javascript - Dynamically generate table in form using Ajax with symfony -


i working symfony project . want generate table in form when select option in form.

this task.

there name options in form ( eg. john , mia , lia .. etc) when select "john" . want display detail "john" using table. table should locate in form .

there example in link http://www.w3schools.com/php/php_ajax_database.asp

but want using symfony.

what best way this.

please mention example.

using javascript , jquery or ajax symfony

the symfony keyword misleading here because generate table client javascript code. in javascript initiate ajax call post request php/symfony application. 1 returns json array of objects, , ajax complete/done method create table using jquery example. here example:

<html> <head>     <script src="https://code.jquery.com/jquery-2.2.0.js"></script>     <script>     $(document).ready (function () {      }) ;      function go () {         $.post ('/index.php/api/mycall',         //$.post ('/php.php',             {                 "param1": "param1",                 "param2": 2             })             .done (function (ret) {                 var tbl =$('#mytable') ;                 tbl.empty () ;                 $(document.createelement('tr'))                     .append ('<th>id</th><th>firstname</th><th>lastname</th>')                     .appendto (tbl) ;                 $.each (ret, function (index, val) {                     $(document.createelement('tr'))                         .append ('<td>' + val.id + '</td><td>' + val.firstname + '</td><td>' + val.lastname + '</td>')                         .appendto (tbl) ;                 }) ;             }) ;     }     </script> </head> <body>     <input type="button" onclick="go()" />     <div id="mytable"></div> </body> </html>

and somewhere in symphony code handle post query /api/mycall

$results =[   (object)[     "id" => 1,     "firstname" => "john",     "lastname" => "doe"   ],   (object)[     "id" => 2,     "firstname" => "jane",     "lastname" => "doe"   ] ] ; return (new jsonresponse ($results, response::http_ok)) ;

hope helps


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