Question

我想显示来自三个单独表格的数据，包括图像/符号上的喜欢计数。目前，结果数组只打印一个结果。当我删除计数时，它会打印我想要的所有5个。

$query = "
    SELECT 
        image_url, image_path, date, location, fname, lname, rating,
        COUNT(DISTINCT likes.sign_id) as 'count'

    FROM 
        signs 
        INNER JOIN users ON signs.user_id=users.user_id
        INNER JOIN likes ON (signs.sign_id=likes.sign_id) 
        WHERE location='Oslo' LIMIT 5";

$result = mysqli_query ( $db_server, $query ) or die ( "Could not find uploads" . mysqli_error ( $db_server ) );
while ( $row = mysqli_fetch_assoc ( $result ) ) {
                            print_r($row);}

我尝试了不同的方法，但问题似乎只与COUNT有关，它确实计算了正确数量的喜欢。因此我不知道为什么它不会输出所有结果。

Answer 1

您缺少group by条款：

    SELECT 
        image_url, image_path, date, location, fname, lname, rating,
        COUNT(DISTINCT likes.sign_id) as 'count'

    FROM 
        signs 
        INNER JOIN users ON signs.user_id=users.user_id
        INNER JOIN likes ON (signs.sign_id=likes.sign_id) 
        WHERE location='Oslo'
        GROUP BY image_url, image_path, date, location, fname, lname, rating
        LIMIT 5;

仅当所有这些列都相同时才会起作用，否则，您需要指定所需的列。

如果您的意思是您希望每行旁边都有一个列，那么您可以使用这样的相关查询：

    SELECT 
        image_url, image_path, date, location, fname, lname, rating,
        (SELECT COUNT(DISTINCT l.sign_id) FROM likes l
         WHERE signs.sign_id=l.sign_id) as `count`
    FROM 
        signs 
        INNER JOIN users ON signs.user_id=users.user_id
        INNER JOIN likes ON (signs.sign_id=likes.sign_id) 
        WHERE location='Oslo'
        LIMIT 5;

JOIN和COUNT只输出一个结果

1 个答案: