title: how to recognize \n in bash
body: <p>Have such script:</p>
  
  <pre><code>#! /bin/bash
  
  typeset -i i END
  let END=500 i=1
  remainder=1
  accum="use yola_test;\n"
  for ((i=1;i&lt;=END;++i)); do 
  #   str1=$( echo "$i" | md5sum | md5sum )
      title="title_$i"
  #   "${str1:2:20}"
      accum="${accum}CALL add_new_enterprise(\"$title\");\n"
      let "remainder=$i % 100"
      if [ $remainder -eq 0 ]; then
          accum="${accum}SELECT count(*) as \`enterprises:\` FROM enterprise;\n"
          mysql --host=l --port=0 --user=r --password='!' --execute="$accum" 
          accum="use yola_test;\n"
      fi
  done
  </code></pre>
  
  <p>But for every \n it gives me  "Pager set to stdout", can i avoid this, i know that when echoing it i must use -e option, yet i read some material about ANSI-C quoting, but nowhere examples how to use it.</p>
  
  <p>I tried to do so </p>
  
  <pre><code>mysql --host=l --port=0 --user=r --password='!' --execute="$( echo -e "$accum" )"
  </code></pre>
  
  <p>but it hasnt effect and i think call for echo will increase runtime.</p>