php - 只返回一条记录

Question

所以我正在尝试检索喜欢的列表。这是我的代码

//Echo total likes of #68
$stmt1 = $con->prepare("SELECT COUNT(*) FROM likes WHERE liked_post_id = :id;");
$stmt1->bindValue(':id', $id, PDO::PARAM_STR);
$stmt1->execute();
$count = $stmt1->rowCount();
echo $count;

我假设这会返回所有喜欢的内容。但我只得到 1 返回。这是结构

+-----------------+------------------+------+-----+---------+----------------+
| Field           | Type             | Null | Key | Default | Extra          |
+-----------------+------------------+------+-----+---------+----------------+
| id              | int(11) unsigned | NO   | PRI | NULL    | auto_increment |
| liked_post_id   | varchar(255)     | YES  |     | NULL    |                |
| liked_post_user | varchar(255)     | YES  |     | NULL    |                |
| liked_post_ip   | varchar(255)     | YES  |     | NULL    |                |
+-----------------+------------------+------+-----+---------+----------------+

快速查询

+----+---------------+-----------------+---------------+
| id | liked_post_id | liked_post_user | liked_post_ip |
+----+---------------+-----------------+---------------+
| 77 | 68            | alex            | 127.0.0.1     |
| 78 | 68            | andrew          | 127.0.0.1     |
+----+---------------+-----------------+---------------+

因此，我希望得到 2 而不是 1，因为有两个用户喜欢它。有任何想法吗？我也没有收到任何错误。

Answer 1

这样做:

$stmt1 = $con->prepare("SELECT COUNT(*) AS count FROM likes WHERE liked_post_id = :id;");
$stmt1->bindValue(':id', $id, PDO::PARAM_STR);
$stmt1->execute();
$row = $stmt1->fetch(PDO::FETCH_ASSOC);
echo $row['count'];

查询返回 1 行，该行的内容是表中匹配行的计数。您需要获取该行以读取该值。