php - 使用ajax使用codeigniter表单上传图像

Question

我需要提交带有表单数据的图像，我已经为此搜索了更好的方法，但不幸的是我找不到更好的方法来实现我的目的，之前使用 this.serialize() 方法完成。但它不适用于图像

这是我的代码

查看

 <?php
          $this->load->helper('form');
          $attributes =  array('method'=>'post','name'=>'create_company','id'=>'create_company');
          echo form_open_multipart('',$attributes);?>
    <label>Code : </label> <?php echo form_input('code');?><br/><br/>
    <label>Name : </label> <?php echo form_input('name');?><br/><br/>
    <label>Logo : </label><input type="file" name="userfile"/><br/><br/>
    <label>URL : </label> <?php echo form_input('url');?><br/><br/>
    <label>Description : </label> <textarea name="description" rows="4" cols="50"> </textarea><br/><br/>
    <input type="submit" name="submit" value="Save"/>
    <?php echo form_close(); ?>     
</div>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
    <script>//no need to specify the language
       $(document).ready(function() {

       $('#create_company').on("submit",function(e) {

            e.preventDefault();

            $.ajax({
                type: "POST",
                url: "<?php echo site_url('site/upload'); ?>",
                data: $(this).serialize(),
                success: function(data){
                    var site_url = "<?php echo site_url('site/create_branch_form'); ?>";
                    var json = $.parseJSON(data);
                    site_url = site_url +"/" + json.results[0].id ;
                    alert(site_url);
                    $("#content").load(site_url);
                    alert(data);
                }
           });            
        });
      });
    </script>

Controller

public function upload(){

        //insert company details
    $this->load->model('company_model');
    $data['results'] = $this->company_model->insert_company();
    $this->output->set_output(json_encode($data));                
    }

型号

function insert_company(){
        //user details
        $username =  $this->session->userdata('username');
        //$query = $this->db->get_where('userdetails', array('username' => $username));

        $query =  $this->db->get_where('userdetails',array('username'=>$username));

        foreach ($query->result() as $function_info) 
        {
            $this->userid = $function_info->id;
        }

    $new_company_insert_data = array(

        'code' => $this->input->post('code'),
                    'name' => $this->input->post('name'),
                    'logo' => $imgpath['file_name'],
                    'url' => $this->input->post('url'),
                    'description' => $this->input->post('description'),
                    'createdat' => date('Y-m-d H:i:s',strtotime('')),
                    'status' => 0,
                    'userid' => $this->userid
             ); 

        $this->db->insert('companydetails',$new_company_insert_data);
        //get id of the last inserted row
        $this->load->database();
        $query = $this->db->query("SELECT MAX(id) AS id FROM companydetails");
        return $query->result();
}

Answer 1

尝试使用 FormData 对象。你的 JS 代码中应该有这样的内容:

 $(document).ready(function() {

   $('#create_company').on("submit",function(e) {
      
        e.preventDefault();
        var formData = new FormData(this);

        $.ajax({
            type: "POST",
            url: "<?php echo site_url('site/upload'); ?>",
            data: formData,
            cache: false,
            contentType: false,
            processData: false
            success: function(data){
                //function success
            },
            error: function(data){
            //error function
        }
       });            
    });
  });