php怎么获取数据库中blob，然后将他显示到datagrid中

        $sql = "select pic from e_user where uid = '1dff5b51f862e6d181577e3ca34248be'";
        $js = get_js_object($sql);
        Header( "Content-type: image/png");
        echo $js->pic;  
        echo '
'; 
?>

 





        
             
                  
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