php怎么获取数据库中blob，然后将他显示到datagrid中

<br /><?php<br />        $sql = "select pic from e_user where uid = '1dff5b51f862e6d181577e3ca34248be'";<br />        $js = get_js_object($sql);<br />        Header( "Content-type: image/png");<br />        echo $js->pic;  <br />        echo '<p>@@##@@</p>
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<br /> <table class="easyui-dategrid" url="../php/testlist.php"><br />        <thead><br />             <tr><br />                  <th field="pic" width="120">图片</th> <br />             </tr><br />        </thead><br /></table><br />