Print out to file using Grep

Question

I'm trying to check reverse lookup of IP address and then write the result to txt file. How I can write the result to file using grep ?

My script:

import sys, os, re, shlex, urllib, subprocess 

cmd = 'dig -x %s @192.1.1.1' % sys.argv[1]

proc = subprocess.Popen(shlex.split(cmd), stdout=subprocess.PIPE)
out, err = proc.communicate()

sys.stdout = open("/tmp/test.txt", "w")
print(out)
sys.stdout.close()

Answer 1

Edited after comment To only print out a line containing the string 'PTR' , first split the output into a list of lines (each line being a str ), then loop over the lines and check if they contain 'PTR' :

import sys, os, re, shlex, urllib, subprocess 

cmd = 'dig -x %s @192.1.1.1' % sys.argv[1]

proc = subprocess.Popen(shlex.split(cmd), stdout=subprocess.PIPE)
out, err = proc.communicate()

# Convert to list of str lines
out = out.decode().split('\n')

# Only write the line containing "PTR"
with open("/tmp/test.txt", "w") as f:
    for line in out:
        if "PTR" in line:
            f.write(line)

Further explanation of the syntax: The out object returned from proc.communicate() is not a str but a bytes object. The out.decode() does the conversion from bytes to str . We then split the string up into lines using .split('\\n') . We can now iterate over these lines using the for loop and ask whether "PTR" is a substring of any of the lines. If it is, write the corresponding line to the file. Note that if multiple lines contain "PTR" , they will all be written to the file, without a separating newline.