How do I modify this MySQL query to only count leadIDs from table leads where column 'Date' contains the newest (youngest) date?
SELECT COUNT(leadID) as accepted FROM leads WHERE change like '%OK%'
The problem is that leadID can have multiple instances in table leads. The original query result is "4" because of one duplicate. The correct result is "3".
The date is stored in this format: 2011-10-26 18:23:52. The result should take hours and minutes into consideration when determining the youngest date.
TABLE leads:
leadID | date | change
1 | 2011-10-26 18:23:52 | BAD
1 | 2011-10-26 17:00:00 | OK
2 | 2011-10-26 19:23:52 | OK
3 | 2011-10-26 20:23:52 | OK
4 | 2011-10-26 21:23:52 | OK
5 | 2011-10-26 22:23:52 | BAD
I think this is what you're looking for:
select count(distinct l1.leadId) as accepted from leads l1
left join leads l2
on l1.leadId = l2.leadId and l1.date < l2.date
where l2.date is null and l1.`change` like '%OK%'
You must decide what you mean by newest date: the single latest? yesterday? today?
if yesterday, then add this to your query clause
select * from mytable where date >= date_sub(now(), interval 1 day)
if you are using oracle database you can use max() function to extract newest date from the table, further to check with the table for this newest date :-
SELECT COUNT(leadID) as accepted FROM leads WHERE change like '%OK%'
and date_col = (select max(date_col) from leads)
I am assuming that with newest date your mean is about newest in the table data..
changes :- as per changes in question and as per mentioned in commends ..
I think you want to take newest date among the records having "change" column value like '%OK%' and want to count distinct leadId
please try the following query-
SELECT COUNT(distinct leadID) as accepted FROM leads WHERE change like '%OK%'
and date_col = (select max(date_col) from leads WHERE change like '%OK%')
You can try (in case your date is a int like return by time() function)
$sql = "SELECT COUNT(leadID) as accepted FROM leads WHERE change like '%OK%' ORDER BY Date DESC LIMIT 1"
You will only extract the newest entry.
Edit: This shouldalso works for your date format YYYY-MM-DD hh:mm:ss
Edit 2: Okay, I did not understood your question.
You have a table lead: leadid date
You want to count the number of row for the newset date.
Like another pointed out you can use the MAX operator:
SELECT COUNT(distinct leadid)
FROM LEAD AS l,
( SELECT MAX(Date) mdate FROM Lead ) AS MaxDate
WHERE l.date = MaxDate.mdate
AND l.change like '%OK%'
Related
I have mysql table called user_log that contains something like:
userid created_at
1 1388514600
2 1391193000
I want to get the record using exact date. For example I have the created_at which is 2013-12-31
SELECT * FROM user_log WHERE created_at = UNIX_TIMESTAMP('2013-31-12');
But record is not selected, I don't know what was the problem in that query. So how can I get the record from unix timestamp column in mysql using date?
To know the exact value in dateformat, you can do :
SELECT FROM_UNIXTIME(created_at) FROM user_log;
That said, the value 1388514600 is not just 2013-12-31 but it's actually 2013-12-31 18:30:00;
So to search by just date, you can try this:
SELECT FROM_UNIXTIME(created_at) AS dt
FROM user_log
HAVING date(dt)='2013-12-31';
Fiddle here
I have three columns User_ID, New_Status and DATETIME.
New_Status contains 0(inactive) and 1(active) for users.
Every user starts from active status - ie. 1.
Subsequently table stores their status and datetime at which they got activated/inactivated.
How to calculate number of active users at the end of each date, including dates when no records were generated into the table.
Sample data:
| ID | New_Status | DATETIME |
+----+------------+---------------------+
| 1 | 1 | 2019-01-01 21:00:00 |
| 1 | 0 | 2019-02-05 17:00:00 |
| 1 | 1 | 2019-03-06 18:00:00 |
| 2 | 1 | 2019-01-02 01:00:00 |
| 2 | 0 | 2019-02-03 13:00:00 |
Format the date time value to a date only string and group by it
SELECT DATE_FORMAT(DATETIME, '%Y-%m-%d') as day, COUNT(*) as active
FROM test
WHERE New_Status = 1
GROUP BY day
ORDER BY day
In MySQL 8 you can use the row_number() window function to get the last status of a user per day. Then filter for the one that indicate the user was active GROUP BY the day and count them.
SELECT date(x.datetime),
count(*)
FROM (SELECT date(t.datetime) datetime,
t.new_status,
row_number() OVER (PARTITION BY date(t.datetime)
ORDER BY t.datetime DESC) rn
FROM elbat t) x
WHERE x.rn = 1
AND x.new_status = 1
GROUP BY x.datetime;
If not all days are in the table you need to create a (possibly derived) table with all days and cross join it.
Find out the last activity status of users whose activity was changed for each day
select User_ID, New_Status, DATE_FORMAT(DATETIME, '%Y-%m-%d')
from activity_table
where not exists
(
select 1
from activity_table at
where at.User_ID = activity_table.User_ID and
DATE_FORMAT(at.DATETIME, '%Y-%m-%d') = DATE_FORMAT(activity_table.DATETIME, '%Y-%m-%d') and
at.DATETIME > activity_table.DATETIME
)
order by DATE_FORMAT(activity_table.DATETIME, '%Y-%m-%d');
This is not the solution yet, but a very very useful information before solution. Note that here not all dates are covered yet and the values are individual records, more precisely their last values on each day, ordered by the date.
Let's get aggregate numbers
Using the query above as a subselect and aliasing it into a table, you can group by DATETIME and do a select sum(new_Status) as activity, count(*) total, DATETIME so you will know that activity - (total - activity) is the difference in comparison to the previous day.
Knowing the delta for each day present in the result
At the previous section we have seen how the delta can be calculated. If the whole query in the previous section is aliased, then you can self join it using a left join, with pairs of (previous date, current date), still having the gaps of dates, but not worrying about that just yet. In the case of the first date, its activity is the delta. For subsequent records, adding the previous day's delta to their delta yields the result you need. To achieve this you can use a recursive query, supported by MySQL 8, or, alternatively, you can just have a subquery which sums the delta of previous days (with special attention to the first date, as described earlier) will and adding the current date's delta yields the result we need.
Fill the gaps
The previous section would already perfectly work (assuming the lack of integrity problems), assuming that there were activity changes for each day, but we will not continue with the assumption. Here we know that the figures are correct for each date where a figure is present and we will need to just add the missing dates into the result. If the results are properly ordered, as they should be, then one can use a cursor and loop the results. At each record after the first one, we can determine the dates that are missing. There might be 0 such dates between two consequent dates or more. What we do know about the gaps is that their values are exactly the same as the previous record, that do has data. If there were no activity changes on a given date, then the number of active users is exactly the same as in the previous day. Using some structure, like a table you can generate the results you have with the knowledge described here.
Solving possible integrity problems
There are several possibilities for such problems:
First, a data item might exist prior to the introduction of this table's records were started to be spawned.
Second, bugs or any other causes might have made a pause in creating records for this activity table.
Third, the addition of user is or was not necessarily generating an activity change, since its popping into existence renders its previous state of activity undefined and subject to human standards, which might change over time.
Fourth, the removal of user is or was not necessarily generating an activity change, since its popping out of existence renders is current state of activity undefined and subject to human standards, which might change over time.
Fifth, there is an infinity of other issues which might cause data integrity issues.
To cope with these you will need to comprehensively analyze whatever you can from the source-code and the history of the project, including database records, logs and humanly available information to detect such anomalies, the time they were effective and figure out what their solution is if they exist.
EDIT
In the meantime I was thinking about the possibility of a user, who was active at the start of the day being deactivated and then activated again by the end of the day. Similarly, an inactive user during a day might be activated and then finally deactivated by the end of the day. For users that have more than an activation at the start of the day, we need to compare their activity status at the start and the end of the day to find out what the difference was.
SELECT
DATE(DATETIME),
COUNT(*)
FROM your_table
WHERE New_Status = 1
GROUP BY User_ID,
DATE(DATETIME)
For MySQL
WITH RECURSIVE
cte AS (
SELECT MIN(DATE(DT)) dt
FROM src
UNION ALL
SELECT dt + INTERVAL 1 DAY
FROM cte
WHERE dt < ( SELECT MAX(DATE(DT)) dt
FROM src )
),
cte2 AS
(
SELECT users.id,
cte.dt,
SUM( CASE src.New_Status WHEN 1 THEN 1
WHEN 0 THEN -1
ELSE 0
END ) OVER ( PARTITION BY users.id
ORDER BY cte.dt ) status
FROM cte
CROSS JOIN ( SELECT DISTINCT id
FROM src ) users
LEFT JOIN src ON src.id = users.id
AND DATE(src.dt) = cte.dt
)
SELECT dt, SUM(status)
FROM cte2
GROUP BY dt;
fiddle
Do not forget to adjust max recursion depth.
Here is what I believe is a good solution for this problem of yours:
SELECT SUM(New_Status) "Number of active users"
, DATE_FORMAT(DATEC, '%Y-%m-%d') "Date"
FROM TEST T1
WHERE DATE_FORMAT(DATEC,'%H:%i:%s') =
(SELECT MAX(DATE_FORMAT(T2.DATEC,'%H:%i:%s'))
FROM TEST T2
WHERE T2.ID = T1.ID
AND DATE_FORMAT(T1.DATEC, '%Y-%m-%d') = DATE_FORMAT(T2.DATEC, '%Y-%m-%d')
GROUP BY ID
, DATE_FORMAT(DATEC, '%Y-%m-%d'))
GROUP BY DATE_FORMAT(DATEC, '%Y-%m-%d');
Here is the DEMO
Im not too mysql savvy and i'm having issues creating a select that fits with my needs.
I have a database table that looks similar to this:
registrar
id balance date
---------------------
1 500.00 2013-01-01
2 402.00 2013-01-01
3 396.00 2013-01-02
4 394.00 2013-01-02
I have a query that I use to pull the data into a script and display the data:
SELECT balance, date FROM $registrar WHERE date BETWEEN '$starting_date' AND '$ending_date'
However, it appears that when querying, mysql is only returning the newest entry for that date. Id like if possible to return the lower balance amount of that date if found multiple rows matching the date criteria.
try ORDER BY
SELECT balance, date FROM $registrar WHERE date BETWEEN '$starting_date' AND '$ending_date' ORDER BY balance DESC
You have to use order by class in the end of query
SELECT balance, date FROM $registrar WHERE date BETWEEN '$starting_date' AND '$ending_date' ORDER BY balance DESC
By default it id , so u have to put balance .
I have a table recording the accumulative total visit numbers of some web pages every day. I want to fetch the real visit numbers in a specific day for all these pages. the table is like
- record_id page_id date addup_number
- 1 1 2012-9-20 2110
- 2 2 2012-9-20 1160
- ... ... ... ...
- n 1 2012-9-21 2543
- n+1 2 2012-9-21 1784
the result I'd like to fetch is like:
- page_id date increment_num(the real visit numbers on this date)
- 1 2012-9-21 X
- 2 2012-9-21 X
- ... ... ...
- N 2012-9-21 X
but I don't want to do this in php, cause it's time consuming. Can I get what I want with SQL directives or with some mysql functions?
Ok. You need to join the table on itself by joining on the date column and adding a day to one side of the join.
Assuming:
date column is a legitimate DATE Type and not a string
Every day is accounted for each page (no gaps)
addup_number is an INT of some type (BIGINT, INT, SMALLINT, etc...)
table_name is substituted for your actual table name which you don't indicate
Only one record per day for each page... i.e. no pages have multiple counts on the same day
You can do this:
SELECT t2.page_id, t2.date, t2.addup_number - t1.addup_number AS increment_num
FROM table_name t1
JOIN table_name t2 ON t1.date + INTERVAL 1 DAY = t2.date
WHERE t1.page_id = t2.page_id
One thing to note is if this is a huge table and date is an indexed column, you'll suffer on the join by having to transform it by adding a day in the ON clause, but you'll get your data.
UPDATED:
SELECT today.page_id, today.date, (today.addup_number - yesterday.addup_number) as increment
FROM myvisits_table today, myvisits_table yesterday
WHERE today.page_id = yesterday.page_id
AND today.date='2012-9-21'
AND yesterday.date='2012-9-20'
GROUP BY today.page_id, today.date, yesterday.page_id, yesterday.date
ORDER BY page_id
Something like this:
SELECT date, SUM(addup_number)
FROM your_table
GROUP BY date
I have a log table with a date field called logTime. I need to show the number of rows within a date range and the number of records per day. The issue is that i still want to show days that do not have records.
Is it possible to do this only with SQL?
Example:
SELECT logTime, COUNT(*) FROM logs WHERE logTime >= '2011-02-01' AND logTime <= '2011-02-04' GROUP BY DATE(logTime);
It returns something like this:
+---------------------+----------+
| logTime | COUNT(*) |
+---------------------+----------+
| 2011-02-01 | 2 |
| 2011-02-02 | 1 |
| 2011-02-04 | 5 |
+---------------------+----------+
3 rows in set (0,00 sec)
I would like to show the day 2011-02-03 too.
MySQL will not invent rows for you, so if the data is not there, they will naturally not be shown.
You can create a calendar table, and join in that,
create table calendar (
day date primary key,
);
Fill this table with dates (easy with a stored procedure, or just some general scripting), up till around 2038 and something else will likely break unitl that becomes a problem.
Your query then becomes e.g.
SELECT logTime, COUNT(*)
FROM calendar cal left join logs l on cal.day = l.logTime
WHERE day >= '2011-02-01' AND day <= '2011-02-04' GROUP BY day;
Now, you could extend the calendar table with other columns that tells you the month,year, week etc. so you can easily produce statistics for other time units. (and purists might argue the calendar table would have an id integer primary key that the logs table references instead of a date)
In order to accomplish this, you need to have a table (or derived table) which contains the dates that you can then join from, using a LEFT JOIN.
SQL operates on the concept of mathematical sets, and if you don't have a set of data, there is nothing to SELECT.
If you want more details, please comment accordingly.
I'm not sure if this is a problem that should be solved by SQL. As others have shown, this requires maintaining a second table that contains the all of the individual dates of a given time span, which must be updated every time that time span grows (which presumably is "always" if that time span is the current time.
Instead, you should use to inspect the results of the query and inject dates as necessary. It's completely dynamic and requires no intermediate table. Since you specified no language, here's pseudo code:
EXECUTE QUERY `SELECT logTime, COUNT(*) FROM logs WHERE logTime >= '2011-02-01' AND logTime <= '2011-02-04' GROUP BY DATE(logTime);`
FOREACH row IN query result
WHILE (date in next row) - (date in this row) > 1 day THEN
CREATE new row with date = `date in this row + 1 day`, count = `0`
INSERT new row IN query result AFTER this row
ADVANCE LOOP INDEX TO new row (`this row` is now the `new row`)
END WHILE
END FOREACH
Or something like that
DECLARE #TOTALCount INT
DECLARE #FromDate DateTime = GetDate() - 5
DECLARE #ToDate DateTime = GetDate()
SET #FromDate = DATEADD(DAY,-1,#FromDate)
Select #TOTALCount= DATEDIFF(DD,#FromDate,#ToDate);
WITH d AS
(
SELECT top (#TOTALCount) AllDays = DATEADD(DAY, ROW_NUMBER()
OVER (ORDER BY object_id), REPLACE(#FromDate,'-',''))
FROM sys.all_objects
)
SELECT AllDays From d