时间:2021-07-01 10:21:17 帮助过:23人阅读
................ $result=mysql_query("select year(addtime) as y,group_concat(jiage) as str from table_name group by year(addtime);"); while($row=mysql_fetch_assoc($result)){ echo ''; echo " '; } ............$row[y] "; foreach(explode(',',$row[str]) as $v) echo "$v ; echo '
------解决方案--------------------
$result=mysql_query("select year(add_date) as y,group_concat(yc_jiage) as str from jiagedata group by year(add_date);") or die(mysql_error());
贴出提示。
------解决方案--------------------
执行#12楼后提示什么信息。