Re: basic output question
- From: Wildemar Wildenburger <lasses_weil@xxxxxxxxxxxxxxxxx>
- Date: Sat, 26 Jan 2008 01:53:34 +0100
John Deas wrote:
Hi, I am very new to Python (1 evening...)Style note: You should probably use int() instead of eval. Not only is this safer (as in security), it is also safer (as in robust), because it will throw an error early if you feed it something other than an int.
I need to process a series of files (toto-1.txt toto-2.txt toto-3.txt
toto-4.txt), and as such I created a small program to go through the
files in a directory. I want to call the script with arguments, like
python script.py toto- 1 1 4
my script is as follow :
import sys
sys.argv
header= sys.argv[1]
start =eval(sys.argv[2])
step =eval(sys.argv[3])
nbit =eval(sys.argv[4])
for i in range(nbit):f.read() spills the text into nothingness. You may want to write its output to a variable. ;)
filename=header+str(start+i*step)+'.txt'
f=open(filename,'r')
f.read()
f.close()
My problem is that f.read() outputs nothing, and should I print
filename, I can check that they are well formed, and the files sits in
the same directory as my script.
/W
.
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- From: John Deas
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