我有一个音乐网站，每次用户点击播放时，我的PHP函数会插入表views
我正在尝试重新排序我的MySQL数据库以优化其性能，因此我尝试在此示例中计算按时间分组的每个track的总视图数。
仅选择表格views中的曲目：
$query = "SELECT `track` FROM `views` GROUP BY `track`";
$query的结果：
|track|
|-----|
|140 |
|125 |
|33 |
|... |
将每个结果的行数计为播放并按日期分组：
$querysel = $this->db->query(sprintf("SELECT COUNT( * ) AS plays , `time`, `track` FROM `views` WHERE `track` = '%s' GROUP BY DATE( `time` )", $tid));
$querysel
|plays |time |track |
|------|--------------------|------|
|82 |2016-12-26 18:20:16 |140 |
|1 |2017-01-10 15:52:55 |140 |
|2 |2017-01-26 13:17:25 |140 |
最终在新表中插入结果：
$this->db->query(sprintf("INSERT INTO views_counts (tid,plays,time) VALUES ('%s','%s','%s')", $newtid, $plays, $time));
这是我的完整功能：
function countViews() {
$query = "SELECT `track` FROM `views` GROUP BY `track`";
$result = $this->db->query($query);
while($row = $result->fetch_assoc()) {
$rows[] = $row;
}
foreach($rows as $row) {
$tid = $row['track'];
$querysel = $this->db->query(sprintf("SELECT COUNT( * ) AS plays , `time`, `track` FROM `views` WHERE `track` = '%s' GROUP BY DATE( `time` )", $tid));
while($rownew = $querysel->fetch_assoc()) {
$rowsnew[] = $rownew;
}
foreach($rowsnew as $rownew) {
$newtid = $rownew['track'];
$plays = $rownew['plays'];
$time = $rownew['time'];
$this->db->query(sprintf("INSERT INTO views_counts (tid,plays,time) VALUES ('%s','%s','%s')", $newtid, $plays, $time));
}
}
}
在我的函数中，我使用GROUP BY DATE(时间)，那么为什么最后一个查询插入到新表views_counts中的同时重复？
这是我的输出：
id | tid | plays | time
----|-----|-------|--------------
1 | 1 | 2 | 2017-01-26 12:43:16
2 | 1 | 1 | 2017-01-27 12:45:24
3 | 1 | 2 | 2017-01-26 12:43:16
4 | 1 | 1 | 2017-01-27 12:45:24
5 | 3 | 30 | 2016-12-26 18:20:16
6 | 1 | 2 | 2017-01-26 12:43:16
7 | 1 | 1 | 2017-01-27 12:45:24
8 | 1 | 2 | 2017-01-26 12:43:16
9 | 1 | 1 | 2017-01-27 12:45:24
10 | 3 | 30 | 2016-12-26 18:20:16
11 | 1 | 2 | 2017-01-26 12:43:16
12 | 1 | 1 | 2017-01-27 12:45:24
13 | 1 | 2 | 2017-01-26 12:43:16
14 | 1 | 1 | 2017-01-27 12:45:24
15 | 3 | 30 | 2016-12-26 18:20:16
如你所见，我同时有多个相同的结果。
这是我的预期：
id | tid | plays | time
----|-----|-------|--------------
1 | 1 | 2 | 2017-01-26 12:43:16
2 | 1 | 1 | 2017-01-27 12:45:24
3 | 3 | 30 | 2016-12-26 18:20:16
更新
这就是我称之为“一次性”功能的方式：
<?php
include("/var/www/html/includes/config.php");
include("/var/www/html/includes/classes.php");
session_start();
$db = new mysqli($CONF['host'], $CONF['user'], $CONF['pass'], $CONF['name']);
if ($db->connect_errno) {
echo "Failed to connect to MySQL: (" . $db->connect_errno . ") " . $db->connect_error;
}
$db->set_charset("utf8");
$feed = new feed();
$feed->db = $db;
$result = $feed->countViews();
mysqli_close($db);
?>
函数countViews()位于classes.php内。
我称这个PHP只是通过网络访问所请求的页面。
答案 0 :(得分：1)
我建议只使用一个查询解决您的问题。
function countViews() {
$query = $this->db->query("SELECT COUNT( * ) AS plays , `time`, `track` FROM `views` GROUP BY `track`,DATE( `time` );");
while($row = $query->fetch_assoc()) {
$this->db->query(sprintf("INSERT INTO views_counts (tid,plays,time) VALUES ('%s','%s','%s')", $row['track'], $row['plays'], $row['time']));
}
}
答案 1 :(得分：0)
试试这个：
function countViews() {
$query = "SELECT `track` FROM `views` GROUP BY `track`";
$result = $this->db->query($query);
while($row = $result->fetch_assoc()) {
$rows[] = $row;
}
$rowsnew = array();
foreach($rows as $row) {
$tid = $row['track'];
$querysel = $this->db->query(sprintf("SELECT COUNT( * ) AS plays , `time`, `track` FROM `views` WHERE `track` = '%s' GROUP BY DATE( `time` )", $tid));
while($rownew = $querysel->fetch_assoc()) {
$rowsnew[] = $rownew;
}
}
foreach($rowsnew as $rownew1) {
$newtid = $rownew1['track'];
$plays = $rownew1['plays'];
$time = $rownew1['time'];
$this->db->query(sprintf("INSERT INTO views_counts (tid,plays,time) VALUES ('%s','%s','%s')", $newtid, $plays, $time));
}
}
答案 2 :(得分：0)
在单个查询中真的可以完成吗？
INSERT INTO views_counts (tid, time, plays)
SELECT track, DATE(`time`), COUNT(*)
FROM `views`
GROUP BY track, DATE(`time`)
请注意，INSERT的行来自SELECT。
我冒昧地使用DATE(time)，否则，选择 time的是任意的。
答案 3 :(得分：-1)
好的，所以只有一个SELECT查询就足够了，具体如下：
$query = "SELECT `track`, `time`, COUNT(*) AS `plays` FROM `views`
GROUP BY `track`, DATE(`time`)
ORDER BY `track`, DATE(`time`);"
$result = $this->db->query($query);
while($row = $result->fetch_assoc()) {
$newtid = $row['track'];
$plays = $row['plays'];
$time = $row['time'];
$this->db->query(sprintf("INSERT INTO views_counts (tid,plays,time)
VALUES ('%s','%s','%s')",
$newtid, $plays, $time));
}
这应该是理想的做法。使用手动创建的样本数据进行简单的查询演示：
create table views (track int, `time` timestamp);
insert into views VALUES(1, '2017-01-26 12:43:16');
insert into views VALUES(1, '2017-01-27 12:45:24');
insert into views VALUES(1, '2017-01-27 12:50:30');
insert into views VALUES(2, '2017-01-25 01:00:00');
insert into views VALUES(2, '2017-01-25 02:00:00');
insert into views VALUES(2, '2017-01-26 11:30:00');
insert into views VALUES(2, '2017-01-26 11:45:00');
insert into views VALUES(2, '2017-01-26 13:45:00');
insert into views VALUES(2, '2017-01-26 15:45:00');
insert into views VALUES(2, '2017-01-27 08:00:00');
insert into views VALUES(3, '2017-01-27 09:00:00');
SELECT `track`, `time`, COUNT(*) AS plays FROM views
GROUP BY `track`, DATE(`time`) ORDER BY `track`, DATE(`time`);
使用此查询生成的输出，这是预期落入views_counts：
track | time | plays
----------------------------------------------
1 | 2017-01-26 12:43:16 | 1
1 | 2017-01-27 12:45:24 | 2
2 | 2017-01-25 01:00:00 | 2
2 | 2017-01-26 11:30:00 | 4
2 | 2017-01-27 08:00:00 | 1
3 | 2017-01-27 09:00:00 | 1
希望有帮助...