Question

我想要做的是,当有人用“一页”表决时,就算选票。我认为,我失去了试图说明如何追踪某一成员投票的情况。我似乎无法在某一成员表决时告诉大家。

//Generate code ID
    $useXID = intval($_GET[ id ]);
    $useXrank = $_GET[ rank ];

    //if($useXrank!=null && $useXID!=null) {
        $rankcheck = mysql_query( SELECT member_id,code_id FROM code_votes WHERE member_id=" .$_MEMBERINFO_ID. " AND WHERE code_id=" .$useXID. " );
            if(!mysql_fetch_array($rankcheck) && $useXrank=="up"){
                $rankset = mysql_query( SELECT * FROM code_votes WHERE member_id=" .$_MEMBERINFO_ID. " );
                $ranksetfetch = mysql_fetch_array($rankset);
                $rankit = htmlentities($ranksetfetch[ ranking ]);
                $rankit+="1";
                mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ( $_MEMBERINFO_ID , $useXID )") or die(mysql_error());
                mysql_query("UPDATE code SET ranking =  ".$rankit."  WHERE ID =  ".$useXID." ");
            }
            elseif(!mysql_fetch_array($rankcheck) && $useXrank=="down"){
                $rankset = mysql_query( SELECT * FROM code_votes WHERE member_id=" .$_MEMBERINFO_ID. " );
                $ranksetfetch = mysql_fetch_array($rankset);
                $rankit = htmlentities($ranksetfetch[ ranking ]);
                $rankit-="1";
                mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ( $_MEMBERINFO_ID , $useXID )") or die(mysql_error());
                mysql_query("UPDATE code SET ranking =  ".$rankit."  WHERE ID =  ".$useXID." ");
            }
            // hide vote links since already voted
            elseif(mysql_fetch_array($rankcheck)){$voted="true";}
    //}

Answer 1

你们使这一局面变得太困难。 2. 对投票使用数字价值,+1,-1,并对表格实施独一无二的限制:

扩大范围:

create table votes (
  pageId    int references pages (pageId)
 ,memberId  int references members (memberId)
 ,value     int check (value in (-1,1))
 ,constraint votes_unique unique (pages,members)
)

现在,你可以“电算元(价值)......获得用户的物品,以及用页等。

Answer 2

象你那样做的事情——只是用所有东西来取代一切形象。

PHP: 星级评级制度概念:

Answer 3

你的第一点是无效的,因为它有两项WHERE条款。

友情链接