title: Pass variable in awk inside command substitution
body: <p>I am new at this and is learning how to get the variable. How do i get print variable parts right inside command substitution? Thanks.</p>
  
  <pre><code>echo "Enter number of parts:"
  read parts
  echo "Enter filename:"
  read filename
  
  LINES=$(wc -l $filename | awk '{print $1/$parts}' OFMT="%0.0f")
  
  echo $LINES
  </code></pre>