Twitter Bootstrap Question posted in

serialized form not sending ajax-Twitter bootstrap

I’m having trouble to send a serialized form through ajax to a php file. I can see the string on the client side, but on the server side I receive an empty array.
I’m trying to save the form data into a database, but a I can’t seem to find a way to separate every input, and show it in my php file after I sent with ajax.

JavaScript

$(function() {
  //twitter bootstrap script
  $("button#guardar").click(function(e) {

    //var info = $('#myform').serialize();
    var info = $('form.contact').serialize();
    $.ajax({
      type: "POST",
      url: "solicitudesProc.php",
      data: info,
      success: function(data) {
        alert(info);
        window.location.href = "solicitudesProc.php";
        //window.location.reload();
        $("#modalnuevo").modal('hide');
      },
      error: function(data) {
        alert("failure");
      }
    });

  });
});

<form class="contact" id="myform" method="post" name='alta'>
  <div class="modal-body">
    <div class="row">
      <div class="col-md-2">
        <label>Solicitante</label>
        <input type="text" class="form-control pull-right" name='solicitante' maxlength="20" required />
      </div>
      <div class="col-md-2">
        <label>Fecha Emision</label>
        <input type="text" class="form-control pull-right" name='fechaEmision' maxlength="20" />
      </div>
    </div>
    <div class="row">
      <div class="col-md-2">
        <label>Area Solicitante</label>
        <input type="text" class="form-control pull-right" name='area' maxlength="20" />
      </div>
    </div>


    <div class="modal-footer">
      <button type="button" class="btn btn-default" data-dismiss="modal">Cerrar</button>
      <button type="submit" id="guardar" name='guardar' class="btn btn-danger pull-right" value="guardar">Generar</button>
    </div>
</form>

server side solicitudesProc.php 

<?php   $info = $_POST;
 echo $_POST["solicitante"]; print_r($_POST);  ?>
Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    I maked it work by doing this changes:

    change the form action to the php file im sending.

    <form action="solicitudesProc.php" class="contact" id="myform" method="post" name='alta' >

    and my ajax changed to:

                    var info = $('#myform').serialize();
        //var info = $('form.contact').serialize();
        $.ajax({
                type: "POST",
                url: form.attr("action"),
                data: $("#myform input").serialize(),
                success: function(data){
                    //console.log(info);
                    window.location.href = "solicitudes.php";
                    //window.location.reload();
                    $("#modalnuevo").modal('hide'); 
                },
                error: function(data){
                    alert("failure");
            }
        });

    });

    });

    Thanks for your help!


  2. 0
    1. Do not change location
    2. Cancel the submit

    I strongly suggest you either remove the form OR wire up the submit event:

    $(function() {
  $("form.contact").on("submit", function(e) {
    e.preventDefault(); // stop the submit
    var info = $(this).serialize();
    $.ajax({
      type: "POST",
      url: "solicitudesProc.php",
      data: info,
      success: function(data) {
        console.log(info);
        $("#modalnuevo").modal('hide');
      },
      error: function(data) {
        alert("failure");
      }
    });
  });
});
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