Send javascript FILE to php

Question

I’ve been reading a lot about it and I’m getting more confused.

I have the form below:

<div class="plano">

  <h1 class="titulos">Cadastro de Plano</h1>

  <form method="post" class="planoCadastrar">

    <input type="text" name="nomePlano" id="nomePlano" class="typeTextMedio" placeholder="Nome"  required /><br /><br />

    <textarea placeholder="Descrição" class="textarea" name="descricao" id="descricao" cols="93" rows="15" required></textarea><br /> <br />

    <div id="multiple_upload">
        <input type="hidden" id="arquivos" name="arquivos" onChange="alert(this.value)" />
        <input type="file" multiple id="fotos"  name="fotos[]" />
        <div id="message">Selecionar fotos</div> 
        <div id="lista"></div>
    </div>          

    <img class="spinner" src="../_img/_bannerImgs/spinner.gif" />          
    <input type="submit" class="btnAcesso" value="Enviar" />

    <label class="resposta"></label>

  </form><br /><br />

  <script> 
    $('.elevate-image').ezPlus({
        zoomType: 'inner',
        cursor: 'crosshair'     
    });
  </script>  

</div>  

In it I have a field FILE for sending of files.

I’m using $.post() of jQuery to send this form one-page php without refresh.

All’s well and I can take php but the field FILE will not.

I’m trying kind of gabiarra but I still had to stop and ask for help.

The jQuery:

// JavaScript Document
$(document).ready(function(e) {

    $("div.conteudo div.plano form.planoCadastrar").on("submit", function() {

        var nomePlano = $("form.planoCadastrar input[type=text]#nomePlano").val();
        var descricao = $("form.planoCadastrar textarea#descricao").val();
        var arquivos  = $("form.planoCadastrar input[type=hidden]#arquivos").val();
        var fotosPost = $("form.planoCadastrar input[type=file]")[0].files;

        fotos = new Array();
             for (var i=0; i < fotosPost.length; i++) { 

                 fotos[i]["name"]     = fotosPost[i].name
                 fotos[i]["type"]     = fotosPost[i].type
                 fotos[i]["size"]     = fotosPost[i].size
                 fotos[i]["tmp_name"] = fotosPost[i].tmp_name

             }

             return false;

        if ( nomePlano == "" || 
             descricao == "") {

            alert("Algum campo está vazio!");

        } 


         $("div.conteudo div.plano form.planoCadastrar input[type=submit].btnAcesso").css('display', 'none');
         $("div.conteudo div.plano form.planoCadastrar img").css('display', 'block');

         $.post ("../_requeridos/cadastraPlano.php", {

             nomePlano   : nomePlano,
             descricao   : descricao,
             arquivos    : arquivos,
             fotos       : fotos

         }, function(retorno){

             $("div.conteudo div.plano form.planoCadastrar input[type=submit].btnAcesso").css('display', 'block');
             $("div.conteudo div.plano form.planoCadastrar img").css('display', 'none');

              if (retorno == 1) {
                resposta = "Plano cadastrado com sucesso!"; 
              } else {
                resposta = "Erro no cadastro do Plano";
              }
             $(".resposta").css("display", "block");
             $(".resposta").html(resposta);     

           }
          );

          return false;

    });

});

I tried to do the one gabiarra with the code below:

     for (var i=0; i < fotosPost.length; i++) { 

         fotos[i]["name"]     = fotosPost[i].name
         fotos[i]["type"]     = fotosPost[i].type
         fotos[i]["size"]     = fotosPost[i].size
         fotos[i]["tmp_name"] = fotosPost[i].tmp_name

     }

But when I get to

         fotos[i]["tmp_name"] = fotosPost[i].tmp_name

I can’t get the number.

Does there exist a recourse for that reason?

Or do I really have to submit the form.

The idea is to take the array of FILE and create a array javascript, send to the php and convert again in array php to see if I can do the upload of filing cabinet.