如何接收来自Codeigniter控制器的Ajax响应? [英] How to receive a ajax response from a codeigniter controller?

问题描述

我有一个将数据发送到模型的控制器，模型将这些数据插入mysql中.

I have a controller that send the data to the model and the model insert this data in the mysql.

我想知道插入的行的最后一个ID，但是我想在我的ajax函数中使用该ID来更新表中的信息.

I want to know the last ID of the row inserted, but I want this ID in my ajax function to use to uptade the table with the information.

这是我拥有的:

模型:

  public function add($nome, $documento)
  {
    $dados = array (
              'nome' => $nome,
              'documento' => $documento
    );

    $this->db->insert('clientes', $dados);
    return $this->db->insert_id();
  }

控制器:

    public function add()
{
    // validar

    $nome = $this->input->post('inputNome');
    $documento = $this->input->post('inputDocumento');
    $this->Cliente_model->add($nome, $documento);
    return "ok";
}

ajax函数:

            $(document).ready(function(){
            $("#salvarCliente").click(function(){
                      $.ajax({
                            url: "cliente/add",
                            type: 'POST',
                            data: $("#formCliente").serialize(),
                            success: function(msg){
                                alert(msg);
                                $("#clienteMensagem").html('Cliente cadastrado com sucesso!');
                                $("#table-clientes tr:last").after('<tr><td>'+msg+'</td><td>' + $('#clienteNome').val() + '</td><td>' + $('#clienteDocumento').val() + '</td><td></td></tr>');
                                $("#clienteNome").val('');
                                $("#clienteDocumento").val('');
                            }
                        });
                    return false;
                });
        });

代码将我的数据添加到mysql中，但是在发送数据之前，我无法从console.log或浏览器的警报中看到来自控制器的确定".

The code add my data to mysql, but I can't see the "ok" from the controller on my console.log or in my alert in the browser before I send the data.

我只想将"$ this-> db-> insert_id()"的结果从我的模型返回到我的控制器，再从我的控制器返回到我的ajax函数.

I only want to return the result of "$this->db->insert_id()" from my model to my controller and from my controller to my ajax function.

推荐答案

更改以下内容:

控制器:

 public function add()
  {
      // validar

     $nome = $this->input->post('inputNome');
     $documento = $this->input->post('inputDocumento');
     $res = $this->Cliente_model->add($nome, $documento);
     echo json_encode($res);
 }

ajax函数:

        $(document).ready(function(){
        $("#salvarCliente").click(function(){
                  $.ajax({
                        url: "cliente/add",//Enter full URL
                        type: 'POST',
                        dataType:'JSON',
                        data: $("#formCliente").serialize(),
                        success: function(msg){
                            alert(msg);
                            $("#clienteMensagem").html('Cliente cadastrado com sucesso!');
                            $("#table-clientes tr:last").after('<tr><td>'+msg+'</td><td>' + $('#clienteNome').val() + '</td><td>' + $('#clienteDocumento').val() + '</td><td></td></tr>');
                            $("#clienteNome").val('');
                            $("#clienteDocumento").val('');
                        }
                    });
                return false;
            });
    });

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