摘要: $tables = $db->query("select table_name from information_schema.tables where table_schema='yuyue' and table_name like 'patient_%' "); for($i=0;$i<1100 阅读全文
posted @ 2018-01-15 15:25 zhangsima 阅读(124) 评论(0) 推荐(0) 编辑