这显示最近10个新结果,一切正常
$query = "SELECT s.name AS song,
SUM(v.hit_shit) AS hit,
COUNT(v.hit_shit) - SUM(v.hit_shit) AS shit,
MAX(date_voted) AS last_vote
FROM songs AS s
INNER JOIN votes AS v ON v.song_id = s.id
WHERE v.date_voted >= CURDATE()
GROUP BY s.id
ORDER BY last_vote DESC
LIMIT 10";
$song = mysql_query($query);
但这显示全部,但我只需要5 :(
$query = "SELECT s.name AS song,
SUM(v.hit_shit) AS hit,
COUNT(v.hit_shit) - SUM(v.hit_shit) AS shit
FROM songs AS s
INNER JOIN votes AS v ON v.song_id = s.id
WHERE v.date_voted >= CURDATE()
AND v.date_voted < CURDATE() + INTERVAL 1 DAY
GROUP BY s.id
ORDER BY hit DESC, shit ASC, song ASC";
$song = mysql_query($query);
P.S。抱歉4我的坏人
答案 0 :(得分:1)
在查询结尾添加限制5。
$query = "SELECT s.name AS song,
SUM(v.hit_shit) AS hit,
COUNT(v.hit_shit) - SUM(v.hit_shit) AS shit,
MAX(date_voted) AS last_vote
FROM songs AS s
INNER JOIN votes AS v ON v.song_id = s.id
WHERE v.date_voted >= CURDATE()
GROUP BY s.id
ORDER BY last_vote DESC
LIMIT 5";
$song = mysql_query($query);
答案 1 :(得分:0)
您应该在查询结尾添加参数“LIMIT”。
$query = "SELECT s.name AS song,
SUM(v.hit_shit) AS hit,
COUNT(v.hit_shit) - SUM(v.hit_shit) AS shit
FROM songs AS s
INNER JOIN votes AS v ON v.song_id = s.id
WHERE v.date_voted >= CURDATE()
AND v.date_voted < CURDATE() + INTERVAL 1 DAY
GROUP BY s.id
ORDER BY hit DESC, shit ASC, song ASC
LIMIT 5";
$song = mysql_query($query);
编辑:修改代码格式。