easyui-php如何获取数据库中blob，然后将他显示到datagrid中

     <?php        $sql = "select pic from e_user where uid = '1dff5b51f862e6d181577e3ca34248be'";        $js = get_js_object($sql);        Header( "Content-type: image/png");        echo $js->pic;          echo '<p><img  src="../php/testlist.php"    style="max-width:90%" alt="easyui-php如何获取数据库中blob，然后将他显示到datagrid中" ></p>
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            <table class="easyui-dategrid" url="../php/testlist.php">                 <thead>                   <tr>                        <th field="pic" width="120">图片</th>                     </tr>            </thead>    </table>