I have this little program(I know there is a lot of errors):
#!/usr/bin/python
import os.path
import sys
filearg = sys.argv[0]
if (filearg == ""):
filearg = input("")
else:
if (os.path.isfile(filearg)):
print "File exist"
else:
print"No file"
print filearg
print "wasn't found"
If i start it by typing python file.py testfile.txt
the output will be always(even if the file doesn't exist):
File exist
If you don't know what iam want from this program, i want to print "File 'filename' wasn't found" if the file isn't exist and if it's exist iam wan't to print "File exist"
Any ideas to solve it?
Thanks
It should be sys.argv[1] not sys.argv[0]:
filearg = sys.argv[1]
From the docs:
The list of command line arguments passed to a Python script. argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string.
The first argument is always the name of the file being executed. This is true for a number of programming languages, you need to use sys.argv[1]
Related
I have written a driver code to take file path from user and use that file path in my functions. Driver code is as below;
import sys
if __name__ == '__main__':
if len(sys.argv) != 2:
print("Usage: %s input_file" % sys.argv[0])
sys.exit()
file_path = sys.argv[1]
connection, color, numOf_Nodes,links = read_problem(file_path)
print(links)
graph_coloring(connection, color, 0, numOf_Nodes)
But when I run this code I got following error:
Usage: C:\ProgramData\Anaconda3\lib\site-packages\ipykernel_launcher.py input_file
An exception has occurred, use %tb to see the full traceback.
SystemExit
Shouldn't it ask for a file path from user? I am new to use a driver code therefore I might skip to write some parts for the driver code to work.
Any help?
The code is expecting that the user pass a file path on the command line. You want to execute your code with something like this:
python ipykernel_launcher.py /path/to/input/file
Upon execution of your code, the variable file_path will get a value of /path/to/input/file.
The variable sys.argv contains the script name in the first position, followed by each of the arguments passed to the script on the command line. That's why 2 is the appropriate expectation here...the first value in the array is the script name. The second value is the file path argument to the program.
I want to create a program that will take two command line arguments. The first being the name of a file to open for parsing and the second the flag -s. If the user provides the wrong number of arguments or the other argument is not -s then it will print the message "Usage: [-s] file_name" and terminate the program using exit.
Next, I want my program to attempt to open the file for reading. The program should open the file read each line and return a count of every float, integer, and other kinds of strings that are not ints or floats. However, if opening the file fails it should raise an exception and print "Unable to open [filename]" and quit using exit.
I've been looking up lots of stuff on the internet about command lines in Python but I've ended up more confused. So here's my attempt at it so far from what I've researched.
from optparse import OptionParser
def command_line():
parser = OptionParser()
parser.add_option("--file", "-s")
options, args = parser.parse_args()
if options.a and obtions.b:
parser.error("Usage: [-s] file_name")
exit
def read_file():
#Try:
#Open input file
#Except:
#print "Unable to open [filename]"
#Exit
from optparse import OptionParser
import sys,os
def command_line():
parser = OptionParser("%prog [-s] file_name")
parser.add_option("-s",dest="filename",
metavar="file_name",help="my help message")
options, args = parser.parse_args()
if not options.filename:
parser.print_help()
sys.exit()
return options.filename
def read_file(fn):
if os.path.isfile(fn):
typecount = {}
with open(fn) as f:
for line in f:
for i in line.split()
try:
t = type(eval(i))
except NameError:
t = type(i)
if t in typecount:
typecount[t] += 1
else:
typecount[t] = 1
else:
print( "Unable to open {}".format(fn))
sys.exit()
print(typecount)
read_file(command_line())
So step by step:
options.a is not defined unless you define an option --a or (preferably) set dest="a".
using the built-in parser.print_help() is better than making your own, you get -h/--help for free then.
you never called your function command_line, therefore never getting any errors, as the syntax was correct. I set the commandline to pass only the filename as a return value, but there are better ways of doing this for when you have more options/arguments.
When it comes to read_file, instead of using try-except for the file I recommend using os.path.isfile which will check whether the file exists. This does not check that the file has the right format though.
We then create a dictionary of types, then loop over all lines and evaluate objects which are separated by whitespace(space,newline,tab). If your values are separated by eg. a comma, you need to use line.split(',').
If you want to use the counts later in your script, you might want to return typecount instead of printing it.
I am trying to run my python module as a command, however I am always getting the error: command not found.
#!/usr/bin/env python
import sys
import re
from sys import stdin
from sys import stdout
class Grepper(object):
def __init__(self, pattern):
self.pattern = pattern
def pgreper(self):
y = (str(self.pattern))
for line in sys.stdin:
regex = re.compile(y)
x = re.search(regex, line)
if x:
sys.stdout.write(line)
if __name__ == "__main__":
print("hello")
pattern = str(sys.argv[1])
Grepper(pattern).pgreper()
else:
print("nope")
I am sure whether it has something to do with the line:
if __name__ == "__main__":
However I just can't figure it out, this is a new area for me, and it's a bit stressful.
Your script name should have a .py extension, so it should be named something like pgreper.py.
To run it, you need to do either python pgreper.py pattern_string or if it has executable permission, as explained by Gabriel, you can do ./pgreper.py pattern_string. Note that you must give the script path (unless the current directory is in your command PATH); pgreper.py pattern_string will cause bash to print the "command not found" error message.
You can't pass the pattern data to it by piping, IOW, cat input.txt | ./pgreper.py "pattern_string" won't work: the pattern has to be passed as an argument on the command line. I guess you could do ./pgreper.py "$(cat input.txt)" but it'd be better to modify the script to read from stdin if you need that functionality.
Sorry, I didn't read the body of your script properly. :embarrassed:
I now see that your pgreper() method reads data from stdin. Sorry if the paragraph above caused any confusion.
By way of apology for my previous gaffe, here's a slightly cleaner version of your script.
#! /usr/bin/env python
import sys
import re
class Grepper(object):
def __init__(self, pattern):
self.pattern = pattern
def pgreper(self):
regex = re.compile(self.pattern)
for line in sys.stdin:
if regex.search(line):
sys.stdout.write(line)
def main():
print("hello")
pattern = sys.argv[1]
Grepper(pattern).pgreper()
if __name__ == "__main__":
main()
else:
print("nope")
Make sure you have something executable here : /usr/bin/env.
When you try to run your python module as a command, it will call this as an interpreter. You may need to replace it with /usr/bin/python or /usr/bin/python3 if you don't have an env command.
Also, make sure your file is executable : chmod +x my_module.py and try to run it with ./my_module.py.
My goal is to compare two data one is from text file and one is from directory and after comparing it this is will notify or display in the console what are the data that is not found for example:
ls: /var/patchbundle/rpms/:squid-2.6.STABLE21-7.el5_10.x86_64.rpm NOT FOUND!
ls: /var/patchbundle/rpms/:tzdata-2014j-1.el5.x86_64.rpm
ls: /var/patchbundle/rpms/:tzdata-java-2014j-1.el5.x86_64.rpm
ls: /var/patchbundle/rpms/:wireshark-1.0.15-7.el5_11.x86_64.rpm
ls: /var/patchbundle/rpms/:wireshark-gnome-1.0.15-7.el5_11.x86_64.rpm
ls: /var/patchbundle/rpms/:yum-updatesd-0.9-6.el5_10.noarch.rpm NOT FOUND
It must be like that. So Here's my python code.
import package, sys, os, subprocess
path = '/var/tools/tools/newrpms.txt'
newrpms = open(path, "r")
fds = newrpms.readline()
def checkrc(rc):
if(rc != 0):
sys.exit(rc)
cmd = package.Errata()
for i in newrpms:
rc = cmd.execute("ls /var/patchbundle/rpms/ | grep %newrpms ")
if ( != 0):
cmd.logprint ("%s not found !" % i)
checkrc(rc)
sys.exit(0)
newrpms.close
Please see the shell script. This script its executing file but because I want to use another language that's why Im trying python
retval=0
for i in $(cat /var/tools/tools/newrpms.txt)
do
ls /var/patchbundle/rpms/ | grep $i
if [ $? != 0 ]
then
echo "$i NOT FOUND!"
retval=255
fi
done
exit $retval
Please see my Python code. What is wrong because it is not executing like the shell executing it.
You don't say what the content of "newrpms.txt" is; you say the script is not executing how you want - but you don't say what it is doing; I don't know what package or package.Errata are, so I'm playing guess-the-problem; but lots of things are wrong.
if ( != 0): is a syntax error. If {empty space} is not equal to zero?
cmd.execute("ls /var/patchbundle/rpms/ | grep %newrpms ") is probably not doing what you want. You can't put a variable in a string in Python like that, and if you could newrpms is the file handle not the current line. That should probably be ...grep %s" % (i,)) ?
The control flow is doing:
Look in this folder, try to find files
Call checkrc()
Only quit with an error if the last file was not found
newrpms.close isn't doing anything, it would need to be newrpms.close() to call the close method.
You're writing shell-script-in-Python. How about:
import os, sys
retval=0
for line in open('/var/tools/tools/newrpms.txt'):
rpm_path = '/var/patchbundle/rpms/' + line.strip()
if not os.path.exists(rpm_path):
print rpm_path, "NOT FOUND"
retval = 255
else:
print rpm_path
sys.exit(retval)
Edited code slightly, and an explanation:
The code is almost a direct copy of the shell script into Python. It loops over every line in the text file, and calls line.strip() to get rid of the newline character at the end. It builds rpm_path which will be something like "/var/patchbundle/rpms/:tzdata-2014j-1.el5.x86_64.rpm".
Then it uses sys.path.exists() which tests if a file exists and returns True if it does, False if it does not, and uses that test to set the error value and print the results like the shell script prints them. This replaces the "ls ... | grep " part of your code for checking if a file exists.
I want to count how many lines of code I have written.
Here is the Python code:
import os
import sys
EXT = ['.c','.cpp','.java','.py']
def main():
l = []
if os.path.isdir(sys.argv[1]):
for root, dirs, files in os.walk(sys.argv[1]):
l.extend([os.path.join(root, name) for name in files])
else:
l.append(sys.argv[1])
params = ["'"+p+"'" for p in l if os.path.splitext(p)[1] in EXT]
result = os.popen("wc -l %s "%" ".join(params)).read()
print result
if __name__ == '__main__':
main()
Before this, it was running as expected. But today, it give me this error:
sh: 1: Syntax error: Unterminated quoted string
I don't know what happened.
Your Python script is missing a shebang line. Add the following to the top of your file:
#!/usr/bin/env python
Then you should be able to run the following, assuming your script is at /path/to/your_script.py and it has the executable bit set:
/path/to/your_script.py arg1 arg2 [...]
Alternatively:
python /path/to/your_script.py arg1 arg2 [...]
Update following comments
I suspect what has changed is that a source file containing a ' in its name has been added to the directory you are checking and the shell is choking on this.
You could add the following function to your program:
def shellquote(s):
return "'" + s.replace("'", "'\\''") + "'"
[Lifted from Greg Hewgill's answer to How to escape os.system() calls in Python? .]
And call it like this:
params = [shellquote(p) for p in l if os.path.splitext(p)[1] in EXT]
#Johnsyweb's updated answer seems to have the correct diagnostic, but the correct fix is to not use a shell to invoke wc. Try something like this instead:
cmd = ['/bin/wc', '-l'] # Need full path!
[cmd.extend(p) for p in l if os.path.splitext(p)[1] in EXT]
result = os.popen2(cmd).read()
Note that the subprocess module is the recommended solution now. Switching to that requires a less intrusive change to your current code, though; see http://docs.python.org/2/library/subprocess.html#replacing-os-popen-os-popen2-os-popen3
Looks like your Python program was parsed like a shell script. Add something like this at the header to indicate where your Python is:
#!/usr/bin/python
or you just run python a.py.