[issue 259] uncover how count is executed in MySQL?

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List of articles

1. Preface

2. Build table

3. count How to execute ？

4. count(1),count(id),count( Non indexed column ),count( Secondary index column ) Analysis of

1. Preface

I believe before that , Many people just remember , I didn't understand , Only know count(*)、count(1) Including all lines , When it comes to statistical results , Column values of... Are not ignored NULL,count( Name ) Only the column name is counted , When it comes to statistical results , Column values of NULL The record of .

Let's analyze it in principle .

2. Build table

Same as before , Use the same table , There are nearly 10W Data

CREATE TABLE demo_info(
    id INT NOT NULL auto_increment,
    key1 VARCHAR(100),
    key2 INT,
    key3 VARCHAR(100),
    key_part1 VARCHAR(100),
    key_part2 VARCHAR(100),
    key_part3 VARCHAR(100),
    common_field VARCHAR(100),
    PRIMARY KEY (id),
    KEY idx_key1 (key1),
    UNIQUE KEY uk_key2 (key2),
    KEY  idx_key3 (key3),
    KEY idx_key_part(key_part1, key_part2, key_part3)
)ENGINE = INNODB CHARSET=utf8mb4;

3. count How to execute ？

We often see such examples ：

When you need to count how much data there is in the table , You will often use the following statements

SELECT COUNT(*) FROM demo_info;

Because the records in clustered index and nonclustered index correspond one-to-one , Instead of columns contained in clustered index records （ Index columns + Primary key id） Is less than the clustered index （ All columns ） Records of the , Therefore, the same number of non clustered index records occupy less storage space than clustered index records .

If we use a nonclustered index to execute the above query , That is, count the non clustered indexes uk_key2 How many records are there in total , It is much less than directly counting the number of records in the clustered index I/O cost . So the optimizer will decide to use nonclustered indexes uk_key2 Execute the above query .

Be careful ： It has been verified here ,uk_key2 Lower cost than other indexes .

Analyze the execution plan

When executing the above query ,server The layer will maintain a name count The variable of , then ：

• server Layer to layer InnoDB First record .

• InnoDB find uk_key2 The first secondary index record , And return it to server layer （ Be careful ： Because only the number of records is counted at this time , So you don't need to go back to the table ）.

• because count The argument to the function is *,MySQL Will * As a constant 0 Handle . because 0 Not at all NULL,server Stratification count Variable plus 1.

• server Layer to layer InnoDB To the next record .

• InnoDB Recorded by secondary index next_record Property to find the next secondary index record , And return it to server layer .

• server Continue to give count Variable plus 1.

• Repeat the process , until InnoDB towards server The layer returns a message that has no record to check .

• server The layer will eventually count The value of the variable is sent to the client .

4. count(1),count(id),count( Non indexed column ),count( Secondary index column ) Analysis of

Let's see count(1)

SELECT COUNT(1) FROM demo_info;

Carry out the plan and count(*) equally

about count(*)、count(1) Or whatever count( constant ) Come on , It doesn't really matter which index record to read , because server The layer only cares about whether the storage engine reads the record , There is no need to extract the specified field from the record to judge whether it is NULL. So the optimizer will use the index that occupies the least storage space to execute the query .

Look again. count(id)：

explain SELECT COUNT(id) FROM demo_info;

about count(id) Come on , because id It's the primary key , Whether it's clustered index records , Or will any secondary index record contain a primary key field , Therefore, in fact, reading records in any index can get id Field , At this time, the optimizer will also select the index that occupies the least storage space to execute the query .

Look again. count( Non indexed column )

explain select count(common_field) from demo_info

about count( Non indexed column ) Come on , The optimizer selects a full table scan , It indicates that only leaf nodes of the clustered index can be scanned sequentially .

Please make sure you understand the full table scan , It scans all leaf nodes of the clustered index in sequence and determines . in addition , More about mysql Interview questions , official account Java selected , reply java interview , Get the latest interview questions , Support online question brushing anytime, anywhere .

For other secondary index columns ,count( Secondary index column ), The optimizer can only select the index containing the column we specify to execute the query , Only non clustered indexes can be specified B+ Tree scan , The cost of index scanning that may lead to the optimizer's selection is not minimal .

in summary

about count(*)、count( constant )、count( Primary key ) Formal count In terms of functions , The optimizer can select the index with the lowest scanning cost to execute the query , So as to improve efficiency , Their execution process is the same , Just judging whether the expression is NULL Choose different ways to judge , The judgment is NULL The cost of the process is negligible , So we can say count(*)、count( constant )、count( Primary key ) The cost is the same .

And for count( Non indexed column ) Come on , The optimizer selects a full table scan , It indicates that only leaf nodes of the clustered index can be scanned sequentially .

count( Secondary index column ) You can only select the index containing the column we specify to execute the query , The cost of index execution that may cause the optimizer to select is not minimal .

In fact, these differences are due to the fact that non clustered index records occupy less storage space than clustered index records , Less and more I/O cost , So the optimizer has the choice of different indexes , That's it .

Copyright notice ： This paper is about CSDN Blogger 「 Brick industry __」 The original article of , follow CC 4.0 BY-SA Copyright agreement , For reprint, please attach the original source link and this statement .

https://liuchenyang0515.blog.csdn.net/article/details/120851980

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