easyui-php如何獲取數據庫中blob，然後將他顯示到datagrid中

php如何獲取數據庫中blob，然後將他顯示到datagrid中

現在php已經將圖片存入到數據庫中，類型blob。

     <?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" width="150"></p>'; 
?>

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

如何修改一下？

最佳回答：

<table class="easyui-dategrid" url="../php/testlist.php">
<thead>
<tr>
    <th field="pic" width="120"><?php echo $js->pic;?></th> 
</tr>
</thead>
</table>