I wrote a program to loop through a folder of text files, and for each one, read it and write its edited contents to a new txt file. When I write to a new file, I add "JSP" to the file name, and so I included an if statement to avoid editing a file with JSP in its name. It gives me an error message that suggests that it tried to do the method writeToFile on a JSP file, and it couldn't be found within the folder. This confuses me because
if it's looping through the files and gets to that specific file, it should exist, and
it shouldn't even enter the if statement if it has "JSP" in its filename.
Any ideas?
import program
import os
def main():
directoryStr = "/Users/Elle/Documents/TMR/txtfiles/untitled folder"
directory = os.fsencode(directoryStr)
for file in os.listdir(directory):
filename = os.fsdecode(file)
if ".txt" in filename and "JSP" not in filename:
storedProcedure = program.StoredProcedure(filename)
storedProcedure.writeToFile()
main()
newFile = open(self.newName + ".txt", "w", encoding="utf16")
FileNotFoundError: [Errno 2] No such file or directory: 'JSP_Pgm_JpgmAPARCustSummary_Ctrl_Pay/Rec_summedbycustid_LtorGr0.txt'
Try doing things this way — as I said in a comment, os.listdir() only gives you a list of filenames, not complete file paths.
import program
import os
def main():
directory = "/Users/Elle/Documents/TMR/txtfiles/untitled folder"
for filename in os.listdir(directory):
if ".txt" in filename and "JSP" not in filename:
filepath = os.path.join(directory, filename)
storedProcedure = program.StoredProcedure(filepath)
storedProcedure.writeToFile()
main()
Related
I want to print all the excel files in my Cleaned_Data folder but when I run the code I get:
FileNotFoundError
The code:
directory = r"C:\Users\andre\OneDrive\Desktop\python_web_scrapper\Cleaned_Data"
for filename in os.listdir(directory):
if filename.endswith(".xlsx"):
pd.read_excel(filename)
filename only contains the name of the file and not its entire path. Your script does not see the file since it's looking for it where you are and not in the directory folder. You need to add the path to the file:
import os
directory = r"C:\Users\andre\OneDrive\Desktop\python_web_scrapper\Cleaned_Data"
for filename in os.listdir(directory):
if filename.endswith(".xlsx"):
pd.read_excel(os.path.join(directory, filename))
I am intending to extract some data stored in a .txt file using python 3, however, when I tried to print out the file content, the program does not display any thing in the console. This is the code snippet I use to read the file:
def get_data(directory):
entries = os.listdir(directory)
#print(entries)
count = 0;
for file in entries:
#print(file)
if file.endswith('.txt'):
with open(file) as curr_file:
#print(curr_file)
#read data and write it to an
#excel worksheet
print(curr_file.readline())
curr_file.close()
What kind of changes am I supposed to make to let the program display contents of the file?
Update: I tried to print out all files saved in entries and the result looks fine. The following is the code snippet I used to unzip files in the directory, I am not sure whether there're anything wrong with it.
def read_zip(path):
file_list = os.listdir(path)
#print(file_list)
#create a new directory and store
#the extracted file there
directory = 'C:/Users/chent/Desktop/Test'
try:
if not os.path.exists(directory):
os.makedirs(directory, exist_ok=True)
print('Folder created')
except FileExistsError:
print ('Directory not created')
for file in file_list:
if file.endswith('.zip'):
filePath=path+'/'+file
zip_file = zipfile.ZipFile(filePath)
for names in zip_file.namelist():
zip_file.extract(names, directory)
get_data(directory)
zip_file.close()
Solution: It turns out that I didn't specify the file path when use with open() statement, which caused the program unable to locate files. To fix it, use with open(file_path, file, "r") as curr_file. See details in my updated code:
def get_data(path):
files = os.listdir(path)
for file in files:
#print(file)
try:
if file.endswith('.txt'):
print(file)
with open('C:/Users/chent/Desktop/Test/' + file, "r", ) as curr_file:
# print(curr_file.readlines())
print(curr_file)
line = curr_file.readline()
print(line)
except FileNotFoundError:
print ('File not found')
path = 'C:/Users/chent/Desktop/Test'
get_data(path)
The problem is that you use curr_file.readline() which only returns the first line.
Use curr_file.read() to get the whole file contents.
In order to open all the files in a specific directory (path). I use the following code:
for filename in os.listdir(path): # For each file inside path
with open(path + filename, 'r') as xml_file:
#Do some stuff
However, I want to read the files in the directory starting from a specific position. For instance, if the directory contains the files f1.xml, f2.xml, f3.xml, ... ,f10.xml in this order, how can I read all the files starting from f3.xml (and ignore f1.xml and f2.xml) ?
Straightforward way
import os
keep = False
first = 'f3.xml'
for filename in os.listdir(path): # For each file inside path
keep = keep or filename == first
if keep:
with open(path + filename, 'r') as xml_file:
#Do some stuff
This current question is building on from this question.
I am trying to create a python script that will loop through all the text files in the specified folder. The text files contain directories to files that will be moved to a different specified folder. When looping through a text file, it takes the file from the file directory on each line of that text file.
The end goal is to have all the files which are referenced in the text file to move into one specified folder (\1855).
import shutil
dst = r"C:/Users/Aydan/Desktop/1855"
with open(r'C:\Users\Aydan\Desktop\RTHPython\Years') as my_folder:
for filename in my_folder:
text_file_name = filename.strip()
with open (text_file_name) as my_file:
for filename in my_file:
file_name = filename.strip()
src = r'C:\Users\Aydan\Desktop' + file_name
shutil.move(src, dst)
One text file (1855.txt) contains:
/data01/BL/ER/D11/fmp000005578/BL_ER_D11_fmp000005578_0001_1.txt
/data01/BL/ER/D11/fmp000005578/BL_ER_D11_fmp000005578_0002_1.txt
/data01/BL/ER/D11/fmp000005578/BL_ER_D11_fmp000005578_0003_1.txt
and another text file (1856.txt) contains:
/data01/BL/ER/D11/fmp000005578/BL_ER_D11_fmp000005578_0004_1.txt
/data01/BL/ER/D11/fmp000005578/BL_ER_D11_fmp000005578_0005_1.txt
/data01/BL/ER/D11/fmp000005578/BL_ER_D11_fmp000005578_0006_1.txt
This is the error I get when I run the above script:
Traceback (most recent call last):
File "<pyshell#11>", line 1, in <module>
with open(r'C:\Users\Aydan\Desktop\RTHPython\Years') as my_folder:
PermissionError: [Errno 13] Permission denied: 'C:\\Users\\Aydan\\Desktop\\RTHPython\\Years'
This script doesn't seem to be moving the files named here to the C:/Users/Aydan/Desktop/1855 destination, even though in the script I'm trying to follow the same logic of iterating through each item in the text file, but applying that logic to a folder instead of inside text file.
Any help to find a solution would be brilliant! If you need any more info about the files just ask.
Thanks!
Aydan.
Since you can't open whole folders with the open method, you can get cycle through every .txt file in that folder like that:
import shutil
import glob
dst = r"C:/Users/Aydan/Desktop/1855"
for filename in glob.glob(r"C:\Users\Aydan\Desktop\RTHPython\Years\*.txt"):
text_file_name = filename.strip()
with open (text_file_name) as my_file:
for filename in my_file:
file_name = filename.strip()
src = r'C:\Users\Aydan\Desktop' + file_name
shutil.move(src, dst)
I am trying to process every files inside a folder line by line. I need to check for a particular string and write into an excel sheet. Using my code, if i explicitly give the file name, the code will work. If I try to get all the files, then it throws an IOError. The code which I wrote is as below.
import os
def test_extract_programid():
folder = 'C://Work//Scripts//CMDC_Analysis//logs'
for filename in os.listdir(folder):
print filename
with open(filename, 'r') as fo:
strings = ("/uri")
<conditions>
for line in fo:
if strings in line:
<conditions>
I think the error is that the file is already opened when the for loop started but i am not sure. printing the file name prints the file name correctly.
The error shown is IOError: [Errno 2] No such file or directory:
if your working directory is not the same as folder, then you need to give open the path the the file as well:
with open(folder+'/'+filename, 'r') as fo
Alternatively, you can use glob
import glob
for filename in glob.glob(folder+'/*'):
print filename
It can't open the path. You should do
for filename in os.listdir(folder):
print folder+os.sep()+filename