如何从表中查找具有不同用户ID的值匹配数

Question

我想要使​​脚本显示与不同用户匹配的值的数量。 例如，在watched表中，我拥有主要用户的id（1）和他观看的电影标题以及其他多个用户的观看电影：

+----+----------+---------+
| id | users_id | watched |
+----+----------+---------+
| 1  | 1        | movie1  |
+----+----------+---------+
| 2  | 1        | movie2  |
+----+----------+---------+
| 3  | 1        | movie3  |
+----+----------+---------+
| 4  | 2        | movie2  |
+----+----------+---------+
| 5  | 2        | movie1  |
+----+----------+---------+
| 6  | 3        | movie1  |
+----+----------+---------+
| 7  | 3        | movie5  |
+----+----------+---------+
| 8  | 3        | movie4  |
+----+----------+---------+

预期产量：

my id = `1`;
matched with id `2` 2 times on titles `movie2`, `movie1`;
matched with id `3` 1 times on titles `movie1`;

或者，如果主要用户的ID为3 ：

my id = `3`;
matched with id `1` 1 times on titles `movie1`;
matched with id `2` 1 times on titles `movie1`;

等等..

我试图做类似的事情：

<?php
session_start();
$connect = mysqli_connect("127.0.0.1", "root", "root", "movie");

$query = mysqli_query($connect, "SELECT * FROM profile INNER JOIN users ON profile.users_id = users.id");

$id = $_SESSION['id'];
echo "my id: " . $id;

$movies = mysqli_query($connect, "SELECT * FROM watched WHERE users_id = '$id'");

$titles = array();
for ($i = 0;$i < $movies->num_rows;$i++)
{
    $result2 = $movies->fetch_assoc();
    $watched = $result2['watched'];
    $find = mysqli_query($connect, "SELECT * FROM watched WHERE watched = '$watched'");

    for ($l = 0;$l < $find->num_rows;$l++)
    {
        $found = $find->fetch_assoc();
        if ($found['watched'] === $result2['watched'] and $found['users_id'] !== $id)
        {
            $titles[] = $found['watched'];

        }
    }
}

$count_titles = array();
foreach ($titles as $a)
{
    @$count_titles[$a]++;
}
$matches = count($count_titles); //num of matches

for ($i = 0;$i < $query->num_rows;$i++)
{
    $result = $query->fetch_assoc();

?>

<div><?php
echo "matched with id " . $result['id'] . " ";

echo $matches . " times";

foreach (array_keys($count_titles) as $key)
{
    echo " on titles " . $key;
}

?></div> 


<?php
} ?>

但是它仅显示整个表中的匹配总数，而不是每个用户的单独匹配总数。 我如何使其像上述那样工作？

三个表：

CREATE TABLE watched (
  `id` INTEGER,
  `users_id` INTEGER,
  `watched` VARCHAR(6)
);

INSERT INTO watched
  (`id`, `users_id`, `watched`)
VALUES
  ('1', '1', 'movie1'),
  ('2', '1', 'movie2'),
  ('3', '1', 'movie3'),
  ('4', '2', 'movie2'),
  ('5', '2', 'movie1'),
  ('6', '3', 'movie1'),
  ('7', '3', 'movie5'),
  ('8', '3', 'movie4');

  CREATE TABLE users (
  `id` INTEGER,
  `name` VARCHAR(355),
  `email` VARCHAR(355)
);

INSERT INTO users
  (`id`, `name`, `email`)
VALUES
  ('1', 'name1', 'email1@mail.com'),
  ('2', 'name2', 'email2@mail.com'),
  ('3', 'name3', 'email3@mail.com');

  CREATE TABLE profile (
  `id` INTEGER,
  `users_id` INTEGER,
  `about` VARCHAR(355)
);

INSERT INTO profile
  (`id`, `users_id`, `about`)
VALUES
  ('1', '1', 'something about me'),
  ('2', '2', 'something about me'),
  ('3', '3', 'something about me');

Answer 1

该查询将为您提供所需的原始数据：

SELECT w1.users_id AS user1, w2.users_id AS user2, GROUP_CONCAT(w2.watched ORDER BY w2.watched) AS movies
FROM watched w1
JOIN watched w2 ON w2.watched = w1.watched AND w2.users_id != w1.users_id
GROUP BY user1, user2

输出（用于样本数据）：

user1   user2   movies
1       2       movie1,movie2
1       3       movie1
2       1       movie1,movie2
2       3       movie1
3       1       movie1
3       2       movie1

dbfiddle上的演示

您可以根据需要优化查询（例如，添加WHERE w1.users_id = 1以仅选择与用户1匹配的内容，或将JOIN WHERE w1.users_id = 1至users和profile表（每个用户ID一次）以获取用户名等）。

要获得您想要的输出，您实际上不需要更多的表，您可以使用类似以下的命令：

$id = $_SESSION['id'];
echo "my id: $id\n";
$movies = mysqli_query($connect, "SELECT w1.users_id AS user1, 
                                         w2.users_id AS user2, 
                                         COUNT(w2.watched) AS num_movies, 
                                         GROUP_CONCAT(w2.watched ORDER BY w2.watched) AS movies
                                  FROM watched w1
                                  JOIN watched w2 ON w2.watched = w1.watched AND w2.users_id != w1.users_id
                                  WHERE w1.users_id = $id
                                  GROUP BY user1, user2");
while ($row = $movies->fetch_assoc()) {
    echo "matched with id {$row['user2']} {$row['num_movies']} times on titles {$row['movies']}\n";
}