TypeError: str, bytes or bytearray expected, not list

Question

So I am trying to pull some IP's from a text file and using that list in the socket connection. My issue is I keep getting the following error:

Here is the code im using:

    import socket

     ips = open('list.txt', 'r').readlines()

     def displayType(sec_type):
       switcher = {
        0: "1",
        1: "2",
        2: "3"
       } 
       print(' type: ' + str(sec_type) + ' (' + switcher.get(sec_type,"Not defined by IETF") +')' )

    try:
       def check(ip,port):
         link = socket.socket(socket.AF_INET,socket.SOCK_STREAM)
         link.connect((ip,int(port)))
         link_ver = link.recv(12)
         print(link_ver)

         link.send(link_ver)
         nb_sec_types = ord(link.recv(1))
         print("types: " + str(nb_sec_types))

    check(ips,"80")

If anyone has an idea on how to use the ip's from the list that would be great.

The error is pretty intuitive.. which line raises the error? — rafaelc
– rafaelc, Commented Apr 21, 2018 at 13:22
sorry about that. its the line that has: check(ips,"5900") & also : link.connect((ip,int(port))) — pylearning
– pylearning, Commented Apr 21, 2018 at 13:24
the object ips Is a list. When you call link.connect, you end up passing a list as an argument, but it says that link.connect expects a string instead. You can fix that by iterating through your list of ips, for example: for ip in ips: check(ip, "80) — rafaelc
– rafaelc, Commented Apr 21, 2018 at 13:28
Awesome thank you for that I made the changes and it is working. — pylearning
– pylearning, Commented Apr 21, 2018 at 13:33

rafaelc · Accepted Answer · 2018-04-21 13:42:31Z

1

Change

check(ips, "80)

for

for ip in ips: 
    check(ip, "80)

link.connect expected a unique ip and you're passing a list of ips.

answered Apr 21, 2018 at 13:42

rafaelc

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Comments

Ganesh Negi · Accepted Answer · 2018-04-21 13:35:45Z

0

The method readlines() reads until EOF and returns a list containing the lines. So you are getting a list in ips. You will need to iterate over this list and call check() on each iteration.

Also, you have used try block without except or finally. This will result into error.

You can put catch() inside the try-except block as:

try:
 catch(ip, port)
except Exception as e:
 print(str(e))

edited Apr 21, 2018 at 13:35

answered Apr 21, 2018 at 13:27

Ganesh Negi

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Comments

amir avira · Accepted Answer · 2022-08-01 16:40:12Z

0

You must first convert the string that contains the IP address into a byte or a string of bytes and then start communicating. According to the code below, your error will be resolved. Make sure your code is working correctly overall.

string = '192.168.1.102'
new_string = bytearray(string,"ascii")
ip_receiver = new_string
s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
s.sendto(text.encode(), (ip_receiver, 5252))

answered Aug 1, 2022 at 16:40

amir avira

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