Question

我的Ajax脚本不会发送到控制器，并且数据也没有上传。

脚本为：

$("#update").click(function(event) {
/* Act on the event */

var chango = $("#update_form").serialize();
alert(chango);
$.ajax({
    type:"POST",
    url:"<?php echo base_url() ?>home/update_profile",
    data:{id:id}
},
    function(data) {
        console.log(data);
        list_user();
    });
event.preventDefault();
});

我的控制器是：

public function update_profile()
{
    try {

        $this->load->library('form_validation');
        $this->form_validation->set_rules("fname", "First Name", 'required|alpha');
        $this->form_validation->set_rules("lname", "Last Name", 'required|alpha');
        $this->form_validation->set_rules("mobile", "Mobile", 'required');
        $this->form_validation->set_rules("dob", "Date of Birth", 'required');

        $this->form_validation->set_rules("gender", "Gender", 'required');

        if ($this->form_validation->run()){
        $this->load->model("user");

        $data=array(
              'fname' => $this->input->post('fname'),
              'lname' => $this->input->post('lname'),
              'mobile' => $this->input->post('mobile'),
              'dob' => $this->input->post('dob'),
              'gender' => $this->input->post('gender')
            );
        if ($this->input->post("update")) {
            $this->user->update_data($data, $this->input->post("id"));
            redirect(base_url() . "home/list_user");
        }
    }
    else
    {

    }
    } 
    catch(Exception $e) {

    }
    $this->load->view('middlepage/update_profile.php');
}
public function update()
{
    $user_id = $this->uri->segment(3);
    $this->load->model("user");
    $data["user_data"] = $this->user->fetch_single_data($user_id);
    $data["fetch_data"] = $this->user->fetch_data1();
    $this->load->view("middlepage/update_profile", $data);
}

我的模型是：

public function fetch_single_data($id)
    {
        $this->db->where("id", $id);
        $query = $this->db->get("user");
        return $query;
    }
    public function update_data($data, $id)
    {
        $this->db->where("id", $id);
        return $this->db->update("user", $data);
    }

我的看法是：

<form method="post" id="update_form" enctype="multipart/form-data" novalidate="true">
                                        <?php if(isset($user_data))
                                        {
                                            foreach ($user_data->result() as $row) {
                                        ?>
                                        <div class="form-group valid-form">
                                            <h4>First Name:</h4>
                                            <input type="text" class="form-control" id="fname" name="fname" value="<?php echo $row->fname ; ?>">
                                            <span class="text-danger"><?php echo form_error("fname");?></span>
                                        </div><br>

                                        <div class="form-group valid-form">
                                            <h4>Last Name:</h4>
                                            <input type="text" class="form-control" id="lname" name="lname" value="<?php echo $row->lname ; ?>">
                                            <span class="text-danger"><?php echo form_error("lname");?></span>
                                        </div><br>

                                        <div class="form-group valid-form">
                                            <h4>Mobile:</h4>
                                            <input type="text" class="form-control" id="mobile" name="mobile" value="<?php echo $row->mobile ; ?>">
                                            <span class="text-danger"><?php echo form_error("mobile");?></span>
                                        </div><br>

                                        <div class="form-group valid-form">
                                            <h4>Date of Birth:</h4>
                                            <input type="date" class="form-control" min="1984-01-01" max="2005-12-31" id="dob" name="dob" value="<?php echo $row->dob ; ?>">
                                            <span class="text-danger"><?php echo form_error("dob");?></span>
                                        </div><br>

                                        <div class="form-group">
                                            <h4>Gender</h4>
                                            <div class="radio">
                                                <label>
                                                  <input type="radio" name="gender" id="male" value="male"<?php if($row->gender == 'male') echo "checked"; ?>><span style="font-weight: bolder; font-size: 15px;">Male</span>
                                                </label>
                                            </div>
                                            <div class="radio">
                                                <label>
                                                <input type="radio" name="gender" id="female" value="female"<?php if($row->gender == 'female') echo "checked"; ?>><span style="font-weight: bolder; font-size: 15px;">Female</span>
                                                </label>
                                            </div>
                                            <span class="text-danger"><?php echo form_error("gender");?></span>
                                        </div><br>

                                        <div class="form-group">
                                            <input type="hidden" name="id" value="<?php echo $row->id ?>">
                                            <input type="submit" name="update" id="update" value="Update" class="btn btn-primary">
                                        </div>
                                        <?php
                                                }
                                            }
                                        ?>
                                    </form>

我无法更新数据库中的数据，也不会列出表 ajax功能无法正常工作，并且无法转到codeigniter的控制器

虽然正在尝试使用这种类似的代码来提交，但效果很好

Answer 1

您从哪里获得ID，并在ID字段中作为数据传递

$.ajax({
    type:"POST",
    url:"<?php echo base_url() ?>home/update_profile",
    data:{id:id} <--
},

ajax不会发送到控制器，数据也不会更新

1 个答案: