Online people counting business is the business logic that we must design when developing web. This article will give several design solutions to analyze the advantages and disadvantages of each solution:
Use Ordered set
This solution can store online users and user online time at the same time, and can perform a lot of aggregation calculations, but the memory consumed is also very considerable.
Use sets
This solution can store online users and can also perform certain aggregation calculations. Compared with ordered sets, it consumes less The memory is smaller, but as the number of users increases, the memory space consumed is also increasing
Use hyperloglog
This solution No matter how many online users are counted, the memory consumed is 12k, but only the statistical information of online users can be given, and the accurate list of online users cannot be obtained
Use bitmap
This solution is still relatively good. It records the status of online users while saving memory space as much as possible, and can perform certain aggregation operations.
Let’s use practical examples below. Note:
# 说 We first have a small number of users of 10 ~ 30W every day, and the user group of 100W will explain the following schemes
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First generate user online record data:
$start_time = mktime(0, 0, 0, 9, 5); //monday
for ($i=0; $i < 6; $i++) {
$day_start_time = $start_time + 86400 * $i; //every day begin time
$day_end_time = $day_start_time + 86400; //every day end time
$online_user_num = mt_rand(100000, 300000); //online user between 100000 and 300000
for ($j=1; $j < $online_user_num; $j++) {
$user_id = mt_rand(1, 1000000);
$redis->zadd('000|online_users_day_'.$i, mt_rand($day_start_time, $day_end_time), $user_id);
}
}//note: 统计每天的在线总人数
for ($i=0; $i < 6; $i++) {
print_r($redis->zsize('000|online_users_day_'.$i). "\n");
}
//note: 统计最近6天都在线的人数
var_dump($redis->zInter('000|online_users_day_both_6',
[
'000|online_users_day_0',
'000|online_users_day_1',
'000|online_users_day_2',
'000|online_users_day_3',
'000|online_users_day_4',
'000|online_users_day_5'
]
));
//note: 统计出近6天中共有多少上线
$redis->zunion('000|online_users_day_total_6', ['000|online_users_day_0', '000|online_users_day_1', '000|online_users_day_2', '000|online_users_day_3', '000|online_users_day_4', '000|online_users_day_5']);
//note: 统计某个时间段总共在线用户
print_r($redis->zcount('000|online_users_day_5', mktime(13, 0, 0, 9, 10), mktime(14, 0, 0, 9, 10)));
//note: 统计某个时间段在线用户名单
print_r($redis->zrangebyscore('000|online_users_day_5', mktime(13, 0, 0, 9, 10), mktime(14, 0, 0, 9, 10),
array('withscores' => TRUE)));PHP video tutorial]
Option 2: Use collections
or First, let’s record data online for users://note set 一般聚合
for ($i=0; $i < 6; $i++) {
$online_user_num = mt_rand(100000, 300000); //online user between 100000 and 300000
for ($j=1; $j < $online_user_num; $j++) {
$user_id = mt_rand(1, 1000000);
$redis->sadd('001|online_users_day_'.$i, $user_id);
}
}//note 判断某个用户是否在线
var_dump($redis->sIsMember('001|online_users_day_5', 100030));
//note 每天在线用户总量的统计
for ($i=0; $i < 6; $i++) {
print_r($redis->ssize('001|online_users_day_'.$i). "\n");
}
//note 对不同时间段的在线用户名单进行聚合
print_r($redis->sInterStore('001|online_users_day_both_4and5', '001|online_users_day_4', '001|online_users_day_5'). "\n");
//note 对指定的时间段的在线用户名单进行统计
print_r($redis->sUnionStore('001|online_users_day_total_4add5', '001|online_users_day_4', '001|online_users_day_5'). "\n");
//note 哪天上线哪天没上线
print_r($redis->sDiffStore('001|online_users_day_diff_4jian5', '001|online_users_day_4', '001|online_users_day_5'). "\n");Option 3: Use hyperloglgo
First come to the user to record data online:// note HyperLogLog 只需要知道在线总人数
for ($i=0; $i < 6; $i++) {
$online_user_num = mt_rand(100000, 300000); //online user between 100000 and 300000
var_dump($online_user_num);
for ($j=1; $j < $online_user_num; $j++) {
$user_id = mt_rand(1, 1000000);
$redis->pfadd('002|online_users_day_'.$i, [$user_id]);
}
}$count = 0;
for ($i=0; $i < 3; $i++) {
$count += $redis->pfcount('002|online_users_day_'.$i);
print_r($redis->pfcount('002|online_users_day_'.$i). "\n");
}
var_dump($count);
//note 3 days total online num
var_dump($redis->pfmerge('002|online_users_day_both_3', ['002|online_users_day_0', '002|online_users_day_1', '002|online_users_day_2']));
var_dump($redis->pfcount('002|online_users_day_both_3'));Option 4: Use bitmap
The author actually likes this solution. It consumes not much memory space but has a lot of statistical information. It is still old. Steps, first generate data://note bitmap 综合前面3个的优缺点
for ($i=0; $i < 6; $i++) {
$online_user_num = mt_rand(100000, 300000); //online user between 100000 and 300000
for ($j=1; $j < $online_user_num; $j++) {
$user_id = mt_rand(1, 1000000);
$redis->setbit('003|online_users_day_'.$i, $user_id, 1);
}
}//note userid today whether online var_dump($userid = mt_rand(1, 1000000)); var_dump($redis->getbit('003|online_users_day_5', $userid)); //note how many user is online var_dump($redis->bitcount('003|online_users_day_5')); //note 6 days both online var_dump($redis->bitop('AND', '003|online_users_day_both_6', '003|online_users_day_0', '003|online_users_day_1', '003|online_users_day_2', '003|online_users_day_3', '003|online_users_day_4', '003|online_users_day_5')); var_dump($redis->bitcount('003|online_users_day_both_6')); //note 6 days total online var_dump($redis->bitop('OR', '003|online_users_day_total_6', '003|online_users_day_0', '003|online_users_day_1', '003|online_users_day_2', '003|online_users_day_3', '003|online_users_day_4', '003|online_users_day_5')); var_dump($redis->bitcount('003|online_users_day_total_6')); //note 6 days only one online var_dump($redis->bitop('XOR', '003|online_users_day_only_one_6', '003|online_users_day_0', '003|online_users_day_1', '003|online_users_day_2', '003|online_users_day_3', '003|online_users_day_4', '003|online_users_day_5')); var_dump($redis->bitcount('003|online_users_day_only_one_6'));
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