How to Display MySQL Database with PHP, and AJAX

Question

Where is the error, console is spitting back ID is not defined, I want the "url: id" to select a button depending on its id to return a result and display it on a div on the index

<script type="text/javascript">
  jQuery(document).ready(function ($) {
    $('button').click(function() {
        $(this).attr("id");
        switch(id) {
            case 1:
                id = 'Reportes/Empresas.php'
                break;
            case 2:
                id = 'Reportes/Departamento.php'
                break;
            case 3:
                id = 'Reportes/Empleados.php'
                break;
        }
      $.ajax({
        type: 'GET',
        url: id,
        data: '',
        datatype: 'html',
        cache: 'false',
        success: function(response) {
          $('div.results').append(response);
          alert('Load was performed.');
        },
        error: function(){
          alert('Nope');
        }
      });

      alert('Fail');
    }); // End onclick
  });
</script>

My coworker says there are three errors, but cant help right now.

Daniel, if the answer below is useful, you can click its checkmark to accept it. — Jose Manuel Abarca Rodríguez
– Jose Manuel Abarca Rodríguez, Commented Jul 15, 2016 at 16:19

Marc B · Accepted Answer · 2016-07-15 16:11:21Z

3

You never defined id in the first place:

        $(this).attr("id");
            switch(id){
                  ^^^^--undefined

Maybe you meant

id = $(this).attr('id');
^^^^^

instead.

answered Jul 15, 2016 at 16:11

Marc B

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Comments

kemalatila · Accepted Answer · 2016-07-15 16:31:56Z

0

be sure from button selector and attribute is true. Usage have to be like this.

<button type="button" id="Reportes/Empresas.php" class="display_db_table">Show Table</button>

$('.display_db_table').click(function() {

}

answered Jul 15, 2016 at 16:31

kemalatila

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