Question

我试图创建排行榜，但我不确定如何进行mysql查询。

我想计算技能表中玩家的所有等级并获得总等级并计算经验表中玩家的所有经验，并获得总Exp以及显示用户列中的人员姓名

有3个表factions_mcmmo_users，factions_mcmmo_experience，factions_mcmmo_skills。

这是我到目前为止所做的，但它不起作用：

    $sql = ("SELECT a.id,
(SELECT COUNT(*) FROM factions_mcmmo_experience WHERE user_id = a.id) as TotalXP,
(SELECT COUNT(*) FROM factions_mcmmo_skills WHERE user_id = a.id) as TotalLevel
FROM (SELECT DISTINCT id FROM factions_mcmmo_users) a LIMIT 10;");

非常感谢任何帮助

编辑：我现在已经开始工作，但我不确定它是否是最有效的做事方式，如果有人能帮助我，如果有更好的方法，那就意味着很多。

我还想知道是否可以用逗号显示总exp和等级，如果数字是数千例如：总等级5,882和总xp 582,882

编辑2： 我已经弄清楚如何格式化数字但仍然不知道我的代码是否有效

$sql = ("SELECT id, user,
(SELECT FORMAT(Sum(taming)+Sum(mining)+Sum(woodcutting)+Sum(repair)+Sum(unarmed)+Sum(herbalism)+Sum(excavation)+Sum(archery)+Sum(swords)+Sum(axes)+Sum(acrobatics)+Sum(fishing)+Sum(alchemy),0) FROM factions_mcmmo_skills b WHERE b.user_id = a.id) as TotalLevel,
(SELECT FORMAT(Sum(taming)+Sum(mining)+Sum(woodcutting)+Sum(repair)+Sum(unarmed)+Sum(herbalism)+Sum(excavation)+Sum(archery)+Sum(swords)+Sum(axes)+Sum(acrobatics)+Sum(fishing)+Sum(alchemy),0) FROM factions_mcmmo_experience c WHERE c.user_id = a.id) as TotalXP
FROM (SELECT id, user FROM factions_mcmmo_users) a group by id ORDER BY TotalLevel DESC, TotalXP DESC LIMIT 10;");

编辑3 更新了scaisEdge中的代码，但是将每个人的级别显示为1，XP显示为1，所以我将count（*）更改为sum，按顺序添加了TotalLevel的顺序，这似乎已经有效但我无法得到它在用户表中显示人员姓名（用户列）？不确定我是否应该改为总和，因为它没有相反的方式。

$sql = ("SELECT a.id, b.TotalXP, c.TotalLevel
   FROM (SELECT DISTINCT id FROM factions_mcmmo_users) a 
   INNER JOIN (
        SELECT user_id, Sum(taming)+Sum(mining)+Sum(woodcutting)+Sum(repair)+Sum(unarmed)+Sum(herbalism)+Sum(excavation)+Sum(archery)+Sum(swords)+Sum(axes)+Sum(acrobatics)+Sum(fishing)+Sum(alchemy) as TotalXP 
        FROM factions_mcmmo_experience
        GROUP By user_id
   ) b on b.user_id = a.id
   INNER JOIN (
       SELECT user_id, Sum(taming)+Sum(mining)+Sum(woodcutting)+Sum(repair)+Sum(unarmed)+Sum(herbalism)+Sum(excavation)+Sum(archery)+Sum(swords)+Sum(axes)+Sum(acrobatics)+Sum(fishing)+Sum(alchemy)  as TotalLevel
       FROM factions_mcmmo_skills
       GROUP by user_id

   ) c on  c.user_id = a.id
   ORDER BY TotalLevel DESC
   LIMIT 10;");

编辑4 一切正常但当我尝试使用内部联接上的＆＃34; FORMAT（Sum（Columns），0）格式化总计时，EXP Total似乎有效但主要的Total Level不显示超过1,000的结果打破排行榜定位，它应该在总水平上排序，但它似乎是随机的，当你删除格式时，0它会恢复工作

如果数字是数千，我希望它显示逗号：总水平：5,532和总EXP 5882,882

查看现场演示：http://mcbuffalo.com/playground/leaderboards/server/factions-mcmmo.php

更新了代码尝试使用格式：

    $sql = ("SELECT a.id, a.user, b.TotalXP, c.TotalLevel
   FROM (SELECT id, user FROM factions_mcmmo_users) a 
   INNER JOIN (
        SELECT user_id, FORMAT(Sum(taming)+Sum(mining)+Sum(woodcutting)+Sum(repair)+Sum(unarmed)+Sum(herbalism)+Sum(excavation)+Sum(archery)+Sum(swords)+Sum(axes)+Sum(acrobatics)+Sum(fishing)+Sum(alchemy), 0) as TotalXP 
        FROM factions_mcmmo_experience
        GROUP By user_id
   ) b on b.user_id = a.id
   INNER JOIN (
       SELECT user_id, FORMAT(Sum(taming)+Sum(mining)+Sum(woodcutting)+Sum(repair)+Sum(unarmed)+Sum(herbalism)+Sum(excavation)+Sum(archery)+Sum(swords)+Sum(axes)+Sum(acrobatics)+Sum(fishing)+Sum(alchemy), 0)  as TotalLevel
       FROM factions_mcmmo_skills
       GROUP by user_id

   ) c on  c.user_id = a.id
   ORDER BY TotalLevel DESC;");

编辑5 使用PHP更改了数字，一切正常

原始图片

Answer 1

你可以使用几个内连接

$sql = ("SELECT a.id, a.name,  b.TotalXP, c.TotalLevel
       FROM (SELECT DISTINCT id, name FROM factions_mcmmo_users) a 
       INNER JOIN (
            SELECT user_id, COUNT(*) as TotalXP 
            FROM factions_mcmmo_experience
            GROUP By user_id
       ) b on b.user_id = a.id
       INNER JOIN (
           SELECT user_id, COUNT(*)  as TotalLevel
           FROM factions_mcmmo_skills
           GROUP by user_id

       ) c on  c.user_id = a.id
       LIMIT 10

MYSQL PHP查询中的多个计数

1 个答案: