POST method Jquery and PHP

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POST method Jquery and PHP

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#1 by (1 votes)
#2 by (1 votes)

0

Dear colleagues.

I have a button in the bootstrap:

<button class="btn btn-xs btn-primary active" name="Botao" value="Ativar" id="ativar">Ativar</button>

I would like the status to be changed to Disable by clicking this button, but without refreshing the page. I'm using the code below to direct to the page that makes the change, but it looks like it's not going. See:

$(document).ready(function() {
    $('#ativar').click(function() {
    var valor = $(this).attr('value');
     $.ajax ({
            type: "POST",
            url: "alterar.php?Key="+valor,
             data:  datastring
      })
    });
  });

The change in PHP is OK when I go directly to the page change.php, but the button does not seem to be able to see the page.

php ajax twitter-bootstrap

asked by anonymous 09.07.2015 / 21:12

2 answers

1

Try this:

    $(document).ready(function() {
            $('#ativar').click(function(e) {
            e.preventDefault();
            var valor = $(this).attr('value');
             $.ajax ({
                    type: "POST",
                    url: "alterar.php",
                    data: {key: valor},
                    cache: false,
                    success: function(result) {
                 }
              })
            });
          });

09.07.2015 / 21:22

Definition of Routes IonicFramework Problem with foreach Shopping Cart

score 1 · Accepted Answer

ARQUIVO JS

var valor = jQuery('button.active').val();

jQuery.ajax({
    url : "alterar.php?Key="+valor,
    dataType : 'json',
    async : false,
    success : function(msg) {
        if(msg.status == 1){
            jQuery('button.active').html('Ativo').val('Ativo');
        }else{
            jQuery('button.active').html('Desativado').val('Desativado');
        }
    }
});

Arquivo PHP

$return = array('status'=>null);
if($_GET['Key'] == 'Ativo'){
    $return['status'] = 0;
}else{
    $return['status'] = 1;
}

die(json_encode($return));