我怎样才能使回车键提交这个数据库
How can i make Enter Key submit this to database?
<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', $this->data);
}
这目前工作时,我点击更改密码按钮,但我想使它与输入键的工作。这是使用Codeigniter,并试图改变它使用标准的html输入类型。但这并没有奏效。如有任何帮助,不胜感激
用表单进行换行,浏览器会处理的。同时使用提交按钮类型:
<?php echo form_open('',array('class'=>'ajaxForm')); ?>
<div class="modal-header">
<a class="modal-close-med" data-dismiss="modal" aria-hidden="true"><img src="<?php echo base_url(); ?>img/close.png"></a>
<h3 class="modal-title no-margin margin-bottom-5p-min">Change Password</h3>
</div>
<div class="modal-body">
<?php if(isset($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();?>
如果您使用这个ajax请求,您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>
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