I am working on a program that clicks a button when it changes color, and at the end of the program, it screenshots the webpage, saves the screenshot, and then moves it to a different directory. Everything seems to be working fine except for moving the file into a different folder. The code I am using to move the file is here:
os.replace("'\\'" + fileName, "'\\'" + saveName + "'\\'" + fileName)
I get the error:
FileNotFoundError: [WinError 3] The system cannot find the path specified: "'\'0.png" -> "'\'saves216'\'0.png"
I don't know how to get the backslash to escape without becoming a double backslash
Remove the extra quotes:
os.replace("\\" + fileName, "\\" + saveName + "\\" + fileName)
You directly escape a \ with another one:
>>> s = "\\" + "filename"
>>> print(s)
\filename
os.replace("\\" + fileName, "\\" + saveName + "\\" + fileName)
extra '' in yours
change the filename variable too if it has ''
Related
I have a folder with the following files:
[11111]Text.txt
[22222]Text.txt
[33333]Text.txt
[44444]Text.txt
I need rename the files to remove the [11111] designation from the beginning of the file name, however that results in duplicate file names.
I wrote a basic script out that will strip the [11111] from the first file, and if any duplication occurs with subsequent files it will name the file [Duplicate]_[#]_text.txt where [#] is a random number
When I ran the code, it renamed the first file correctly, and renamed the second file with the required string, but it did not continue with the other files, and instead presented the following error:
FileExistsError: [WinError 183] Cannot create a file when that file already exists: 'Destination/[33333]Text.txt' -> 'Destination/[Duplicate]_[1]Text.txt'
The code below is what I have currently, though i have tried several iterations also
Location = (Destination_Folder)
Dupe_Counter = random.randint(0,255)
for filename in os.listdir(Location):
try:
if filename.startswith("["):
os.rename(Location + filename, Location + filename[7:])
except:
os.rename(Location + filename, Location +'[Duplicate]_' + '[' + str(Dupe_Counter) +']' + filename[7:])
I'm assuming that it's not actually picking up the Dupe_Counter when creating new files, however I'm not 100% sure where i'm going wrong.
Any help appreciated.
In your Dupe_Counter you are generating a random number that can collide with the results sometimes. But on top of that, you are generating the random Dupe_Counter once only.
Try to generate a random number for each iteration.
Location = (Destination_Folder)
for filename in os.listdir(Location):
Dupe_Counter = random.randint(0,255)
try:
if filename.startswith("["):
os.rename(Location + filename, Location + filename[7:])
except:
os.rename(Location + filename, Location +'[Duplicate]_' + '[' + str(Dupe_Counter) +']' + filename[7:])
But I would recommend generating an increasing sequence for renaming the files and better understanding.
Something Like this:
Location = (Destination_Folder)
for filename in os.listdir(Location):
Dupe_Counter = 101
try:
if filename.startswith("["):
os.rename(Location + filename, Location + filename[7:])
except:
os.rename(Location + filename, Location +'[Duplicate]_' + '[' + str(Dupe_Counter) +']' + filename[7:])
Dupe_Counter += 1
Hope I've been of some help.
I am stuck from a couple of days on an issue in my micro Address Book project. I have a function that writes all records from a SQLite3 Db on file in order to open in via OS module, but as soon as I try to open the file, Python gives me the following error:
Error while opening tempfile. Error:startfile: filepath should be string, bytes or os.PathLike, not _io.TextIOWrapper
This is the code that I have to write records on file and to open it:
source_file_name = open("C:\\workdir\\temp.txt","w")
#Fetching results from database and storing in result variable
self.cur.execute("SELECT id, first_name, last_name, address1, address2, zipcode, city, country, nation, phone1, phone2, email FROM contacts")
result = self.cur.fetchall()
#Writing results into tempfile
source_file_name.write("Stampa Elenco Contatti\n")
for element in result:
source_file_name.write(str(element[0]) + "|" + str(element[1]) + "|" + str(element[2]) + "|" + str(element[3]) + "|" + str(element[4]) + "|" + str(element[5]) + "|" + \
str(element[6]) + "|" + str(element[7]) + "|" + str(element[8]) + "|" + str(element[9]) + "|" + str(element[10]) + "|" + str(element[11]) + "\n")
#TODO: Before exiting printing function you MUST:
# 1. filename.close()
# 2. exit to main() function
source_file_name.close()
try:
os.startfile(source_file_name,"open")
except Exception as generic_error:
print("Error while opening tempfile. Error:" + str(generic_error))
finally:
main()
Frankly I don't understand what this error means, in my previous code snippets I've always handled text files without issues, but I realize this time it's different because I am picking my stream from a database. Any ideas how to fix it?
Thanks in advance, and sorry for my english...
Your problem ultimately stems from poor variable naming. Here
source_file_name = open("C:\\workdir\\temp.txt","w")
source_file_name does not contain the source file name. It contains the source file itself (i.e., a file handle). You can't give that to os.startfile(), which expects a file path (as the error also says).
What you meant to do is
source_file_name = "C:\\workdir\\temp.txt"
source_file = open(source_file_name,"w")
But in fact, it's much better to use a with block in Python, as this will handle closing the file for you.
It's also better to use a CSV writer instead of creating the CSV manually, and it's highly advisable to set the file encoding explicitly.
import csv
# ...
source_file_name = "C:\\workdir\\temp.txt"
with open(source_file_name, "w", encoding="utf8", newline="") as source_file:
writer = csv.writer(source_file, delimiter='|')
source_file.write("Stampa Elenco Contatti\n")
for record in self.cur.fetchall():
writer.writerow(record)
# alternative to the above for loop on one line
# writer.writerows(self.cur.fetchall())
I have this part code which i don't understand the for loop part:
also what does file path holds and how it is different than all file path?
all_files_path = glob.glob("ADNI1_Screening_1.5T/ADNI/*/*/*/*/*.nii")
for file_path in all_file_path:
print(os.system("./runROBEX.sh " + file_path + " stripped/" +file_path.split("/")[-1]))
I'm trying to right a script that will walk through a given directory and rename, move, and delete the original directory. Each directory has a *.mp4 file and a subfolder containing *.jpg files. I'm having trouble deleting the original directory when I'm done renaming and moving the files. What so far is:
import datetime, os, re, shutil, logging, smtplib
logging.basicConfig(level = logging.INFO, format = ' %(asctime)s - %(levelname)s - %(message)s')
#logging.disable(logging.info)
path = '\\\\RT-N66U\\sda2\\Documents'
desired_text1 = re.compile(r'^\d{3}_[a-zA-Z]+\d{3}.jpg')
desired_text2 = re.compile(r'^\d{3}_[a-zA-Z]+_\d{3}.jpg')
renamed_files = ''
for folderName, subfolders, filenames in os.walk(path):
if folderName.startswith(path + '\\GEO-'):
logging.info('The current folder is ' + folderName)
for subfolder in subfolders:
if subfolder.startswith('Pics'):
pic_folder = subfolder
for filename in filenames:
if filename.endswith('.mp4'):
changed_date = datetime.datetime.strptime(filename[3:17], '%b %d, %Y').strftime('%y.%m.%d') # Changes the name of the date to the desired format
year = datetime.datetime.strptime(filename[3:17], '%b %d, %Y').strftime('%Y')
name = filename[20:-25]
name = name.lower()
modified_name = '.'.join(re.findall('\S+', name))
final_name = 'geo.' + changed_date + '.' + modified_name
if not os.path.exists('\\\\RT-N66U\\sda2\\Me\\GEO\\%s\\%s\\%s' % (year, changed_date[3:5], final_name)):
os.makedirs('\\\\RT-N66U\\sda2\\Me\\GEO\\%s\\%s\\%s' % (year, changed_date[3:5], final_name))
if not os.path.exists('\\\\RT-N66U\\sda2\\Me\\GEO\\%s\\%s\\%s\\%s' % (year, changed_date[3:5], final_name, pic_folder)):
os.makedirs('\\\\RT-N66U\\sda2\\Me\\GEO\\%s\\%s\\%s\\%s' % (year, changed_date[3:5], final_name, pic_folder))
if filename.endswith('.mp4'):
logging.info(filename)
os.rename(folderName + '\\' + filename, '\\\\RT-N66U\\sda2\\Me\\GEO\\%s\\%s\\%s\\%s.mp4' % (year, changed_date[3:5], final_name, final_name))
# This part will logg and rename the files by adding a '_'
mo1 = desired_text1.search(filename)
mo2 = desired_text2.search(filename)
if mo1:
logging.info(folderName + '\\' + filename)
os.rename(folderName + '\\' + filename, '\\\\RT-N66U\\sda2\\Me\\GEO\\%s\\%s\\%s\\%s\\%s_%s' % (year, changed_date[3:5], final_name, pic_folder, filename[:-7], filename[-7:]))
renamed_files += filename + ' -> ' + filename[:-7] + '_' + filename[-7:] + '\n'
elif mo2:
logging.info(folderName + '\\' + filename)
os.rename(folderName + '\\' + filename, '\\\\RT-N66U\\sda2\\Me\\GEO\\%s\\%s\\%s\\%s\\%s' % (year, changed_date[3:5], final_name, pic_folder, filename))
renamed_files += filename + '\n'
# This part will check if the directory is empty and delete it if so
if os.listdir(folderName) == []:
logging.info(folderName)
shutil.rmtree(folderName)
What ends up happening is that all sub-directories will be deleted if they're empty and leaving the primary directory empty. I would like the primary directory to be removed as well, since all the files have been moved. Is there a way to accomplish this within the for statement?
If you mean deleting the directory known as folderName inside your main for loop, I think it's not really possible since you're iterating over an object that contains it. Like if inside a for i in this_list: you would like to remove an object from this_list. This is not possible before you're out of the loop.
In order to achieve what you want, I would write a function to recurse into the folder tree. The idea would be something like:
def process_folder(folder_name):
do some stuff on files
get the subfolders of folder_name
for f in subfolders:
process_folder(subfolder_name)
if subfolder_name is empty, delete it
maybe do some other stuff on files
and in the main script, call process_folder(path). (This would leave only the top folder there even if it's empty though, so be sure you don't want to delete it; or to call process_folder at a higher level, or to add an instruction to delete it too if it's empty).
I am using appjs * and I want to execute a command to open a folder.
What I have
var path = __dirname + '/folder to open/';
// path = C:\Program Files\myapplication/folder to open/
require("child_process").exec("start " + path);
Error
Could not find file C:\Program
What I tried
I already tried to escape the spaces, that didn't work.
var path = __dirname + '/folder to open/';
path = path.replace(' ', '\ ');
// path = C:\Program Files\myapplication/folder to open/
require("child_process").exec("start " + path);
When I put the path between quotes, No folder is opened, only another prompt.
var path = "\"" + __dirname + "/folder to open/\"";
path = path.replace(' ', '\ ');
// path = "C:\Program Files\myapplication/folder to open/"
require("child_process").exec("start " + path);
Related bug https://github.com/isaacs/npm/pull/2479
Does anyone has a fix or a workaround?
* link removed
To open a path than contains spaces, you must replace with a double backslash.
In your code you escaped the space character:
"\ "
What you need to do is escape the backslash character so it makes it into the output string:
"\\ "
Try this:
var path = __dirname + '/folder to open/';
// Notice the double-backslashes on this following line
path = path.replace(/ /g, '\\ ');
require("child_process").exec("start " + path);
Well, I fixed it.
Or something like it.
Instead of using
"start " + path
I used
"%SystemRoot%\\explorer.exe \"" + path + "\""
Notice the quotes and the forward slashes.
In my case it is fixed by adding double quotes for the path except the first drive name or letter.
import * as path from 'path'; // npm module
const filePath = 'C:/Program Files/my application/file to open/test.txt';
const rootName = path.parse(filePath).root; // "C:/"
const filePathTo = `${rootName}"${filePath.replace(rootName, '')}"`; // C:/"Program Files/my application/file to open/test.txt"
require("child_process").exec(`start ${filePathTo}`);
The text file will be opened.
this works for me
f= file.replace(/ /g,"\\\ ")
You can also use the old-style 8-character-max/no-space names for each path.
The one I always used was always coding c:\PROGRA~1 instead of c:\Program Files, although this only works on english systems.
If the first 8 characters of any path with more chars are unique, I expect you can do something like newPath = origPath.sub(/\W/g, '').substr(0, 6) + "~1"
Don't have a windows system here, just going by memory.