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Use Redis bitmaps to count active users

齐天大圣
齐天大圣Original
2020-05-20 07:19:212230browse

First, let’s look at a scenario: a website needs to count users who have logged in continuously within a week, and users who have logged in within a month.

If it is implemented using a traditional database such as Mysql, it will be difficult to achieve. But if you use Redis to do it, it is very simple. Redis's collection type and Bitmap type can be easily achieved. Today, we will mainly talk about how to use Bitmaps to implement the function of counting active users.

Bitmaps

In a computer system, the smallest unit of information is a byte. 1 byte is equal to 8 bits. Each bit It can only be 0 or 1 (the computer only knows these two numbers). Bitmaps allow direct manipulation of bits.

Bigmaps can be regarded as an array. Each bit in the array can only be 0 or 1. The subscript of the array is regarded as an offset here.

Let’s introduce a few commands related to Bitmaps:

setbit

##setbit key offset value: Set the value for the corresponding bit

For example, if users 3, 8, 23, and 32 visited the website today, then

setbit user:view:2020-5-17 3 1
setbit user:view:2020-5-17 8 1
setbit user:view:2020-5-17 23 1
setbit user:view:2020-5-17 32 1

Development tips : Many application IDs do not start from 1, but many start from a specified number, such as 1001, 10001. For these, we can subtract the initial value when setting to prevent wasting space

getbit

##getbit key offset gets the value of the specified bit

If I want to know whether user No. 8 and user No. 45 have logged in today, then

127.0.0.1:6379> getbit user:view:2020-5-17 8
(integer) 1
127.0.0.1:6379> getbit user:view:2020-5-17 45
(integer) 0

I can see that user No. 8 has logged in today, But user No. 45 has not logged in today.

bitcount

bitcount key [start] [end] Get the number in the specified range of 1

I want to know how many users have logged in today, then

127.0.0.1:6379> bitcount user:view:2020-5-17
(integer) 4

Operation between Bitmaps

bitop op destkey key [key ...]

The bitop command can perform intersection (and), union (or), non (not), exclusive or (xor) on multiple bitmaps, and the operation results Stored in destkey.

If you want to know the number of users who logged in for three consecutive days, that is, the number of users who logged in on May 17th, 18th, and 19th.

The login situation in the past three days is as follows:

    On May 17, 3, 8, 23, and 32 users logged in
  • Users No. 3, 23, 43, and 54 logged in on May 18th
  • Users No.3, 5, 23, 32, 56, and 78 logged in on May 19th
  • 127.0.0.1:6379> bitop and three:and user:view:2020-5-17 user:view:2020-5-18 user:view:2020-5-19
    127.0.0.1:6379> bitcount three:and
    (integer) 2
  • If you want to know how many users have logged in in the past three days.
127.0.0.1:6379> bitop or three:or user:view:2020-5-17 user:view:2020-5-18 user:view:2020-5-19
(integer) 10
127.0.0.1:6379> bitcount three:or
(integer) 9

As you can see, a total of 9 users have logged in in the past three days.

Practical combatAfter finishing the above knowledge, we can complete the desired requirement: we need to count the number of consecutive logins within a week users, and users who have logged in within a month.

First simulate the user login situation within 30 days. The pseudo code is as follows:

for ($i = 0; $i < 20000; $i++) {
    $userId = mt_rand(1, 10000);
    $date   = time() - 86400 * mt_rand(0, 30);
    $key   = &#39;userlogin_&#39;.date(&#39;Ymd&#39;, $date);
 
    $redis->setBit($key, $userId, 1);
}

Get the users who have logged in within a week. Of course, we will not fetch them all at once, but we want to fetch them all at once like paging. A certain number, the pseudo code is as follows:

for ($i = 1; $i <= 7; $i ++) {
    $key = "userlogin_".date(&#39;Ymd&#39;, time() - (86400*$i));
 
    if ($i == 1) {
        $redis->bitOp(&#39;and&#39;, &#39;week_logined&#39;, $key);
    } else {
        $redis->bitOp(&#39;and&#39;, &#39;week_logined&#39;, &#39;week_logined&#39;, $key);
    }
}
 
// 获取前50个用户
$userIds = [];
for ($i=1; $i<=10000; $i++) {
    $ret = $redis->getBit(&#39;week_logined&#39;, $i);
    $ret && $userIds[] = $i;
 
    if (count($userIds) >=50) break;
}

There is a point to note, and it is also an error-prone point. During the bitop, the first time, because week_logined does not exist yet, so the op is performed. There is only one key. When starting from the second time, there are 2 keys for op.

To obtain the users who logged in within a month, the idea is basically the same as above, except that the and is changed to or

for ($i = 1; $i <= 3; $i ++) {
    $key = "userlogin_".date(&#39;Ymd&#39;, time() - (86400*$i));
    $redis->bitOp(&#39;or&#39;, &#39;month_loginOnce&#39;, &#39;month_loginOnce&#39;, $key);
}
 
// 获取一个月内登陆过的用户
$userIds = [];
for ($i=1; $i<=10000; $i++) {
    $ret = $redis->getBit(&#39;month_loginOnce&#39;, $i);
    $ret && $userIds[] = $i;
}

As you can see, there are still some differences between doing or and and. of. or, there is no need to judge the first time. The reason is for everyone to understand.

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