[python] Does Python have an argc argument?

I have written the same program (open text file and display contents) in C and C++. Now am doing the same in Python (on a Linux machine).

In the C programs I used the code:

if (argc != 2) {
    /* exit program */
}

Question: What is used in Python to check the number of arguments

#!/usr/bin/python
import sys
try:
    in_file = open(sys.argv[1], "r")
except:
    sys.exit("ERROR. Did you make a mistake in the spelling")
text = in_file.read()
print text
in_file.close()

Current output:

./python names.txt = Displays text file (correct)
./python nam = error message: stated from the sys.ext line (correct)
./python = error message: stated from the sys.ext line (wrong: want it to be a
separate error message stating *no file name input*)

This question is related to python linux file-io error-handling arguments

The answer is


I often use a quick-n-dirty trick to read a fixed number of arguments from the command-line:

[filename] = sys.argv[1:]

in_file = open(filename)   # Don't need the "r"

This will assign the one argument to filename and raise an exception if there isn't exactly one argument.


dir(sys) says no. len(sys.argv) works, but in Python it is better to ask for forgiveness than permission, so

#!/usr/bin/python
import sys
try:
    in_file = open(sys.argv[1], "r")
except:
    sys.exit("ERROR. Can't read supplied filename.")
text = in_file.read()
print(text)
in_file.close()

works fine and is shorter.

If you're going to exit anyway, this would be better:

#!/usr/bin/python
import sys
text = open(sys.argv[1], "r").read()
print(text)

I'm using print() so it works in 2.7 as well as Python 3.


You're better off looking at argparse for argument parsing.

http://docs.python.org/dev/library/argparse.html

Just makes it easy, no need to do the heavy lifting yourself.


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