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="php如何获取数据库中blob，然后将他显示到datagrid中" ></p>'; ?>

 <table class="easyui-dategrid" url="../php/testlist.php">        <thead>             <tr>                  <th field="pic" width="120">图片</th>              </tr>        </thead></table>