Mysql commonly used SQL statement collection

mysql commonly used sql statement collection

basics

// query time, friendly prompt
$sql = "select date_format(create_time, '%Y-%m-%d') as day from table_name";
//int timestamp type
$sql = " select from_unixtime(create_time, '%Y-%m-%d') as day from table_name";
//One sql returns multiple totals
$sql = "select count(*) all, " ;
$sql .= " count( case when status = 1 then status end) status_1_num, ";
$sql .= " count(case when status = 2 then status end) status_2_num ";
$sql .= " from table_name";
//Update Join / Delete Join
$sql = "update table_name_1 ";
$sql .= " inner join table_name_2 on table_name_1.id = table_name_2.uid ";
$sql .= " inner join table_name_3 on table_name_3.id = table_name_1.tid ";
$sql .= " set *** = *** ";
$sql .= " where *** ";

//delete join same as above.
//The statement to replace the content of a field
$sql = "update table_name set content = REPLACE(content, 'aaa', 'bbb') ";
$sql .= " where (content like '%aaa%')";
/ /Get the data that a field in the table contains a string
$sql = "SELECT * FROM `table name` WHERE LOCATE('keyword', field name) ";
//Get the first 4 digits in the field
$sql = "SELECT SUBSTRING(field name,1,4) FROM table name";
//Find redundant duplicate records in the table
//Single field
$sql = "select * from table name where field name in ";
$sql .= "(select field name from table name group by field name having count(field name) > 1 )";


$sql .= "(select field 1, field 2 from table name group by field 1, field 2 having count(*) > 1 )";
//delete redundant duplicate records in the table (with the smallest id)
//single field
$sql = "delete from table name where field name in ";
$sql .= "(select field name from table name group by field name having count(field name) > 1) ";
$sql .= "and primary key ID not in ";
$sql .= "(select min(primary key ID) from table name group by field name having count(field name)>1) ";
//Multiple fields
$sql = "delete from table name alias where (alias .field1, alias.field2) in ";
$sql .= "(select field1,field2 from table name group by field1,field2 having count(*) > 1) ";
$sql .= "and Primary key ID not in ";
$sql .= "(select min(primary key ID) from table name group by field 1, field 2 having count(*)>1) ";
Business articles

Continuous range problem

//Create a test table
CREATE TABLE `test_number` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `number` int(11) unsigned NOT NULL DEFAULT '0' COMMENT 'number',
  PRIMARY KEY (`id`)
) ENGINE= innodb DEFAULT CHARSET=utf8
//Create test data
insert into test_number values(1,1);
insert into test_number values(2,2);
insert into test_number values(3,3);
insert into test_number values(4,5);
insert into test_number values(5,7) ;
insert into test_number values(6,8);
insert into test_number values(7,10);
insert into test_number values(8,11);
Experiment goal: find the continuous range of numbers.

Based on the data above, the range should be obtained.

1-3
5-5
7-8
10-11
//Execute Sql
select min(number) start_range,max(number) end_range
from
(
    select number,rn,number-rn diff from
    (
        select number,@number:=@number+1 rn from test_number,(select @number:=0) as number
    ) b
) c group by diff;
continuous range of numbers

Sign question

// Create a reference table (needed for simulation data)
CREATE TABLE `test_nums` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  PRIMARY KEY (`id` )
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='Reference table';
//Simulate data, insert 1-10000 continuous data.
//Create test table
CREATE TABLE `test_sign_history` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT ,
  `uid` int(11) unsigned NOT NULL DEFAULT '0' COMMENT 'User ID',
  `create_time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP COMMENT 'Check-in time',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT ='Sign-in history table';
//Create test data
insert into test_sign_history(uid,create_time)
select ceil(rand()*10000),str_to_date('2016-12-11','%Y-%m-%d' )+interval ceil(rand()*10000) minute
from test_nums where id<31;
//Statistics for hourly user check-in every day
select
    h,
    sum(case when create_time='2016-12-11' then c else 0 end ) 11Sign,
    sum(case when create_time='2016-12-12' then c else 0 end) 12Sign,
    sum(case when create_time='2016-12-13' then c else 0 end) 13Sign,
    sum(case when create_time='2016-12-14' then c else 0 end) 14Sign,
    sum(case when create_time='2016-12-15' then c else 0 end) 15Sign,
    sum(case when create_time='2016-12-16' then c else 0 end) 16Sign,
    sum(case when create_time='2016-12-17' then c else 0 end) 17Sign
from
(
    select
        date_format(create_time,'%Y-%m-%d') create_time,
        hour(create_time) h,
        count(*) c
    from test_sign_history
    group by
        date_format(create_time,'%Y-%m-%d'),
        hour(create_time)
) a
group by h with rollup; each day//Count the hourly user check-in situation of each day (when there is no data for a certain hour, display 0)
Count the hourly user check-in situation of




select
    h ,
    sum(case when create_time='2016-12-11' then c else 0 end) 11Sign,
    sum(case when create_time='2016-12-12' then c else 0 end) 12Sign,
    sum(case when create_time='2016-12-13' then c else 0 end) 13Sign,
    sum(case when create_time='2016-12-14' then c else 0 end) 14Sign,
    sum(case when create_time='2016-12-15' then c else 0 end) 15Sign,
    sum(case when create_time='2016-12-16' then c else 0 end) 16Sign,
    sum(case when create_time='2016-12-17' then c else 0 end) 17Sign
from
(
    select b.h h,c.create_time,c.c from
     (
        select id-1 h from test_nums where id<=24
     ) b
     left join
     (
        select
         date_format(create_time,'%Y-%m-%d') create_time,
         hour(create_time) h,
         count(*) c
        from test_sign_history
        group by
         date_format(create_time,'%Y-%m-%d'),
         hour(create_time)
      ) c on (bh=ch)
) a
group by h with rollup;
Count the daily user check-in situation every hour (when there is no data for a certain hour, display 0)



// Count the daily user check-in data and every day of incremental data
select
        type,
        sum(case when create_time='2016-12-11' then c else 0 end) 11Sign,
        sum(case when create_time='2016-12-12' then c else 0 end) 12Sign,
        sum(case when create_time='2016-12-13' then c else 0 end) 13Sign,
        sum(case when create_time='2016-12-14' then c else 0 end) 14Sign,
        sum(case when create_time='2016-12-15' then c else 0 end) 15Sign,
        sum(case when create_time='2016-12-16' then c else 0 end) 16Sign,
        sum(case when create_time='2016-12-17' then c else 0 end) 17Sign
from
(
        select b.create_time,ifnull(b.c-c.c,0) c,'Increment' type from
        (
            select
             date_format(create_time,'%Y-%m-%d') create_time,
             count(*) c
            from test_sign_history
            group by
             date_format(create_time,'%Y-%m-%d')
        ) b
        left join
        (
            select
             date_format(create_time,'%Y-%m-%d') create_time,
             count(*) c
            from test_sign_history
            group by
             date_format(create_time,'%Y-%m-%d')
        ) c on(b.create_time=c.create_time+ interval 1 day)
    union all
        select
         date_format(create_time,'%Y-%m-%d') create_time,
         count(*) c,
         'Current'
        from test_sign_history
        group by
         date_format(create_time,'%Y-%m-%d')
) a
group by type
order by case when type='Current' then 1 else 0 end desc;
count daily user check-in data and daily incremental data



//simulate different users check-in for different days
insert into test_sign_history(uid, create_time)
select uid,create_time + interval ceil(rand()*10) day from test_sign_history,test_nums
where test_nums.id <10 order by rand() limit 150;
//Count the number of users with the same sign-in days
select
    sum(case when day =1 then cn else 0 end) 1Day,
    sum(case when day=2 then cn else 0 end) 2Day,
    sum(case when day=3 then cn else 0 end) 3D ay,
    sum(case when day=4 then cn else 0 end) 4Day,
    sum(case when day=5 then cn else 0 end) 5Day,
    sum(case when day=6 then cn else 0 end) 6Day,
    sum(case when day=7 then cn else 0 end) 7Day,
    sum(case when day=8 then cn else 0 end) 8Day,
    sum(case when day =9 then cn else 0 end) 9Day,
    sum(case when day=10 then cn else 0 end) 10Day
from
(
    select c day,count(*) cn
    from
    (
        select uid,count(*) c from test_sign_history group by uid
    ) a
    group by c
) b;


Count the number of users with the same number of check-in days



//Count the continuous check-in time of each user
select * from (
    select d.*,
    @ggid := @cggid,
    @cggid := d.uid,
    if (@ggid = @cggid, @grank := @grank + 1, @grank := 1) grank
    from
    (
        select uid,min(c.create_time) begin_date ,max(c.create_time) end_date,count(*) count from
        (
            select
            b.*,
            @gid := @cgid,
            @cgid := b.uid,
            if(@gid = @cgid, @rank := @rank + 1, @rank := 1) rank,
            b.diff-@rank flag from (
                select
                distinct
                uid,
                date_format(create_time,'%Y-%m-%d') create_time,
                datediff(create_time,now()) diff
                from test_sign_history order by uid,create_time
            ) b, (SELECT @gid := 1, @cgid := 1, @rank := 1) as a
        ) c group by uid,flag
        order by uid,count(*) desc
    ) d,(SELECT @ggid := 1, @cggid := 1, @grank := 1) as e
)f
where grank=1;
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