title: Sed : print all lines after match
body: <p>I got my research result after using <code>sed</code> :</p>
  
  <pre><code>zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | cut -f 1 - | grep "pattern"
  </code></pre>
  
  <p>But it only shows the part that I cut. How can I print all lines after a match ?</p>
  
  <p>I'm using zcat so I cannot use awk.</p>
  
  <p>Thanks.</p>
  
  <p><strong>Edited :</strong></p>
  
  <p>This is my log file :</p>
  
  <pre><code>[01/09/2015 00:00:47]       INFO=54646486432154646 from=steve   idfrom=55516654455457       to=jone       idto=5552045646464 guid=100021623456461451463   n
  um=6    text=hi my number is 0 811 22 1/12   status=new      survstatus=new
  </code></pre>
  
  <p>My aim is to find all users that spam my site with their telephone numbers (using <code>grep "pattern"</code>) then print all the lines to get all the information about each spam. The problem is there may be matches in INFO or id, so I use <code>sed</code> to get the text first.</p>