php如何获取数据库中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><img src="../php/testlist.php" style="max-width:90%" alt="php怎么获取数据库中blob，然后将他显示到datagrid中" ></p>'; <br />?><br />
<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 />
