什么检索图像从数据库使用JSON AJAX函数codeigniter

Question

这是我第一次试图检索使用AJAX/JSON格式的图像，所有其他变量显示除了图像，因为我不知道如何将语法集成在AJAX， 当我打开浏览器上的检查元素的图像名称正在显示，并将其保存到图像文件夹以及正确 我只需要你们帮助我在没有功能的窗体里面显示图像 Thanks in推进

这是我的AJAX功能

//Ajax Load data from ajax 
 
    $.ajax({ 
 
     url : "<?php echo site_url('person/ajax_edit/')?>/" + firstname, 
 
     type: "GET", 
 
     dataType: "JSON", 
 
     success: function(data) 
 
     { 
 

 
      $('[name="id"]').val(data.id); 
 
      $('[name="firstName"]').val(data.firstname); 
 
      $('[name="lastName"]').val(data.lastname); 
 
      $('[name="gender"]').val(data.gender); 
 
      $('[name="address"]').val(data.address); 
 
\t \t \t $('[name="gender"]').val(data.gender); 
 
\t \t \t $('[name="telephone"]').val(data.telephone); 
 
\t \t \t $('[name="level"]').val(data.level); 
 
\t \t \t $('name="image"').val(data.image) 
 
\t \t \t $('[name="religion"]').val(data.religion); 
 
\t \t \t $('[name="entrance"]').val(data.entrance); 
 
\t \t \t $('[name="graduate"]').val(data.graduate); 
 
\t \t \t $('[name="parents"]').val(data.parents); 
 
      $('[name="dob"]').datepicker('update',data.dob); 
 
      $('#modal_form').modal('show'); // show bootstrap modal when complete loaded 
 
      $('.modal-title').text('Edit Person'); // Set title to Bootstrap modal title 
 

 
     }, 
 
     error: function (jqXHR, textStatus, errorThrown) 
 
     { 
 
      alert('Error get data from ajax'); 
 
     } 
 
    });

这是控制器

public function ajax_edit($firstname) 
 
\t { 
 
\t \t $data = $this->person->get_by_id($firstname); 
 
\t \t $data->dob = ($data->dob == '0000-00-00') ? '' : $data->dob; // if 0000-00-00 set tu empty for datepicker compatibility 
 
\t \t 
 
\t \t echo json_encode($data); 
 
\t }

这是我的模型

public function get_by_id($firstname) 
 
\t { 
 
\t \t /* 
 
\t \t $this->db->from($this->table); 
 
\t \t $this->db->where('id',$id); 
 
\t \t $query = $this->db->get(); 
 

 
\t \t return $query->row(); 
 
\t \t */ 
 
\t \t 
 
\t \t 
 
\t \t $this->db->select("e.*,edu.*"); 
 
\t \t $this->db->from("student_details e"); 
 
\t \t $this->db->join("images edu", "edu.firstname = e.firstname", 'left'); 
 
\t \t $this->db->where('e.firstname', $firstname); 
 
\t \t $result = $this->db->get(); 
 
\t \t return $result->row(); 
 
\t \t 
 
\t }

这是我的看法

<div class="form-group"> 
 
          <label class="control-label col-md-3">Religion</label> 
 
          <div class="col-md-9"> 
 
           <select name="religion" class="form-control"> 
 
            <option value="">--Select Religion--</option> 
 
            <option value="Christain">Christain</option> 
 
            <option value="Muslim">Muslim</option> 
 
            </select> 
 
           <span class="help-block"></span> 
 
          </div> 
 
         </div> 
 
         
 
         <div class="form-group"> 
 
         <label class="control-label col-lg-4">Image</label> 
 
         <div class=""> 
 
          <div class="fileupload fileupload-new" data-provides="fileupload"> 
 
           <div class="fileupload-new thumbnail" style="width: 200px; height: 150px;"><img src="assets/img/demoUpload.jpg" alt="" /></div> 
 
           <div class="fileupload-preview fileupload-exists thumbnail" style="max-width: 200px; max-height: 150px; line-height: 20px;"></div> 
 
           <div align="center"> 
 
            <span class="btn btn-file btn-primary"><span class="fileupload-new">Select image</span><span class="fileupload-exists">Change</span><input type="file" name="userfile" ></span> 
 
            
 
            <a href="#" class="btn btn-danger fileupload-exists" data-dismiss="fileupload">Remove</a> 
 
            <span class="help-block"></span> 
 
           </div> 
 
          </div> 
 
         </div> 
 
         </div> 
 
        

Answer 1

数据库里面的数据字段是图像和文件的名称是userfile的不是图像...我做了Ajax的功能，第二我有内部错误纠正它BT仍然无法正常工作