0
我有這樣的形式提交個人信息笨提交阿賈克斯警報成功
這裏是我的看法
<?php $attributes=a rray('autocomplete'=>"off", 'id' => 'theForm');?>
<?php echo form_open_multipart('employee/personaldetails', $attributes); ?>
<div class="row">
<div class="span3 offset1">
<label>First Name</label>
<input onkeypress="return fnAllowAlpha(event);" maxlength="50" type="text" value="<?php echo $info['first_name']; ?>" name="first_name" disabled>
<input type="hidden" value="<?php echo $info['employee_image']; ?>" name="employee_image" disabled>
</div>
<div class="span3">
<label>Middle Name</label>
<input onkeypress="return fnAllowAlpha(event);" maxlength="50" type="text" value="<?php echo $info['middle_name']; ?>" name="middle_name" disabled>
</div>
<div class="span3">
<label>Last Name</label>
<input onkeypress="return fnAllowAlpha(event);" maxlength="50" type="text" value="<?php echo $info['last_name']; ?>" name="last_name" disabled>
</div>
<div class="span1">
<div class="down1">
<button type="submit" class="button-orange" id="btnSave" style="display:none;">Save</button>
</div>
</div>
控制器
$this->load->model('mdl_employee','emp');
$id = $this->session->userdata('id');
$P1 = $this->input->post('first_name');
$P2 = $this->input->post('middle_name');
$P3 = $this->input->post('last_name');
$this->emp->update_myinfo($id, $P1, $P2, $P3);
模型
public function update_myinfo($id, $P1, $P2, $P3){
$employee_data = array(
'first_name' => $P1,
'middle_name' => $P2,
'last_name' => $P3);
$this->db->where('employee_id', $id);
$this->db->update('employee', $employee_data);
}
我不知道如何使用ajax。 如果成功與否，我想提醒ajax。在更新我的數據庫後..即時新程序員