look at the string format for the strftime function. Also have a look at the open function and what you can do with files

                        
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