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Saving a csv with current date and time

I am running some code and I would like to save a csv file which include the current date and time in its name.
For example: I run some code now (12:24, Jan 15) and I would like to have something like
name_1224_01152021.csv
Can you tell me how to print/save this information, please?

The following code should format the name as per your requirement:
import datetime
name = f'name_{datetime.datetime.now().strftime("%H%M_%m%d%Y")}.csv'
print(name)
# prints 'name_0628_01152021.csv'

Here is the code according to your question :
from datetime import datetime
filename = datetime.now().strftime('filename_%H%M_%m%d%Y.csv')
with open(filename, "w+") as f_output:
csv_output = csv.writer(f_output)
csv_output.writerow(["row1", "row2"])
in filename you have to write your file name the output of this will be shown as
filename_0620_01152021_.csv

Something like this might be what you want :
from datetime import date
today = date.today()
# dd/mm/YY
d1 = today.strftime("%d/%m/%Y")
fname = "name_1224" + d1 + ".csv"
#fname = "name_1224" + str(d1) + ".csv"

import datetime
val_time =datetime.time.now()
now you can append name and val_time and save file with that name

You can format your date with strftime.
import datetime
filename = datetime.datetime.now().strftime('name_%H%_%d%m%Y.csv')
then you can do something like
open(filename, "w").write("blahblah")

How do I create different txt.name in other execute [duplicate]

This question already has answers here:
Create file but if name exists add number
(16 answers)
Closed 3 years ago.
i = 1
while i <=10:
f = open("txtfile.txt",+str(i) "a+")
f.write("111\n")
f.write("222\n")
i = i + 1
f.close()
I want to create txt in many times , But the top code is not in my mind.
I want to create a txt.file1 , and if it exist , the next time execute name is txt.file2.

Assuming, on every run new file (txtfileNN.txt) with incremented number need to be created ( NN is one or two digit number), try below code:
import os
import re
file_base_name = 'txtfile'
r = re.compile(file_base_name+'\d{0,2}.txt')
all_files_in_dir=sorted([i for i in os.listdir() if r.match(i)])
print('Existing files in directory: {}'.format(all_files_in_dir))
# Existing files in directory: ['txtfile.txt', 'txtfile1.txt', 'txtfile10.txt']
if not all_files_in_dir:
# File does not exist yet
out_file = file_base_name + '.txt'
else:
highest_file=all_files_in_dir[-1]
# 'txtfile10.txt'
int_portion = highest_file.replace('.txt', '').split(file_base_name)[-1]
if not int_portion:
# no integer in file, so it it txtfile.txt
next_int = 1
else:
next_int = int(int_portion) + 1
out_file = file_base_name + str(next_int) + '.txt'
print('Next file name : {}'.format(out_file))
# Next file name : txtfile11.txt
# Now write text in new file
f = open(out_file, 'a')
f.write("111\n")
f.write("222\n")
f.close()

Here is a simple way to create 10 files in Python 3.6+, named from file.txt01 to file.txt10:
from pathlib import Path
for i in range(1, 11):
f = Path(f'file.txt{i:02d}')
f.write_text('111\n222\n')
If you want to create a new file on every run, sequentially numbered ad infinitum, do this:
from pathlib import Path
i = 1
while True:
if Path(f'file.txt{i}').exists():
i += 1
else:
Path(f'file.txt{i}').write_text('111\n222\n')
break
But that is very inefficient though.
So maybe this is a better solution:
from pathlib import Path
source = Path('/home/accdias/temp')
prefix = 'file.txt'
slots = set([int(_.name.replace(prefix, '')) for _ in source.glob(f'{prefix}*')])
slot = min(set(range(1, max(slots, default=1) + 1)) - slots, default=max(slots, default=1) + 1)
filename = source / f'{prefix}{slot}'
filename.write_text('111\n222\n')
The solution above is nice because it take into account any gaps that may exist and pick the next lowest slot number available.

Put all files with same name in a folder

I'm new to Python and I need a program that copies files from the same day into a new folder.
Example files:
20120807_first_day_pic.jpg
20120807_first_day_sheet.jpg
20120807_first_day_sheet2.jpg
20120907_second_day_pic.jpg
20120907_second_day_sheet.jpg
20120907_second_day_sheet2.jpg
This is what I have so far, but every file gets a folder and not the whole day.
import os, re, shutil
tfolder = 'D:/Testing/src/'
os.chdir(tfolder)
re_year19xxxxxx = re.compile('(19[0-9][0-9][0-9][0-9])')
re_year20xxxxxx = re.compile('(20[0-9][0-9][0-9][0-9])')
re_ed = re.compile('(ED[0-9])')
destPath = 'D:/Testing/Dest/'
def analyse_file_name(fname):
filePath, coords = os.path.split(fname) #the new folders will be named according to the first 4 characters of the original file name
coordsFolder = coords[:53]
coordsFname = coords[:53]
coordsExt = os.path.splitext(fname)
year = 'year' #create variable year
ed = 'ed' #create variable ed to store the edition number if necessary
bname = fname #the original file name
for re_year in (re_year19xxxxxx, re_year20xxxxxx):
rx = re_year.search(fname) #search for regex in the file name and store it in rx
if rx:
year = rx.group(1) #if the regex is found, store the year
bname.replace(year, ' ')
res = re_ed.search(fname)
if res:
ed = res.group(1)
bname.replace(ed, ' ')
os.chdir(destPath)
if year is 'year':
fname2 = os.path.join(destPath, coordsFolder) + '\\' + coordsFname + coordsExt[1]
else:
fname2 = os.path.join(destPath, coordsFolder,year,ed) + '\\' + coordsFname + coordsExt[1]
print('%s -> %s' % (fname, fname2)) #debug print
dirn, _ = os.path.split(fname2)
if not os.path.exists(dirn):
os.makedirs(dirn)
shutil.copy(fname, fname2)
for root, dirs, files in os.walk(tfolder):
for name in files:
fn = os.path.join(root, name)
analyse_file_name(fn)

If you just want to copy files that start with a known date string format, how about something like this?
def copyfile(filepath, target_dir):
p, filename = os.path.split(filepath)
# get date component of name
date_component = filename.split("_", 1)[0]
# try to parse out the date
try:
d = datetime.datetime.strptime(date_component, "%Y%m%d")
except ValueError:
print "Could not place: ", filename
return
target_date_dir = os.path.join(target_dir, str(d.year), str(d.month), str(d.day))
os.makedirs(target_date_dir)
shutil.copy(filepath, target_date_dir)

First, create a dict (a defaultdict was even more convenient here) that will gather the files for a date (it's good to use re, but given the names of your files using split was easier):
>>> import os
>>> import re
>>> pat = r'(\d+)(?:_\d+)?_(\w+?)[\._].*'
>>> from collections import defaultdict
>>> dict_date = defaultdict(lambda : defaultdict(list))
>>> for fil in os.listdir(path):
if os.path.isfile(os.path.join(path, fil)):
date, animal = re.match(pat, fil).groups()
dict_date[date][animal].append(fil)
>>> dict_date['20120807']
defaultdict(<type 'list'>, {'first': ['20120807_first_day_pic.jpg', '20120807_first_day_sheet.jpg', '20120807_first_day_sheet2.jpg']})
Then for each date, create a subfolder and copy the corresponding files there:
>>> from shutil import copyfile
>>> for date in dict_date:
for animal in dict_date[date]:
try:
os.makedirs(os.path.join(path, date, animal))
except os.error:
pass
for fil in dict_date[date][animal]:
copyfile(os.path.join(path, fil), os.path.join(path, date, animal, fil))
EDIT: took into account OP's new requirements, and Khalid's remark.

Regex day :)
What about trying to match the filename with
pattern=r'(?P<filedate>(?P<year>\d{4})(?P<month>\d{2})(?P<day>\d{2}))\_(?P<bloodyrestofname>.*)'
Complete date, year, etc. may be retrieved from the respective named groups in the match.

import os, shutil
src_path = "D:\\Testing\\Src\\"
dest_path = "D:\\Testing\\Dest\\"
for file in os.listdir(src_path):
if not os.path.isdir(dest_path + file.split("-")[0]):
os.mkdir(dest_path + file.split("-")[0])
shutil.copy(src_path + file, dest_path + file.split("-")[0])

Naming multiple files in python and scrapy

I'm trying to save files to a directory after scraping them from the web using scrapy. I'm extracting a date from the file and using that as the file name. The problem I'm running into, however, is that some files have the same date, i.e. there are two files that would take the name "June 2, 2009". So, what I'm looking to do is somehow check whether there is already a file with the same name, and if so, name it something like "June 2, 2009.1" or some such.
The code I'm using is as follows:
def parse_item(self, response):
self.log('Hi, this is an item page! %s' % response.url)
response = response.replace(body=response.body.replace('<br />', '\n'))
hxs = HtmlXPathSelector(response)
date = hxs.select("//div[#id='content']").extract()[0]
dateStrip = re.search(r"([A-Z]*|[A-z][a-z]+)\s\d*\d,\s[0-9]+", date)
newDate = dateStrip.group()
content = hxs.select("//div[#id='content']")
content = content.select('string()').extract()[0]
filename = ("/path/to/a/folder/ %s.txt") % (newDate)
with codecs.open(filename, 'w', encoding='utf-8') as output:
output.write(content)

You can use os.listdir to get a list of existing files and allocate a filename that will not cause conflict.
import os
def get_file_store_name(path, fname):
count = 0
for f in os.listdir(path):
if fname in f:
count += 1
return os.path.join(path, fname+str(count))
# This is example to use
print get_file_store_name(".", "README")+".txt"

The usual way to check for existence of a file in the C library is with a function called stat(). Python offers a thin wrapper around this function in the form of os.stat(). I suggest you use that.
http://docs.python.org/library/stat.html
def file_exists(fname):
try:
stat_info = os.stat(fname)
if os.S_ISREG(stat_info): # true for regular file
return True
except Exception:
pass
return False

one other solution is you can append time with date, for naming file like
from datetime import datetime
filename = ("/path/to/a/folder/ %s_%s.txt") % (newDate,datetime.now().strftime("%H%M%S"))

The other answer pointed me in the correct direction by checking into the os tools in python, but I think the way I found is perhaps more straightforward. Reference here How do I check whether a file exists using Python? for more.
The following is the code I came up with:
existence = os.path.isfile(filename)
if existence == False:
with codecs.open(filename, 'w', encoding='utf-8') as output:
output.write(content)
else:
newFilename = ("/path/.../.../- " + '%s' ".1.txt") % (newDate)
with codecs.open(newFilename, 'w', encoding='utf-8') as output:
output.write(content)
Edited to Add:
I didn't like this solution too much, and thought the other answer's solution was probably better but didn't quite work. The main part I didn't like about my solution was that it only worked with 2 files of the same name; if three or four files had the same name the initial problem would occur. The following is what I came up with:
filename = ("/Users/path/" + " " + "title " + '%s' + " " + "-1.txt") % (date)
filename = str(filename)
while True:
os.path.isfile(filename)
newName = filename.replace(".txt", "", filename)
newName = str.split(newName)
newName[-1] = str(int(newName[-1]) + 1)
filename = " ".join(newName) + ".txt"
if os.path.isfile(filename) == False:
with codecs.open(filename, 'w', encoding='utf-8') as output:
output.write(texts)
break
It probably isn't the most elegant and might be kind of a hackish approach, but it has worked so far and seems to have addressed my problem.

Write file to directory based on variable in Python

The script will generate multiple files using the year and id variable. These files need to be placed into a folder matching year and id. How do I write them to the correct folders?
file_root_name = row["file_root_name"]
year = row["year"]
id = row["id"]
path = year+'-'+id
try:
os.makedirs(path)
except:
pass
output = open(row['file_root_name']+'.smil', 'w')
output.write(prettify(doctype, root))

If I understand your question correctly, you want to do this:
import os.path
file_name = row['file_root_name']+'.smil'
full_path = os.path.join(path, file_name)
output = open(full_path, 'w')
Please note that it's not very common in Python to use the + operator for string concatenation. Although not in your case, with large strings the method is not very fast.
I'd prefer:
file_name = '%s.smil' % row['file_root_name']
and:
path = '%i-%i' % (year, id)