Question

 <div class="modal-header">
     <a class="modal-close-med" data-dismiss="modal" aria-hidden="true"><img src="<?php echo base_url(); ?>img/close.png"></a>
    <h3 class="modal-title no-margin margin-bottom-5p-min">Change Password</h3>
</div>

<div class="modal-body"> 
    <?php if(isset($message))echo '<span class="text-success txt-upper" style="margin-left:2rem;">'. $message .'</span>';?>

    <?php echo form_open('',array('class'=>'ajaxForm')); ?>
      <fieldset class="table "> 
        <div class="form-group">
        <?php $class = form_error('newpassword')?"input-error":"" ?>
          <div  class="col-md-12" style="margin: 10px 0"><?php echo form_password('newpassword','','class="form-control margin-both-0 '. $class.'" id="newpassword"  placeholder="New Password" autocomplete="off"'); ?><?php echo form_error('newpassword'); ?></div>            
        </div>
        <div class="form-group">
        <?php $class = form_error('conpassword')?"input-error":"" ?>
          <div class="col-md-12 " style="margin: 10px 0"><?php echo form_password('conpassword','','class="form-control  margin-both-0 '. $class.'" id="conpassword"  placeholder="Confirm Password" autocomplete="off"'); ?><?php echo form_error('conpassword'); ?></div>           
        </div>  

</div>
<div class="modal-footer">
    <form method="post" action="" id="myform">
        <?php echo form_submit('submit_btn', 'Change Password', 'class="submit btn btn-success margin-left-4p pad-1-rem margin-bottom-10"'); ?>
    </form>
     </fieldset>
    <?php echo form_close();?>  
  </div>

控制器：

public function change_password ()
{
    if($this->input->post('submit_btn')){
        $this->session->set_flashdata('success','Password Changed Successfully');
        redirect('dashboard');
    }   
    // Set up the form
    $rules = array(

        'newpassword' => array(
            'field' => 'newpassword', 
            'label' => 'New Password', 
            'rules' => 'trim|required|min_length[6]|xss_clean|check_pass'
        ), 
        'conpassword' => array(
            'field' => 'conpassword', 
            'label' => 'Confirm Password', 
            'rules' => 'trim|required|min_length[6]|matches[newpassword]|xss_clean|check_pass'
        ), 
    );


    $this->form_validation->set_rules($rules);
    $this->form_validation->set_message('required', 'this field is required');

    // Process the form
    if ($this->form_validation->run() == TRUE) {
        $password = $this->input->post("newpassword");
        $userid = $this->session->userdata("id");               
        if($this->user_m->change_password($password, $userid)){
            $this->data['message'] = 'password changed successfully';
            //$this->data['subview'] = 'home/index';
            //$this->load->view('_layout_main_1', $this->data);
        }else{
            $this->data['message'] = 'password must have at least one uppercase letter and a number';
        }               
        $this->data['refresh'] = true;
    }

    // Load view

    $this->load->view('home/change_password', $this->data);
}

当我点击更改密码按钮时，这当前有效，但我想使用回车键。这是使用Codeigniter并尝试使用标准html输入类型更改它。但它不起作用。任何帮助将不胜感激

Answer 1

使用表单进行换行，浏览器将处理此问题。还可以使用提交按钮类型：

<?php echo form_open('',array('class'=>'ajaxForm')); ?>
<div class="modal-header">
    <a class="modal-close-med" data-dismiss="modal" aria-hidden="true"><img src="<?php echo base_url(); ?>img/close.png"></a>
    <h3 class="modal-title no-margin margin-bottom-5p-min">Change Password</h3>
</div>
<div class="modal-body">
    <?php if(isset($message))echo '<span class="text-success txt-upper" style="margin-left:2rem;">'. $message .'</span>';?>

    <fieldset class="table ">
        <div class="form-group">
            <?php $class = form_error('newpassword')?"input-error":"" ?>
            <div  class="col-md-12" style="margin: 10px 0"><?php echo form_password('newpassword','','class="form-control margin-both-0 '. $class.'" id="newpassword"  placeholder="New Password" autocomplete="off"'); ?><?php echo form_error('newpassword'); ?></div>
        </div>
        <div class="form-group">
            <?php $class = form_error('conpassword')?"input-error":"" ?>
            <div class="col-md-12 " style="margin: 10px 0"><?php echo form_password('conpassword','','class="form-control  margin-both-0 '. $class.'" id="conpassword"  placeholder="Confirm Password" autocomplete="off"'); ?><?php echo form_error('conpassword'); ?></div>
        </div>
    </fieldset>
</div>
<div class="modal-footer">
    <?php echo form_submit('submit_btn', 'Change Password', 'class="submit btn btn-success margin-left-4p pad-1-rem margin-bottom-10"'); ?>
</div>
<?php echo form_close();?>

如果您将此用于ajax请求，则可以禁用发送表单本身，但输入仍然有效：

<script>
    $(function() {
        $('.ajaxForm').on('submit', function(e) {
            // ex.: $.post(...
            console.log(e);
            e.preventDefault();
        })
    })
</script>

如何让Enter Key将此提交到数据库？

1 个答案: