php - want to write code for popup window from my given code -


i have simple website written php code. have detect button on page , given below code writen function problem need popup window when detect link clicked. have tired set code in created code not able .kindly please me solve problem.

<?php  /*echo "permessi".$permessi; echo "<br>id".$id_nome;*/  if($action=='canc'){?>     <h1>are sure want delect course?</h1> <form action="<?=$_server['php_self']?>" method="post" name="canc1" style="float: left; margin-left:25px;"> <input type="hidden" name="idcorsocanc" value="<?=$canc_id?>"> <input type="hidden" name="action" value=""> <input type="submit" name="ok" value="si,cancella" class="puls_invia"> </form>  <form action="<?=$_server['php_self']?>" method="post" name="canc2" style="float: left; margin-left:25px;"> <input type="hidden" name="action" value=""> <input type="submit" name="ok" value="no" class="puls_invia"> </form> <?php }

try this

<script> function open_win() { window.open("url here") } </script>  <input type="button" value="open window" onclick="open_win()">

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