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I created an agenda with fullcalendar using this project website, but when I use PHP to get the events from the database, they are not displayed in the calendar, but if I call a json shows the event normally. In the PHP file I understood that a json is generated, but I don’t know how to save it to use. If anyone knows what is wrong, I appreciate the help. Follow the code below:
Calendar.php
  <script>
           $(document).ready(function() {   
            //CARREGA CALENDÁRIO E EVENTOS DO BANCO
            $('#calendario').fullCalendar({
                header: {
                    left: 'prev,next today',
                    center: 'title',
                    //right: 'month,agendaWeek,agendaDay'
                },
                //defaultDate: '2016-01-12',
                editable: true,
                eventLimit: true, 
                events: 'eventos.php',          
                eventColor: '#dd6777'
            }); 
       }); 
    </script>
    <style>
        #calendario{
            position: relative;
            width: 70%;
            margin: 0px auto;
        }        
    </style>
</head>
<body>    
    <div id='calendario'>
    </div>
</body>
php events.
<?php
  $conexao = mysqli_connect('', '', '', ''); 
    $consulta = $conexao->query("SELECT event_id, title, start, end FROM eventos"); 
    while ($linha = $consulta->fetch(PDO::FETCH_ASSOC)){
        $vetor[]= "{id:{$linha['event_id']}, title: {$linha['title']}, start: {$linha['start']}, end:{$linha['end']}}";
     }
    //Passando vetor em forma de json
    echo json_encode($vetor);
I do not understand where this created json goes, I would like to save it paa use in the agenda, but while searching I did not find how to do this
– R.Gasparin
The json "goes" to the client when it requests, in your first code, you define
events: 'eventos.php',, soon it calls this page. Your server executes the query, mounts JSON and returns tofullCalendar, which is on the client’s side and executes thefullCalendar. The problem with your code is the error when generating JSON, this is what I mentioned in the first line.– Inkeliz
But before using like this, I had tried the way you suggested, because the site I used base is used like this. It doesn’t work the way you suggested
– R.Gasparin
You changed the query so that Mysql informs the
idwith the value of? Se não você vai criar um json contendoevent_id=1eid=1, em resumo. No seu código original você tenta utilizarid, por isso pressuponho que ele queira oide nãoevent_id`.– Inkeliz
Yes, I changed the id na in the query call
– R.Gasparin
I can test this later, but I can’t see any other problems. See in F12 > Console if any problem is pointed out and also in F12 > Network if JSON is returned correctly.
– Inkeliz
Okay, I’ll test to see if it’s returning.
– R.Gasparin