title: Python submit and display on one click
body: <p>I'm new to app engine. With python.
  My question is: I want to submit and display same record on add button click. As I'm using lightbox to display my records.</p>
  
  <p>My html file: where add button is there.</p>
  
  <pre><code>&lt;td&gt;&lt;a href='/display?patients_id={{patient_display.key.id}}&amp;key={{patient_display.key}}' class='display' &gt;&lt;input type="submit"  name="submitButtonName" value="Add" id="submitButtonName"&gt;&lt;/a&gt;&lt;/td&gt;
  </code></pre>
  
  <p>My main.py file: </p>
  
  <pre><code>class DisplayAddHandler(webapp.RequestHandler):
      def get(self):
          self.response.out.write("worksss")
  
          data_key_display = self.request.get('patients_id')
          key = self.request.get('key')
          patient_print_display = PatientInfo.get_by_id(int(data_key_display),parent=None)
          results_print_display = db.GqlQuery("SELECT * FROM PatientMeds WHERE patientinfo_ID=" + data_key_display)
          results_patientalerts_print_display = db.GqlQuery("SELECT * FROM PatientAlerts WHERE patientinfo_ID=" + data_key_display)
  
          template_values = {
          'patient_display': patient_print_display,
          'meds_display': results_print_display,
          'alert_display': results_patientalerts_print_display,
  
          }
  
          path = os.path.join(os.path.dirname(__file__), 'display.html')        
          self.response.out.write(template.render(path, template_values))
  </code></pre>
  
  <p>Now I want when I save record. It will saved successfully and after saving it will display the record which I have added on one click. (I mean on submit button it will save record + display).</p>