title: How to find greater than value of column 4
body: <p>How can I find greater than value of column 4 and print the whole line ? 
  My output is a status report of a storage box.
  eg: </p>
  
  <pre><code>Source                     Destination                   State          Lag         Status
  host1:SystemState          backup1:/vol/xxxxx/xxxxxx     Snapvaulted    11:24:21    Idle
  host2:/vol/xxxxx/xxxxxx    backup2:/vol/xxxxx/xxxxxx     Snapvaulted    1898:58:16  Idle
  host3:/vol/xxxxx/xxxxxx    backup3:/vol/xxxxx/xxxxxx     Snapvaulted    18:58:02    Idle
  host4:/vol/xxxxx/xxxxxx    backup4:/vol/xxxxx/xxxxxx     Snapvaulted    19:46:24    Idle
  host5:/vol/xxxxx/xxxxxxr   backup5:/vol/xxxxx/xxxxxxr    Snapvaulted    5009:22:26  Idle
  </code></pre>
  
  <p>I have tried the following command :</p>
  
  <p><code>ssh backupFiler snapvault status | awk '{print $4}'| cut -d ':' -f 1 | awk '{ if ( $1 &gt; 24 ) print $0 }'</code></p>
  
  <ul>
  <li>Problem is I get only the 4th column as output</li>
  </ul>
  
  <p><code>ssh backupFiler snapvault status | awk '{ if ( $4 &gt; 24 ) print $0 }'</code></p>
  
  <ul>
  <li>Output is not accurate as the column is being considered as a string.</li>
  </ul>
  
  <p>I need my output to show only the lines that have Lag greater than 24hrs.</p>