title: Use date command within awk arguments
body: <p>I've a file that looks like this</p>
  
  <pre><code>server1-adm.test.com,/var,dir,29987,2007-12-03 15:52:43,root,root
  server2.fs.com,/DATA_File.out,file,299076487,2008-10-15 05:12:23,user1,group1
  server3-prd.mod.com,/opt,dir,29987,2009-05-03 00:13:23,user1,group1
  server4,/var/tmp/xxz.zip,file,400,2007-12-03 15:52:43,root,root
  server1-adm.test.com,/usr,dir,34299876,2006-12-03 16:52:43,root,root
  server3-prd.mod.com,/local/home,dir,400,2009-05-03 12:13:23,user2,group1
  </code></pre>
  
  <p>The 5th column is <code>date + time</code> value that I want to modify. I tried using multiple awk statements together, but it looked too confusing, prone to mistakes. 
  Instead of using multiple statements to do date format conversion, how can I use the <code>date</code> system command within awk to convert to below results. </p>
  
  <pre><code>server1-adm.test.com,/var,dir,29987,2007-12-03 03:52:43 PM,root,root
  server2.fs.com,/DATA_File.out,file,299076487,2008-10-15 05:12:23 AM,user1,group1
  server3-prd.mod.com,/opt,dir,29987,2009-05-03 00:13:23 AM,user1,group1
  server4,/var/tmp/xxz.zip,file,400,2007-12-03 03:52:43 PM,root,root
  server1-adm.test.com,/usr,dir,34299876,2006-12-03 04:52:43 PM,root,root
  server3-prd.mod.com,/local/home,dir,400,2009-05-03 12:13:23 PM,user2,group1
  </code></pre>
  
  <p>Something in lines of <code>date -d "2007-12-03 15:52:43" +%Y/%m/%d:%H%M%S</code>. I don't know how we can get AM/PM in <code>date</code> command.</p>
  
  <p>I already have multiple awk statements running together as part of a script to perform other text modification on the same incoming file, so I would like to use an awk + date statement to do it.</p>