鉴于代码:
$counters = $db->query("SELECT COUNT(film_id) FROM review WHERE film_id = $film_id ");
我需要将对象$ counter转换为整数。仅供参考:查询返回单个int,即4(取决于film_ID)
答案 0 :(得分:0)
为存储COUNT结果的列指定别名,然后像使用任何其他SELECT查询一样使用它。
$counters = $db->query("SELECT COUNT(film_id) AS counter FROM review WHERE film_id = $film_id "); $counter = $counters->fetch_assoc(); echo $counter['counter'];
答案 1 :(得分:0)
$ counter不是您假设的结果字符串,而是资源对象。
我的建议是更改代码如下:
$res = $db->query("SELECT COUNT(film_id) as counter FROM review WHERE film_id = $film_id ");
$row = $res->fetch_assoc();
$counter = (int) $row["counter"];