title: Sed expression not working in cron job
body: <p>I have a sh file with this script:</p>
  
  <pre><code>#!/bin/bash
  echo "Scripts para limpiar los JS"
  find . -type f  -name "*.js" -print0 | xargs -0 sed -i "s/var _0xaae8=.*//"
  echo "Finalizo la limpieza de JS"
  </code></pre>
  
  <p>If I run this line isolated in the server, it works perfectly:</p>
  
  <pre><code>find . -type f  -name "*.js" -print0 | xargs -0 sed -i "s/var _0xaae8=.*//"
  </code></pre>
  
  <p>But when I run the script in the cron-job I get this error message:</p>
  
  <pre><code>Scripts para limpiar los JS
  sed: -e expression #1, char 19: unknown option to `s'
  sed: -e expression #1, char 19: unknown option to `s'
  sed: -e expression #1, char 19: unknown option to `s'
  sed: -e expression #1, char 19: unknown option to `s'
  find: `./.checkers': Permission denied
  sed: -e expression #1, char 19: unknown option to `s'
  Finalizo la limpieza de JS
  </code></pre>
  
  <p>What I'm doing wrong?</p>