I have a text file which is not suitable for splitting and I want to print out the path and filename from each file.
Lines can look like this:
"sometextC:\folder\folder\filename.exesometext"
"sometext C:\folder\filename.exe sometext"
and so on...
I want something like this:
for line in inputFile.readlines():
hit = find everything between C:\ and .exe
print hit
output should look like this:
C:\folder\folder\file.exe
C:\folder\file.exe
WHERE folder and file is changed with whatever is found in the text. Sorry if my first post was a bit unclear. Number of parent dirs etc can vary a lot.
Thanks!
As a first approximation, you could use re.finditer to find things that might be paths. The example shown below might also match things which are not valid paths (and I'm not strong on the rules for paths on Windows, so you should test that this does what you want thoroughly):
contents = inputFile.read()
for path in re.finditer("(C:\\.*?\.exe)", contents):
print path
You can use
for path in re.finditer("(C:\Users\Default\thanks.exe)", contents):
print path
to print the path
Related
This problem puzzled me.
Maybe the problem is in the code, I hope you take a look
with open(training_images_labels_path,'r') as file:
lines = file.readlines()
He says that the file does not exist
FileNotFoundError: [Errno 2] No such file or directory: '\\Desktop\\project\\data\\generated\\training_images_labels.txt'
Though the file exists
I need solutions
If it says that the file does not exist though the file exists, it means the path has been not given properly. Try giving the path correctly.
Method 1:
Giving correct path 'C:\\Users\\Public\\Desktop\\project\\data\\generated\\training_images_labels.txt' or
'C:\\Users\\<insert your username>\\Desktop\\project\\data\\generated\\training_images_labels.txt' is your path if I guess correctly
Method 2:
Using os module ( Recommended )
mydir = 'C:/Users/Public/Desktop/project/data/generated'
myfile = 'training_images_labels.txt'
training_images_labels_path = os.path.join(mydir, myfile)
with open(training_images_labels_path,'r') as file:
lines = file.readlines()
Method 3:
You can also try changing the working directory to the location where your data is present. ie Desktop>project>data>generated here and open the file with file name. ie
with open('training_images_labels.txt','r') as file:
lines = file.readlines()
I had the same problem with importing an excel file, which of course exists in the same directory with my .py file. The chosen solution above did not help me, and actually I didn't understand those three methods, as I am working on mac OS.
This method worked for me: in Spyder, I usually run the file by pressing the keys "Shift + Enter", which in this case made the problem. So, my solution was, to press on the "Run file" button (or the fn+F5 key) instead.
Maybe someone wants to explain the difference.
Looks like its a windows path you are working and i believe path really thrown in the error is wrong when compared to the actual where the txt file resides.. just cross check once, if that's the case try to pass the correct path in to the variable "training_images_labels_path"
Can you tell how you created this path.Some advise.
use path separator library to generate path to avoid this error.
training_images_labels_path
further try to navigate parent directory using python and print pth.may be some new line or linux/windwos convereted path or other special character in path. navigating parent directory and listing will solve
if still not solve try to navigatep parent-parent dir and print path
try hard
Watch your path if its correct or not. I had the same problem and it turned out i didnt had the good path set
I have a main file which uses(from the main I do a source) a properties file with variables pointing to paths.
The properties file looks like this:
TMP_PATH=/$COMPANY/someProject/tmp
OUTPUT_PATH=/$COMPANY/someProject/output
SOME_PATH=/$COMPANY/someProject/some path
The problem is SOME_PATH, I must use a path with spaces (I can't change it).
I tried escaping the whitespace, with quotes, but no solution so far.
I edited the paths, the problem with single quotes is I'm using another variable $COMPANY in the path
Use one of these threee variants:
SOME_PATH="/mnt/someProject/some path"
SOME_PATH='/mnt/someProject/some path'
SOME_PATH=/mnt/someProject/some\ path
I see Federico you've found solution by yourself.
The problem was in two places. Assignations need proper quoting, in your case
SOME_PATH="/$COMPANY/someProject/some path"
is one of possible solutions.
But in shell those quotes are not stored in a memory,
so when you want to use this variable, you need to quote it again, for example:
NEW_VAR="$SOME_PATH"
because if not, space will be expanded to command level, like this:
NEW_VAR=/YourCompany/someProject/some path
which is not what you want.
For more info you can check out my article about it http://www.cofoh.com/white-shell
You can escape the "space" char by putting a \ right before it.
SOME_PATH=/mnt/someProject/some\ path
should work
If the file contains only parameter assignments, you can use the following loop in place of sourcing it:
# Instead of source file.txt
while IFS="=" read name value; do
declare "$name=$value"
done < file.txt
This saves you having to quote anything in the file, and is also more secure, as you don't risk executing arbitrary code from file.txt.
If the path in Ubuntu is "/home/ec2-user/Name of Directory", then do this:
1) Java's build.properties file:
build_path='/home/ec2-user/Name\\ of\\ Directory'
Where ~/ is equal to /home/ec2-user
2) Jenkinsfile:
build_path=buildprops['build_path']
echo "Build path= ${build_path}"
sh "cd ${build_path}"
I'm new to Python and i'd like to build a script (Python 3) to test electronic modules and save log files.
I'd like to save the logfiles in the following format:
201410log.txt (yearmonthlog.txt)
This is done with the code:
import os
logfile=open(time.strftime('%Y%mlog.txt'), 'a')
logfile.write('This is a test\n\n\n')
This way, every month a new log file is created.
However, i'd like the logfiles to be in in a subdirectory (\logs).
I tried approaches like
logfile=open(time.strftime('\logs\%Y%mlog.txt'), 'a')
and similar things but i couldnt get any of them to work.
I searched trough other questions on stackoverflow (for example: Relative paths in Python ) and elsewhere in the internet, but i couldnt find the right solution.
Could someone point me in the right direction?
(sorry for any mistakes/spelling errors, i'm no native english speaker)
Remove the leading backslash. It makes the path absolute. Beside that, you need to escape a backslash.
logfile = open(time.strftime('logs\\%Y%mlog.txt'), 'a')
or use r'raw string literal':
logfile = open(time.strftime(r'logs\%Y%mlog.txt'), 'a')
For your current path string literal, it does not make problem. But paths like 'a\nb' will not work because \n is interpreted as newline instead of a literal backslash and n.
I have the following code:
with open('EcoDocs TK pdfs.csv', 'rb') as pdf_in:
pdflist = csv.reader(pdf_in, quotechar='"')
for row in pdflist:
if row[1].endswith(row[2]):#check if file type is appended to file name
pathname = ''.join(row[0:2])
else:
pathname = ''.join(row)
if os.path.isfile(pathname):
filehash = md5.md5(file(pathname).read()).hexdigest()
It reads in file paths, file names and file types from a csv file. It then checks to see if the file type is appended to the file name, before joining the file path and file name. It then checks to see if the file exists, before doing something with the file. There are about 5000 file names in the csv file, but isfile only returns True for about half of these. I've manually checked that some of those isfile returns False for exist. As all the data is read in, there shouldn't be any problems with escape characters or single backslashes, so I'm a bit stumped. Any ideas? An example of the csv file format is below, as well as an example of some of the pathnamethat isfile can't find.
csv file-
c:\2dir\a. dir\d dir\lo dir\fu dir\wdir\5dir\,5_l B.xls,.xls
c:\2dir\a. dir\d dir\lo dir\fu dir\wdir\5dir\,5_l A.pdf,.pdf
pathname created-
c:\2dir\a. dir\d dir\lo dir\fu dir\wdir\5dir\5_l B.xls
c:\2dir\a. dir\d dir\lo dir\fu dir\wdir\5dir\5_l A.pdf
Thanks.
You can safely assume that os.path.isfile() works correctly. Here is my process to debug issues like this:
Add a print(pathname) before I use it.
Eyeball the output. Does anything look suspicious?
Copy the output into the clipboard -> Win+RcmdReturndirSpace" + paste into new command prompt + "Return
That checks whether the path is really correct (finds slight mistakes that eyeballing will miss). It also helps to validate the insane DOS naming conventions which are still enforced even on Windows.
if this also works, the next step is to check file and folder permissions: Make sure the user that runs the script actually has permissions to see and read the file.
EDIT Paths on Windows are ... complicated. An important detail, for example, is that "." is a very, very special character. The name "a.something very long" isn't valid in the command prompt because it demands that you have at most three characters after the last "." in a file name! You're just lucky that it doesn't demand that the name before the last dot is at most 8 characters.
Conclusion: You must be very, very, very careful with "strange characters" in file names and paths on Windows. The only characters which are safe are listed in this document.
I'm using os.listdir() to get all the files from a directory and dump them out to a txt file. I'm going to use the txt file to import into access to generate hyperlinks. The problem I'm having is getting the correct path. So when the script is ran it uses whatever directory you are in. Here is an example. Right now it half works, it create links.txt, but there is nothing in the text file.
myDirectory = os.listdir("links")
f.open("links.txt", "w")
f.writelines([os.getcwd %s % (f) for f in myDirectory])
This line of yours:
f.writelines([os.getcwd %s % (f) for f in myDirectory])
is invalid Python syntax and it's very hard to guess what you had in mind for it -- for example, why would you care about the current directory when myDirectory lists, not files in the current directory, but rather files in subdirectory "links"?
Trying to read your mind is always a difficult and generally unrewarding exercise, but assuming you do mean to use the current directory, you might want
f.writelines(os.path.join(os.getcwd(), f) for f in myDirectory)
You have to call os.getcwd() with the trailing parens.
What you probably actually want here though is os.path.join()
os.getcwd is a function you need to call... also I'm not sure what you're doing with the string escape % - but they only work inside strings... I'm guessing you want something like this:
f.writelines([os.path.join(os.getcwd(),f) for f in myDirectory])
[Edit: os.path.join from Alex Martelli's better answer]