Question

我有一个类似这样的数据库：

user_id    photo_id
1          1
1          2
1          3
1          4
2          5
2          6

我想从中获取最受欢迎用户的列表。像这样：

Popular Users: 1 (4) & 2 (2)

我将如何在PHP中使用mysql进行此操作？

谢谢，库尔顿

PS：我对mysql命令了解很多，所以你不必愚弄它。谢谢！

Answer 1

基本查询将是：

select user_id, count(user_id) as cnt
from yourtable
group by user_id
order by cnt desc

显示结果，如：

$results = array()
while($row = mysql_fetch_assoc($query_result)) {
   $results[] = "{$row['user_id']} ({$row['cnt']})"
   // repeat for however many results you want
}
echo "Popular user: ", implode(" &amp; ", $results);

Answer 2

select user_id, count(user_id) as count from table order by count desc group by user_id

无论如何......

Answer 3

这可以仅使用SQL命令来完成。这就是我要做的事情：

SELECT user_id, count(user_id) uid_count
FROM <<table>>
GROUP BY user_id
ORDER BY uid_count DESC
LIMIT 5;

GROUP BY收集所有具有相同user_id的行，ORDER BY ... DESC按降序对结果进行排序，因此第一行代表最受欢迎的用户。 LIMIT为您提供前5名结果。

Answer 4

数据库查询看起来像这样：

select user_id, count(photo_id) as c 
from table group by user_id 
order by c desc limit 5;

在PHP中，它看起来像这样：

$sql = 'select user_id, count(photo_id) as c from table group by user_id order by c desc limit 5';
$result = mysql_query($sql, $link);

if (!$result) {
    echo "DB Error, could not query the database\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}

while ($row = mysql_fetch_assoc($result)) {
    echo $row['user_id'];
    echo $row['c'];
}

Answer 5

SELECT COUNT(a.category_id) as cnt,b.category,b.image FROM bookings as a 
INNER JOIN service_category as b ON a.category_id=b.category_id 
GROUP BY a.category_id ORDER BY cnt DESC LIMIT 6

获取mysql数据库中前5名最受欢迎的项目？

5 个答案: