基本的MySQL查询

Question

提供下一个数据库

我需要提出几个问题，当我尝试：

时，我遇到了麻烦

1. 显示具有船长等级和每个星球上战斗次数的士兵的所有行星列表。

   ID_PLANET | PLANET_NAME | CAPTAINS COUNT | BATTLES COUNT

   SELECT id_planet，planet_name，count（rank） 来自星球 INNER JOIN士兵在planet_id = id_planet 等级=＆＃39;船长＆＃39;;

2. 在他们自己星球的战争中仅的所有士兵的名单。

   ID_SOLDIER | NAME_SOLDIER

   SELECT id_soldier，name 来自士兵 INNER JOIN planet ON planet_id = id_planet INNER JOIN战斗在id_planet = id_planet_battle 在哪里planet_id = id_planet_battle;

3. 包括下一个在内的士兵名单：

   * NAME |排名| PLANET_FROM |来自家庭行星的士兵数量| BATTLES *

   继续努力。

我的尝试是一场灾难，所以经过两天的努力，我在这里寻求帮助。

Answer 1

您可以尝试查询 http://sqlfiddle.com/#!9/839d2/1

查询1

所有拥有船长的行星的清单。它显示了每个行星上的id_planet，planet_name，总队长以及发生在那里的战斗数量（如果有的话）。

var textHide = document.querySelectorAll(".hide");

您可以使用此结果获得相同的结果：

SELECT DISTINCT aa.id_planet, aa.planet_name, _aa.captains_count, _bb.battles_count
FROM planet AS aa
INNER JOIN soldier AS bb
ON aa.id_planet = bb.planet_id
INNER JOIN (
    SELECT planet_id, COUNT(*) AS captains_count
    FROM soldier
    WHERE rank = 'captain'
    GROUP BY planet_id
) AS _aa
ON aa.id_planet = _aa.planet_id
LEFT JOIN (
    SELECT id_planet_battle, COUNT(*) AS battles_count
    FROM battle
    GROUP BY id_planet_battle
) AS _bb
ON aa.id_planet = _bb.id_planet_battle
WHERE bb.rank = 'captain';

查询3

SELECT DISTINCT aa.id_planet, aa.planet_name,
(
    SELECT COUNT(*)
    FROM soldier AS _aa
    WHERE _aa.rank = 'captain' AND aa.id_planet = _aa.planet_id
    GROUP BY _aa.planet_id
) AS captains_count,
(
    SELECT COUNT(*)
    FROM battle AS _bb
    WHERE aa.id_planet = _bb.id_planet_battle
    GROUP BY _bb.id_planet_battle 
) AS battles_count
FROM planet AS aa
INNER JOIN soldier AS bb
ON aa.id_planet = bb.planet_id
WHERE bb.rank = 'captain';

在这里，我没有使用Joins来计算number_of_soldiers和number_of_battles，就像我在查询1.1中所做的那样，因为那将是一个相关的子查询，因此它无法访问外部查询（https://dev.mysql.com/doc/refman/5.5/en/from-clause-subqueries.html）。

查询错误：

SELECT aa.name, aa.rank, bb.planet_name AS planet_from, (
    SELECT COUNT(*) 
    FROM soldier AS _aa 
    WHERE _aa.planet_id = aa.planet_id
) AS number_of_soldiers, 
(
    SELECT COUNT(*)
    FROM battle AS _bb
    WHERE _bb.id_planet_battle = aa.planet_id
) AS number_of_battles
FROM soldier AS aa
INNER JOIN planet AS bb
ON aa.planet_id = bb.id_planet;

因此上述查询错误并产生错误：'where子句'中的未知列'aa.id_planet'。

至于您要求的第二个查询，我希望其他人可以尝试一下。