Question

我一直在使用这个查询试图让它使用五个表输出嵌套结果，但一次又一次失败我认为我想使用子查询但我不明白它们足以放它们在我的代码中使用。

我一直在打我的头，有些请告诉我如何得到我正在寻找的结果。

我需要输出我的10个朋友的问题，其中有5个他们尊重的答案以及1个类别和8个类别标签

以下是10个输出应该是什么样的：

朋友： John Doe 1
- 类别：数学
- 标签： tag1 tag2，tag3等...
问题：什么是1 + 1？
数学问题答案1
数学问题答案2
数学问题答案3
数学问题答案4
数学问题答案5
等...

为了实现上述目标，我使用了以下MYSQL语句和PHP代码段

<?php 

$query = $db->prepare(
    "SELECT
        question.post,          
        question.id,     
        question.user_id,

        question_responses.id response_id,
        question_responses.response,

        category.id category_id,
        category.category_name,
        category.category_posts_id cpid,

        tags.tag_name tn,
        tags.id tagid,

        friends.id myid,
        friends.logged_username,

    FROM (SELECT * FROM question ORDER BY question.id DESC LIMIT 10) AS question

    LEFT JOIN friends ON friends.friend_id = question.user_id    
    LEFT JOIN (SELECT * FROM question_responses LIMIT 5) AS question_responses ON question_responses.question_response_id = question.id
    LEFT JOIN category ON category.user_id = question.user_id
    LEFT JOIN (SELECT * FROM tags LIMIT 8) AS tags ON tags.top_tags = category.category_id

    WHERE friends.user_id = ?");

    $id = "1";

$query->bindValue(1, $id, PDO::PARAM_INT);

try {

    $query->execute();
    $questions = array();
    $question_responses = array();
    $tag = array();

    while($row = $query->fetch(PDO::FETCH_ASSOC)) {
        $question_id = $row['id'];
        $tags_id = $row['tagid'];

        $questions[$question_id] = $row;
        $tags[$tags_id][] = $row;
        $question_responses[$question_id][] = $row;
    }

    foreach ($questions as $question_id => $row) { 
        foreach ($tag as $tags_id => $row) {
            echo $row['tn'];
        }
        echo "<b>".$row['post']."</b></br></br>";
        foreach ($question_responses[$question_id] as $response_id => $row) {    
            echo $row['response']."</br></br>";
        }

    }

} catch (PDOException $e) {
    echo $e->getMessage();
    exit();
}
?>

提前感谢您的帮助。

Answer 1

使用大量用户变量执行此操作的可能方法。这可能很慢。它有子查询，可以在添加计数的情况下返回结果，然后丢弃ON子句中不需要的结果。

SELECT
    question.post,          
    question.id,     
    question.user_id,

    question_responses.id response_id,
    question_responses.response,

    category.category_id,
    category.category_name,
    category.cpid,

    category.tn,
    tagstagid,

    friends.id myid,
    friends.logged_username

FROM  friends
INNER JOIN 
(   
    SELECT category.user_id,
            category.id AS category_id,
            category.category_name,
            category.category_posts_id AS cpid,
            tags.tag_name AS tn,
            tags.id AS tagid,
            @category_user_id_cnt:=IF(@category_user_id = user_id, @category_user_id_cnt + 1, 1) AS category_user_id_cnt,
            @tag_cnt:=IF(@category_user_id = user_id, @tag_cnt + 1, 1) AS tag_cnt,
            @category_user_id := user_id
    FROM category
    CROSS JOIN (SELECT @category_user_id=0, @category_user_id_cnt:=0, @tag_cnt:=0) sub0
    LEFT OUTER JOIN tags
    ON tags.top_tags = category.category_id
) AS category
ON category.user_id = friends.user_id
AND category.category_user_id_cnt = 1
AND category.tag_cnt <= 8
LEFT JOIN
(
    SELECT post,          
            id,     
            user_id,
            @question_user_id_cnt:=IF(@question_user_id = user_id, @question_user_id_cnt + 1, 1) AS question_user_id_cnt,
            @question_user_id := user_id
    FROM
    (
        SELECT post,          
                id,     
                user_id 
        FROM question 
        ORDER BY user_id, id DESC 
    ) sub_question
    CROSS JOIN (SELECT @question_user_id=0, @question_user_id_cnt:=0) sub0
) AS question
ON friends.friend_id = question.user_id    
AND question.question_user_id_cnt <= 10
LEFT JOIN 
(
    SELECT id response_id, 
            question_responses.response,  
            @question_response_cnt:=IF(@question_response_id = question_response_id, @question_response_cnt + 1, 1) AS question_response_cnt,
            @question_response_id := question_response_id
    FROM question_responses 
    CROSS JOIN (SELECT @question_response_id=0, @question_response_cnt:=0) sub0
) AS question_responses ON question_responses.question_response_id = question.id AND question_responses.question_response_cnt <= 5

WHERE friends.user_id = ?

使用GROUP_CONCAT和SUBSTRING_INDEX的技巧可以使用子查询来获取最新的X ID，并将这些X ID连接到表中以获取其余数据。然而，这样做我很难挣扎，因为我不知道朋友是否只能回答一个问题。

MYSQL使用四个LEFT JOINS和LIMIT CLAUSES显示嵌套结果

1 个答案: