Currently, I'm trying to determine how to return a list of filenames in the directory containing a particular string...
here's how i began:
def searchABatch(directory, extension, searchString):
for file in os.listdir(directory):
if fnmatch.fnmatch(file, extension):
return file
print(searchABatch("procedural", ".py", "foil"))
I expected it to print simply the files with the extension ".py" in my "procedural" directory but I am getting the following error:
Traceback (most recent call last):
File "pitcher20aLP2.py", line 38, in <module>
print(searchABatch("procedural", ".py", "foil"))
File "pitcher20aLP2.py", line 34, in searchABatch
for file in os.listdir(directory):
FileNotFoundError: [Errno 2] No such file or directory: 'procedural'
You're trying to print contents of directory that doesn't exist in you current working directory. You should check if provided directory is actually a directory before calling os.listdir(), by using os.path.isdir()
def searchABatch(directory, extension, searchString):
if os.path.isdir(directory):
for file in os.listdir(directory):
if fnmatch.fnmatch(file, extension):
return file
print(searchABatch("procedural", ".py", "foil"))
Kevin is right, you are trying to find procedural in /home/2020/pitcher20a/procedural directory.
Check your working directory with os.getcwd() and change it needed by using os.chdir(path). Also, you can check if the directory is even a directory by using os.path.isdir() before using os.listdir().
Here is the python os module documentation - https://docs.python.org/2/library/os.html
Also, try to handle the errors gracefully.
Related
dir_path = os.path.dirname(os.path.realpath(__file__))
from os.path import isfile, join
onlyfiles = [f for f in listdir(dir_path) if isfile(join(dir_path, f))]
print(onlyfiles);
with open("config.json", 'r') as jsondata:
config = json.loads(jsondata.read())
Running this code, somehow, triggers a non existing error despite the file being listed during
print(onlyfiles);
Here is the full output log from the console.
Traceback (most recent call last):
['builder.py', 'builder.spec', 'builderloader2.rb', 'config.json',
'Run.bat', 'Run.bat.lnk', 'test.json']
File "C:/Users/cody.jones/Desktop/Builder Generator Release/builder.py",
line 26, in <module>
with open("config.json", 'r') as jsondata:
FileNotFoundError: [Errno 2] No such file or directory: 'config.json'
Process finished with exit code 1
provide full path to open() instead of just file name as by default it will look for file in same directory
try:
open(r"C:/Users/cody.jones/Desktop/Builder Generator Release/config.json", "r")
The script will look for config.json in the current working directory - which presumably is not the same as the folder that the script resides in.
Update your open call to include the path you've already generated.
with open(os.path.join(dir_path, "config.json"), 'r')) as jsondata:
By doing it this way (rather than just including the absolute path) this script will still work if you move it to a different directory or computer so long as you keep the script and config together.
I am trying to use the following code to unzip all the zip folders in my root folder; this code was found on this thread:
Unzip zip files in folders and subfolders with python
rootPath = u"//rootdir/myfolder" # CHOOSE ROOT FOLDER HERE
pattern = '*.zip'
for root, dirs, files in os.walk(rootPath):
for filename in fnmatch.filter(files, pattern):
print(os.path.join(root, filename))
zipfile.ZipFile(os.path.join(root, filename)).extractall(os.path.join(root, os.path.splitext(filename)[0]))
but I keep getting this error that says FileNotFoundError saying the xlsx file does not exist:
Traceback (most recent call last):
File "//rootdir/myfolder/Python code/unzip_helper.py", line 29, in <module>
zipfile.ZipFile(os.path.join(root, filename)).extractall(os.path.join(root, os.path.splitext(filename)[0]))
File "//rootdir/myfolder/Python\Python36-32\lib\zipfile.py", line 1491, in extractall
self.extract(zipinfo, path, pwd)
File "//myaccount/Local\Programs\Python\Python36-32\lib\zipfile.py", line 1479, in extract
return self._extract_member(member, path, pwd)
File "//myaccount/Local\Programs\Python\Python36-32\lib\zipfile.py", line 1542, in _extract_member
open(targetpath, "wb") as target:
FileNotFoundError: [Errno 2] No such file or directory: '\\rootdir\myfolder\._SGS Naked 3 01 WS Kappa Coated and a very long very long file name could this be a problem i dont think so.xlsx'
My question is, why would it want to unzip this excel file anyways?!
And how can I get rid of the error?
I've also tried using r instead of u for rootPath:
rootPath = r"//rootdir/myfolder"
and I get the same error.
Any help is truly appreciated!
Some filenames and directory names may have extra dots in their names, as a consequence the last line, unlike Windows filenames can have dots on Unix:
zipfile.ZipFile(os.path.join(root, filename)).extractall(os.path.join(root, os.path.splitext(filename)[0]))
this line fails. To see how that happens:
>>> filename = "my.arch.zip"
>>> root = "/my/path/to/mydir/"
>>> os.path.join(root, os.path.splitext(filename)[0])
'/my/path/to/mydir/my.arch'
With or without extra dots, problems will still take place in your code:
>>> os.path.join(root, os.path.splitext(filename)[0])
'/my/path.to/mydir/arch'
If no '/my/path.to/mydir/arch' can be found, FileNotFoundError will be raised. I suggest that you be explicit in you path, otherwise you have to ensure the existence of those directories.
ZipFile.extractall(path=None, members=None, pwd=None)
Extract all members from the archive to the current working directory. path specifies a different directory to extract to...
Unless path is an existent directory, FileNotFoundError will be raised.
import os
for filename in os.listdir("."):
if not filename.startswith("renamefilesindir"):
for filename2 in os.listdir(filename):
if filename2.startswith("abcdefghij"):
newName = "[abcdefghij.com][abcde fghij][" + filename + "][" + filename2[11:16] + "].jpg"
print(filename2)
print(newName)
os.rename(filename2, newName)
I have a folder with a few hundred other folders inside of it. Inside each secondary folder is a number of files all similarly named. What I want to do is rename each file, but I get the following error whenever I run the above program.
abcdefghij_88741-lg.jpg
[abcdefghij.com][abcde fghij][3750][88741].jpg
Traceback (most recent call last):
File "C:\directory\renamefilesindir.py", line 9, in <module>
os.rename(filename2, newName)
FileNotFoundError: [WinError 2] The system cannot find the file specified: 'abcdefghij_88741-lg.jpg' -> '[abcdefghij.com][abcde fghij][3750][88741].jpg'
I don't know what this means. It prints the existing file name, so I know it's found the file to be changed. Am I renaming the file wrong? What can't it find?
os.listdir contains just the names of the files and not the full paths. That's why your program actually tried to rename a file inside your current directory and it failed. So you could do the following:
import os.path
os.rename(os.path.join(filename, filename2), os.path.join(filename, newName))
since file with name filename2 is inside directory with name filename.
The code below is part of a program I am writing that runs a method on every .py, .sh. or .pl file in a directory and its folders.
for root, subs, files in os.walk("."):
for a in files:
if a.endswith('.py') or a.endswith('.sh') or a.endswith('.pl'):
scriptFile = open(a, 'r')
writer(writeFile, scriptFile)
scriptFile.close()
else:
continue
When writing the program, it worked in the directory tree I wrote it in, but when I moved it to another folder to try it there I get this error message:
Traceback (most recent call last):
File "versionTEST.py", line 75, in <module>
scriptFile = open(a, 'r')
IOError: [Errno 2] No such file or directory: 'enabledLogSources.sh'
I know something weird is going on because the file is most definitely there...
You'll need to prepend the root directory to your filename
scriptFile = open(root + '/' + a, 'r')
files contains only the file names, not the entire path. The path to the file can be obtained by joining the file name and the root:
scriptFile = open(os.path.join(root, a), "r")
You might want to have a look at
https://docs.python.org/2/library/os.html#os.walk
I'm having trouble figuring out how to move all .log and .txt files in a certain folder and it's subdirectories to a new folder. I understand how to move one file with shutil. But, I tried to use a loop, unsuccessfully, to move all. Can someone help me with this? Thanks ....
import os, os.path
import re
def print_tgzLogs (arg, dir, files):
for file in files:
path = os.path.join (dir, file)
path = os.path.normcase (path)
defaultFolder = "Log_Text_Files"
if not defaultFolder.endswith(':') and not os.path.exists('c:\\Extracted\Log_Text_Files'):
os.mkdir('C:\\Extracted\\Log_Text_Files')
if re.search(r".*\.txt$", path) or re.search(r".*\.log$", path):
os.rename(path, 'C:\\Extracted\\Log_Text_Files')
print path
os.path.walk('C:\\Extracted\\storage', print_tgzLogs, 0)
Below is the trace back error:
Traceback (most recent call last):
File "C:\SQA_log\scan.py", line 20, in <module>
os.path.walk('C:\\Extracted\\storage', print_tgzLogs, 0)
File "C:\Python27\lib\ntpath.py", line 263, in walk
walk(name, func, arg)
File "C:\Python27\lib\ntpath.py", line 263, in walk
walk(name, func, arg)
File "C:\Python27\lib\ntpath.py", line 263, in walk
walk(name, func, arg)
File "C:\Python27\lib\ntpath.py", line 259, in walk
func(arg, top, names)
File "C:\SQA_log\scan.py", line 16, in print_tgzLogs
os.rename(path, 'C:\\Extracted\\Log_Text_Files')
WindowsError: [Error 183] Cannot create a file when that file already exists
According to the traceback, the log-files are already existing. The Python docs to the os.rename say:
On Windows, if dst already exists, OSError will be raised [...].
Now you can either:
delete the files manually or
delete the files automatically using os.remove(path)
If you want the files to be automatically deleted, the code would look like this (notice that I replaced your regular expression with the python endswith as suggested by utdemir):
import os, os.path
def print_tgzLogs (arg, dir, files):
for file in files:
path = os.path.join (dir, file)
path = os.path.normcase (path)
defaultFolder = "Log_Text_Files"
if not defaultFolder.endswith(':') and not os.path.exists('c:\\Extracted\Log_Text_Files'):
os.mkdir('C:\\Extracted\\Log_Text_Files')
if path.endswith(".txt") or path.endswith(".log"):
if os.path.exists('C:\\Extracted\\Log_Text_Files\\%s' % file):
os.remove('C:\\Extracted\\Log_Text_Files\\%s' % file)
os.rename(path, 'C:\\Extracted\\Log_Text_Files\\%s' % file)
print path
os.path.walk('C:\\Extracted\\storage', print_tgzLogs, 0)
It looks like are trying to use
os.rename(path, 'C:\\Extracted\\Log_Text_Files')
to move the file path into the directory C:\Extracted\Log_Text_Files, but rename doesn't work like this: it's going to try to make a new file named C:\Extracted\Log_Text_Files. You probably want something more like this:
os.rename(path, os.path.join('C:\\Extracted\\Log_Text_Files',os.path.basename(path))