如何防止PHP中另一個foreach循環內的foreach循環重復？
[英]How to prevent duplicates on a foreach loop inside another foreach loop in PHP?
問題描述
我有一個音樂網站，每次用戶單擊播放時，我的PHP函數就會插入表格views
我試圖對MySQL數據庫進行重新排序以優化其性能，因此在此示例中，我嘗試按時間對每個track的視圖總數進行計數。
僅選擇表格views軌道：
$query = "SELECT `track` FROM `views` GROUP BY `track`";
$query的結果：
|track|
|-----|
|140 |
|125 |
|33 |
|... |
計算每個結果作為播放的行數，並按日期分組：
$querysel = $this->db->query(sprintf("SELECT COUNT( * ) AS plays , `time`, `track` FROM `views` WHERE `track` = '%s' GROUP BY DATE( `time` )", $tid));
$querysel的結果
|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 |
最后在新表中插入結果：
$this->db->query(sprintf("INSERT INTO views_counts (tid,plays,time) VALUES ('%s','%s','%s')", $newtid, $plays, $time));
這是我的完整功能：
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));
}
}
}
在我的函數中，我使用GROUP BY DATE( time ) ，那么為什么最終查詢同時插入到新表中views_counts重復呢？
這是我的輸出 ：
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
如您所見，我同時具有多個相同的結果。
這是我所期望的：
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
UPDATE
這就是我所說的“一次”功能：
<?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);
?>
函數countViews()在classes.php 。
我稱此PHP僅用於通過Web訪問請求的頁面。
4 個解決方案
解決方案1
1 已采納 2017-02-02 18:46:36
我建議僅使用一個查詢來解決您的問題。
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']));
}
}
解決方案2
0 2017-02-02 01:28:06
嘗試這個：
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));
}
}
解決方案3
0 2017-02-03 01:47:37
真的不能在單個查詢中完成嗎？
INSERT INTO views_counts (tid, time, plays)
SELECT track, DATE(`time`), COUNT(*)
FROM `views`
GROUP BY track, DATE(`time`)
注意， INSERT的行來自SELECT 。
我隨意使用DATE(time) ，否則，選擇哪個 time是任意的。
解決方案4
-1 2017-02-02 01:28:22
好的，因此只有一個SELECT查詢就足夠了，如下所示：
$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));
}
理想情況下，這應該可以解決問題。 使用手動創建的示例數據的簡單查詢演示：
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`);
使用此查詢產生的輸出，這將是在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
希望有幫助...
如何在 foreach 循環中獲取特定值並在 PHP（codeigniter）中的另一個 foreach 循環中匹配它
[英]How to get a specific value inside a foreach loop and match it in another foreach loop in PHP (codeigniter)
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