title: Bash awk print matched delimiter
body: <p>Is there a way to print the currently matched delimiter with awk?</p>
  
  <p>For example:</p>
  
  <pre><code>awk -F '["RESTART" | "FAILURE" | "WARNING" | [:blank:]]{2}' 'FNR &gt; 4 { for (i=1; i&lt;=NF; i++) print $i;}' file
  </code></pre>
  
  <p>Example Input</p>
  
  <pre><code>XX      XXXX   RESTART 6666  XX X
  XXXX    XXXX   WARNING 8888  YYY YYY
  XXX     XXXX   INFORM  7777  XXXX XX
  </code></pre>
  
  <p>Example Output (must)</p>
  
  <pre><code>XX
  XXXX
  RESTART
  6666
  XX X
  XXXX
  XXXX
  WARNING
  8888
  YYY YYY
  XXX
  XXXX
  INFORM
  7777
  XXXX XX
  </code></pre>
  
  <p>Example Output (now)</p>
  
  <pre><code>XX
  XXXX
  6666
  XX X
  XXXX
  XXXX
  8888
  YYY YYY
  XXX
  XXXX
  INFORM
  7777
  XXXX XX
  </code></pre>
  
  <p>I use more than 2 white spaces as column delimiter, but there are some cases (<code>RESTART 6666</code>) or (<code>WARNING 8888</code>) where two columns are not separated by two whitespaces, that is why I have to use the content (<code>RESTART</code>, <code>WARNING</code>) as delimiter, but if I use the content as delimiter, it doesn't get displayed, so I want to display/print the used delimiter (in case it is content and not whitespaces)</p>
  
  <p>The main problem is to differentiate between one white space used as column separator and one white space used as word separator in one column. I cannot affect the file I have to deal with.</p>