title: Jasper Report -- Resource not found
body: <p>I am developing a jasper report in PHP 
  I have tried embedded jasper library 0.9d -- PhpJasperLibrary-master
  After fixing part of the code in the library,
  It works normally but it cannot display table format which is very important for me to develop Jasper report so I give up in this method and build up an jasper server.</p>
  
  <p>I have installed the PHP Client(v2.0) by composer and using the sample code in the following link</p>
  
  <p><a href="http://community.jaspersoft.com/wiki/php-client-sample-code" rel="nofollow">http://community.jaspersoft.com/wiki/php-client-sample-code</a>
  <a href="http://community.jaspersoft.com/project/php-client/releases" rel="nofollow">http://community.jaspersoft.com/project/php-client/releases</a></p>
  
  <p>I am not sure why there is a lot of bugs in the sample code like getjob(s) function and passing variable to class client.</p>
  
  <p>Anyway, I fixed it and the server info can be called.
  In the last stage, I cannot run the report and the error message shown that the resource of my report not found.</p>
  
  <pre><code>Jaspersoft\Exception\RESTRequestException
  
  Resource /20150604 not found. 
  </code></pre>
  
  <p>I could like you to help me and I hope I can release a prefect version after all parts done.Thanks</p>
  
  <p>And the below is my report information.</p>
  
  <p>I have put my report under the folder /reports
  and it can run in the jasper server</p>
  
  <p>I have checked the URL that calling the report is like below</p>
  
  <pre><code>localhost:9090/jasperserver/rest_v2/reports/20150604.html
  </code></pre>
  
  <p>I have concern about the URL but I cant find any reference to it.
  So please support me to fix the problem.
  Thanks</p>
  
  <p>additional: here is the code I call the report</p>
  
  <pre><code>$report = $c-&gt;reportService()-&gt;runReport('/20150604', 'html');
  </code></pre>