I have this code:
from sys import argv
script, filename = argv
txt = open(filename)
print "Here's your file %r:" % filename
print txt.read()
print "Type the filename again:"
file_again = raw_input("> ")
txt_again = open(file_again)
print txt_again.read()
When I run it in terminal the text file won't open. Why?
When you use the follow two lines, what you're saying is asking the interpreter to read 2 arguments:
from sys import argv
script, filename = argv
So when you save your code as test.py and you try to run the code below you get a ValueError, because it only $ python test.py only have one argument.
alvas@ubi:~$ python test.py
Traceback (most recent call last):
File "test.py", line 2, in <module>
script, filename = argv
ValueError: need more than 1 value to unpack
Let's say you have a textfile call in.txt and it only contains the text Hello World, and you try to call the script with the in.txt as second argument, it runs:
alvas@ubi:~$ echo "Hello World" > in.txt
alvas@ubi:~$ python test.py in.txt
Here's your file 'in.txt':
Hello World
Type the filename again:
>
Note: The first line is just a fancy way to create in.txt file with text Hello World
See http://www.tutorialspoint.com/python/python_command_line_arguments.htm for a nice tutorial on using argv =)
It maybe happening because of exceptions. The file may not be existing or you argument maybe an invalid string file name or the file maybe password protected.
you could try and catch the exception by using a try block and then carry on.
