MySQL: Select All Dates In a Range Even If No Reco

问题:

I have a database of users. I would like to create a graph based on userbase growth. The query I have now is:

SELECT DATE(datecreated), count(*) AS number FROM users 
WHERE DATE(datecreated) > \'2009-06-21\' AND DATE(datecreated) <= DATE(NOW())
GROUP BY DATE(datecreated) ORDER BY datecreated ASC

This returns almost what I want. If we get 0 users one day, that day is not returned as a 0 value, it is just skipped and the next day that has at least one user is returned. How can I get something like (psuedo-response):

date1 5
date2 8
date3 0
date4 0
date5 9
etc...

where the dates with zero show up in sequential order with the rest of the dates?

Thanks!

回答1:

I hope you will figure out the rest.

select  * from (
select date_add(\'2003-01-01 00:00:00.000\', INTERVAL n5.num*10000+n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) as date from
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n1,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n2,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n3,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n4,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n5
) a
where date >\'2011-01-02 00:00:00.000\' and date < NOW()
order by date

With

select n3.num*100+n2.num*10+n1.num as date

you will get a column with numbers from 0 to max(n3)*100+max(n2)*10+max(n1)

Since here we have max n3 as 3, SELECT will return 399, plus 0 -> 400 records (dates in calendar).

You can tune your dynamic calendar by limiting it, for example, from min(date) you have to now().

回答2:

This question asks the same thing I think. Generally the accepted answer seems to be that you either do it in your application logic (read in what you have into an array, then loop through the array and create the missing dates), or you use temporary tables filled with the dates you wish to join.

回答3:

This is better to do as:

-- 7 Days:
set @n:=date(now() + interval 1 day);
SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
    right join (
        select (select @n:= @n - interval 1 day) day_series from tbl1 limit 7 ) as qb 
    on date(qa.Timestamp) = qb.day_series and 
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 7 day) order by qb.day_series asc

-- 30 Days:
set @n:=date(now() + interval 1 day);
SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
    right join (
        select (select @n:= @n - interval 1 day) day_series from tbl1 limit 30 ) as qb 
    on date(qa.Timestamp) = qb.day_series and 
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 30 day) order by qb.day_series asc;

or without variable like this:

SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
right join (
    select curdate() - INTERVAL a.a day as day_series from(
        select 0 as a union all select 1 union all select 2 union all 
        select 3 union all select 4 union all 
        select 5 union all select 6 union all select 7
    ) as a ) as qb
on date(qa.Timestamp) = qb.day_series and
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 7 day) order by qb.day_series asc;

回答4:

Do a right outer join to a table, call it tblCalendar, that is pre-populated with the dates you wish to report on. And join on the date field.

Paul

回答5:

On further thought, something like this should be what you want:

CREATE TEMPORARY TABLE DateSummary1 ( datenew timestamp ) SELECT DISTINCT(DATE(datecreated)) as datenew FROM users;

CREATE TEMPORARY TABLE DateSummary2 ( datenew timestamp, number int ) SELECT DATE(datecreated) as datenew, count(*) AS number FROM users 
WHERE DATE(datecreated) > \'2009-06-21\' AND DATE(datecreated) <= DATE(NOW())
GROUP BY DATE(datecreated) ORDER BY datecreated ASC;

SELECT ds1.datenew,ds2.number FROM DateSummary1 ds1 LEFT JOIN DateSummary2 ds2 on ds1.datenew=ds2.datenew;

This gives you all the dates in the first table, and the count summary data in the second table. You might need to replace ds2.number with IF(ISNULL(ds2.number),0,ds2.number) or something similar.

回答6:

Query is:

SELECT qb.dy as yourday, COALESCE(count(yourcolumn), 0) as yourcount from yourtable qa 
right join (
    select curdate() as dy    union
    select DATE_SUB(curdate(), INTERVAL 1 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 2 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 3 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 4 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 5 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 6 day) as dy        
    ) as qb 
on qa.dates = qb.dy 
and qa.dates > DATE_SUB(curdate(), INTERVAL 7 day)
order by qb.dy asc;

and the result is:

+------------+-----------+
| yourday    | yourcount |
+------------+-----------+
| 2015-06-24 | 274339    |
| 2015-06-25 |      0    |
| 2015-06-26 |      0    |
| 2015-06-27 |      0    |
| 2015-06-28 | 134703    |
| 2015-06-29 |  87613    |
| 2015-06-30 |      0    |
+------------+-----------+