title: bash awk first 1st column and 3rd column with everything after
body: <p>I am working on the following bash script:</p>
  
  <pre><code># contents of dbfake file
  1 100% file 1
  2 99%  file name 2
  3 100% file name 3
  
  #!/bin/bash
  
  # cat out data
  cat dbfake |
  
  # select lines containing 100%
  grep 100% |
  
  # print the first and third columns
  awk '{print $1, $3}' |
  
  # echo out id and file name and log
  xargs -rI % sh -c '{ echo %; echo "%" &gt;&gt; "fake.log"; }'
  
  exit 0
  </code></pre>
  
  <p>This script works ok, but how do I print everything in column $3 and then all columns after?</p>