解决方案_解决当distinct和join同时存在distinct失效问题

$sql = "select distinct(ontopid),gb.id as id,f.id as fid,g.id as gid,g.*,gb.*,f.* from pk_groupbuy gb

        left join pk_ontop as o on o.ontopid=gb.id

        left join pk_goods g on gb.goodsid=g.id and g.status=2 and g.invalid>UNIX_TIMESTAMP()

        left join pk_fastgroupbuy f on gb.fastgroupbuyid=f.id

        where gb.id in (".$arr_str.") and (gb.status="2" or gb.status="3")

        and gb.endtime>UNIX_TIMESTAMP() and gb.starttime

        group by onid limit $start,$num";

return TableSystem::query($sql);

变量说明：$arr_str是一个由pk_groupbuy中主键 id组成的一个数组，经过explode函数得到的字符串，

          $start,$num分别是查询的开始记录数，和要查询的记录数。

问题说明：pk_ontop表中ontopid在不能记录中有重复现象

比如：我只需要查询出来pk_ontop中当天置顶的ontopid，即商品id，不需要其他的商品信息，查询出来的有ontopid就算有重复现象，这时我可以通过去除数组重复元素解决问题，但是如果我要查询出相应商品id并查询其他相关联表中的信息，并按照ontop表中starttime，status，paixu字段进行排序等操作时，就需要join pk_ontop表，所以之前解决的重复问题就又会出现，无法处理，特别是在api中，是不允许出现重复的，这要怎么办呢？我也不会额，别人教我这样弄，请大家参考下：

 

    $sql = "SELECT DISTINCT(ontopid),starttime,paixu FROM pk_ontop ORDER BY starttime DESC,STATUS ASC,paixu ASC LIMIT 17";
    $arr = TableSystem::query($sql);
    foreach($arr as $key=>$val){
     $topids[$key] = $val["ontopid"];
    }
    $arr_str = implode(",",$topids);
    $arr1 = TableSystem::query($sql);

    $sql = "select gb.local,f.phone,f.shopname as fshopname,gb.maxnum,gb.intro,gb.buynum,g.pic,f.googleaddress,gb.goodsclassid,gb.sellerid,f.img,gb.province,gb.city,gb.id,gb.title,g.pic,gb.starttime,
      gb.endtime,gb.price,gb.goodsprice from pk_groupbuy gb
      left join  pk_goods g on gb.goodsid=g.id and g.status=2 and g.invalid > UNIX_TIMESTAMP()
      left JOIN  pk_fastgroupbuy f ON f.id=gb.fastgroupbuyid
      where (gb.status="2" or gb.status="3") and gb.endtime > UNIX_TIMESTAMP()
      and gb.starttime < UNIX_TIMESTAMP()  AND gb.id in (".$arr_str.")";
   
    $arr2 = TableSystem::query($sql);
    foreach($arr2 as $key=>$val){
     $local[$val["id"]] = $val["local"];
     $phone[$val["id"]] = $val["phone"];
     $fshopname[$val["id"]] = $val["fshopname"];
     $maxnum[$val["id"]] = $val["maxnum"];
     $intro[$val["id"]] = $val["intro"];
     $buynums[$val["id"]] = $val["buynum"];
     $fgoogleaddresss[$val["id"]] = $val["googleaddress"];
     $goodsclassid[$val["id"]] = $val["goodsclassid"];
     $sellids[$val["id"]] = $val["sellerid"];
     $provices[$val["id"]] = $val["province"];
     $citys[$val["id"]] = $val["city"];
     $titles[$val["id"]]= $val["title"];
     $pics[$val["id"]] = $val["pic"] ? $val["pic"] : $val["img"];
     $starttimes[$val["id"]] = $val["starttime"];
     $endtimes[$val["id"]] = $val["endtime"];
     $prices[$val["id"]] = $val["price"];
     $goodsprices[$val["id"]] = $val["goodsprice"];
    }
    unset($arr2);
    foreach($arr1 as $key=>$val){
     $list[$key]["id"] = $val["ontopid"];
     $list[$key]["province"] = $provices[$val["ontopid"]];
     $list[$key]["city"] = $citys[$val["ontopid"]];
     $list[$key]["title"] = $titles[$val["ontopid"]];
     $list[$key]["pic"] = $pics[$val["ontopid"]];
     $list[$key]["starttime"] = $starttimes[$val["ontopid"]];
     $list[$key]["endtime"] = $endtimes[$val["ontopid"]];
     $list[$key]["price"] = $prices[$val["ontopid"]];
     $list[$key]["goodsprice"] = $goodsprices[$val["ontopid"]];
     $list[$key]["sellerid"] = $sellids[$val["ontopid"]];
     $list[$key]["fgoogleaddress"] = $fgoogleaddresss[$val["ontopid"]];
     $list[$key]["goodsclassid"] = $goodsclassid[$val["ontopid"]];
     $list[$key]["buynum"] = $buynums[$val["ontopid"]];
     $list[$key]["intro"] = $intro[$val["ontopid"]];
     $list[$key]["maxnum"] = $maxnum[$val["ontopid"]];
     $list[$key]["fshopname"] = $fshopname[$val["ontopid"]];
     $list[$key]["fphone"] = $phone[$val["ontopid"]];
     $list[$key]["local"] = $local[$val["ontopid"]];
    }
    return $list;

 

解决方案_解决当distinct和join同时存在distinct失效问题

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