How to find a word in a line of a file (bash)?

Question

bash

There is the following task: there is a log file which contains records about the users who used the server. Each line contains a record of one session (date, time, ip, user, etc.).
How can I extract the username from this file? it is known that all lines are ordered, i.e. the username is exactly between the 6th and 7th spaces.
log example:

Sun Oct 12 11:23:54 2016 101.45.x.x User  ........... .. .  . .

how to implement it as a script in bash?
UPD: did this:
#!/bin/bash
cut -d " " -f 6

Answer 1

Alternatively, make a similar script using AWK:
cat filename.log | gawk 'BEGIN { FS=" " } // { print $7 }'

Answer 2

Use
cut -d "<separator, e.g. space>" -f <number of desired column"