SQL 使用 COUNT() 和 LEFT JOIN 合并两个或三个表

Question

我确定之前有人问过这个问题，但我无法让我的代码工作:(

用户提出了多个问题。我需要显示用户进行某种用户搜索，然后显示每个用户问了多少问题。

表 1（用户）： ID， 用户名， 密码， 头像， 注册日期， is_banned (ETC...)

表 2（问题）： ID， questioner_id, 问题， 回答1， 回答2， 答案 3（等...）

我需要将两者合并，然后显示用户名等普通用户信息，但还要计算 questioner_id 行并显示用户提出了多少问题。

这是我到目前为止所拥有的，但它只是吐出一个结果并计算所有内容：

SELECT 
    users.id,
    users.username,
    users.avatar_location,
    users.datetime,
    users.last_action,
    users.last_action_description,
    users.is_banned,
    COUNT(questions.questioner_id)
FROM 
    `users` 
LEFT JOIN 
    `questions` 
ON 
    users.id = questions.questioner_id
ORDER BY
    datetime 
    ASC 

我很抱歉我完全“笨拙”。我已经找了一两个小时来了解如何做到这一点，但我无法解决它。

非常感谢

编辑：

非常感谢您的帮助！我最后的附加问题是我现在还有三分之一的桌子，这次我想计算回答的问题，因为我已经提出了问题。

表 3（questions_answered）： ID， question_id, 用户身份， 正确（ETC...）

我已尝试将其添加到我的查询中，但不是显示回答的问题数，而是重复提出问题的结果。

这是我更新后的查询（再次抱歉，我只是在努力解决这个问题）

SELECT 
    users.id,
    users.username,
    users.avatar_location,
    users.datetime,
    users.last_action,
    users.last_action_description,
    users.is_banned,
    COUNT(questions_answered.question_id),
    COUNT(questions.questioner_id)
FROM 
    `users` 
LEFT JOIN 
    `questions` 
ON 
users.id = questions.questioner_id
LEFT JOIN
    `questions_answered`
ON
    users.id = questions_answered.user_id
GROUP BY
    users.id,
    users.username,
    users.avatar_location,
    users.datetime,
    users.last_action,
    users.last_action_description,
    users.is_banned
ORDER BY
    datetime 
    ASC 

非常感谢您的帮助！

Answer 1

第一部分（两张表）

我在查询字符串中有我的 cmets：

SELECT 
    users.id,
    users.username,
    users.avatar_location,
    users.datetime,
    users.last_action,
    users.last_action_description,
    users.is_banned,
    COUNT(questions.id) as number_of_questions #count each question
FROM 
    `users` 
LEFT JOIN 
    `questions` 
ON 
    users.id = questions.questioner_id
GROUP BY
    users.id # you need to have unique user id in each row
ORDER BY
    questions.datetime # sort by question date right? 
    DESC 

关于更新（三张表）：

您需要嵌套选择，您不能同时对两列执行计数。我们把第一个select 当作users_questions_count 像一张桌子，那么一切又和两张桌子一样。错误的部分可能会在名称上产生歧义。

SELECT 
    users_questions_count.id,
    users_questions_count.username,
    users_questions_count.avatar_location,
    users_questions_count.datetime,
    users_questions_count.last_action,
    users_questions_count.last_action_description,
    users_questions_count.is_banned,
    users_questions_count.number_of_questions,
    COUNT(questions_answered.id) as number_of_answers # make sure you are counting the correct field!
FROM 
    (SELECT 
        users.id as id,
        users.username as username,
        users.avatar_location as avatar_location,
        users.datetime as datetime,
        users.last_action as last_action,
        users.last_action_description as last_action_description,
        users.is_banned as is_banned,
        COUNT(questions.id) as number_of_questions #count each question
    FROM 
        `users` 
    LEFT JOIN 
        `questions` 
    ON 
        users.id = questions.questioner_id
    GROUP BY
        users.id 
    ) as users_questions_count
LEFT JOIN 
    `questions_answered` 
ON 
    users_questions_count.id = questions_answered.user_id
GROUP BY
    users_questions_count.id 

三个表的总体思路（table1、table2、table3 加入table1.id、table2.item_id、table3.item_id）：

SELECT 
    table1_table2_count.id,
    table1_table2_count.counter_1,
    COUNT(table3.id) as counter_2
FROM 
    (SELECT 
        table1.id as id,
        COUNT(table2.id) as counter_1
    FROM 
        `table1` 
    LEFT JOIN 
        `table2` 
    ON 
        table1.id = table2.item_id
    GROUP BY
        table1.id 
    ) as table1_table2_count
LEFT JOIN 
    `table3` 
ON 
    table1_table2_count.id = table3.item_id
GROUP BY
    table1_table2_count.id 

但最好找到一种解决方案来避免在实际庞大的数据中进行此类选择。如果您可以更新一张表并在其上放置counter_1 和counter_2，则写入过程会变慢，但读取（搜索）会更快。

Answer 2

将GROUP BY 添加到您的查询中：

SELECT 
    users.id,
    users.username,
    users.avatar_location,
    users.datetime,
    users.last_action,
    users.last_action_description,
    users.is_banned,
    COUNT(questions.questioner_id)
FROM 
    `users` 
LEFT JOIN 
    `questions` 
ON 
    users.id = questions.questioner_id
GROUP BY
    users.id,
    users.username,
    users.avatar_location,
    users.datetime,
    users.last_action,
    users.last_action_description,
    users.is_banned
ORDER BY
    datetime 
    ASC