Random string generation I/O python

Question

I'm creating a python script which should create a new file every time it generates a new string (everytime the script is executed it prints out the string to file), and also compares this random string with a string which is a text like: THIS IS A STRING, which is located on another already created file.

For example, the string THIS IS A STRING is located on file called file.dat, and the random generated string should be written to a file, in this case I call it newfile.txt.

However, I have this code:

import string
import random

file = open('file.dat', 'r')
file=file.read()
print file

def id_generator(size=28, chars=string.ascii_uppercase + string.digits):
    return ''.join(random.choice(chars) for _ in range(size))
file = open("newfile.txt", "w")
file.write(id_generator())
file.close()

This code, simply reads the file.dat archive, prints it to console, then generates a random string and stores it on a file called newfile.txt, but it doesn't compares anything, so, to accomplish that I've modified the code like this:

import string
import random

file = open('bullshit.dat', 'r')
file=file.read()
print file

def id_generator(size=28, chars=string.ascii_uppercase + string.digits):
    return ''.join(random.choice(chars) for _ in range(size))
with open("newfile.txt", "r") as f: stored = f.readline() 
if stored == id_generator(): 
    print('success!')

Now, my problem is, that this code just reads an already created file, which is newfile.txt, I need to create a new one like the code I had before, but comparing the strings.

I tried modifying the last three lines like this:

with open("newfile.txt", "w") as f: stored = f.readline() 
if stored == id_generator(): 
    print('success!')

But it throws me:

Traceback (most recent call last):
File "string2.py", line 20, in <module>
with open("newfile.txt", "w") as f: stored = f.readline() 
IOError: File not open for reading

How can I accomplish something like the first version, but comparing the strings as the second one?

when using the open call, you are using w flag which means open for writing...see documentation ..docs.python.org/2/library/functions.html#open — Jay Rajput
– Jay Rajput, Commented Oct 31, 2016 at 2:18
Thank You, I know, but what I'm trying to do, is to create the file everytime I run the script so it can be actually random, if I read it, then it is a file somehow created before, so the random code doesn't make any sense, I don't know if my approach is correct to accomplish this though. — NeoVe
– NeoVe, Commented Oct 31, 2016 at 2:23

thesonyman101 · Accepted Answer · 2016-10-31 02:18:23Z

1

Solution you had "w" which can only write to the file so you have to open it as "r" which reads it.

import string
import random

file = open('test.dat', 'r')
file=file.read()
print file

def id_generator(size=28, chars=string.ascii_uppercase + string.digits):
    return ''.join(random.choice(chars) for _ in range(size))
with open("newfile.txt", "r") as f: stored = f.readline()
if stored == id_generator():
    print('success!')

answered Oct 31, 2016 at 2:18

thesonyman101

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3 Comments

NeoVe Over a year ago

Thank You very much, but this file should be created, not readed, if I read it, then it is the same file created before, so, should I delete it and run the script again maybe? Just to be sure, thank You. But if I delete then it is not random though.

thesonyman101 Over a year ago

Deleting it and rewriting it would be pretty random.

thesonyman101 Over a year ago

Thanks for applying my answer. :)

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