I am trying to combine two MYSQL Queries into one. What I want to do is select the first and last row added for each day and subtract the last column for that day from the first column of that day and output that. What this would do is give me a net gain of XP in this game for that day.
Below are my two queries, their only difference is ordering the date by DESC vs ASC. the column in the database that i want to subtract from each other is "xp"
$query = mysql_query("
SELECT * FROM (SELECT * FROM skills WHERE
userID='$checkID' AND
skill = '$skill' AND
date >= ".$date."
ORDER BY date DESC) as temp
GROUP BY from_unixtime(date, '%Y%m%d')
");
$query2 = mysql_query("
SELECT * FROM (SELECT * FROM skills WHERE
userID='$checkID' AND
skill = '$skill' AND
date >= ".$date."
ORDER BY date DESC) as temp
GROUP BY from_unixtime(date, '%Y%m%d')
");
SELECT FROM_UNIXTIME(date, '%Y%m%d') AS YYYYMMDD, MAX(xp)-MIN(xp) AS xp_gain
FROM skills
WHERE userID = '$checkID'
AND skill = '$skill'
AND date >= $date
GROUP BY YYYYMMDD
This assumes that XP always increases, so it doesn't need to use the times to find the beginning and ending values.
If that's not a correct assumption, what you want is something like this:
SELECT first.YYYYMMDD, last.xp - first.xp
FROM (subquery1) AS first
JOIN (subquery2) AS last
ON first.YYYYMMDD = last.YYYYMMDD
Replace subquery1 with a query that returns the first row of each day, and subquery2 with a query that returns the last row of each day. The queries you posted in your question don't do this, but there are many SO questions you can find that explain how to get the highest or lowest row per group.
Related
I have a table that stores shift records for employees.
Simply, there's the following data:
id = Shift ID
employeenum = Employee Number
start = unix timestamp of shift start time
end = unix timestamp of shift end time
date = YYYY-mm-dd description of date the shift starts on
status = shift status (numeric status identifier)
I am currently determining conflicts through a looping php script but it's far too slow. I've searched other questions and can't quite find the answer I'm looking for.
I am trying to come up with a query that will basically give me a list of employeenums that have conflicting shifts within a given time period.
i.e. for the period 2016-07-03 to 2016-07-10, which employees have overlapping start and end timestamps for shifts with a status value of 1 or 7.
Any help would be appreciated.
Thank you!
EDIT
This is essentially the table structure.
id is a primary auto increment key. The table is full of numeric data.
ID is an autoincremented number, employeenum is a 6 digit number, start and end are unix timetamps, date is YYYY-mm-dd date format, overridden is 1 or 0, status is 1,2,3,4,5,6, or 7.
Current loop works by querying:
SELECT * FROM schedule WHERE overridden =0 AND date >=$startdate AND date <= $enddate AND (status = 1 OR status = 7) AND employeenum != 0 ORDER BY date ASC
It then loops through all of those returned shifts to test whether or not another one conflicts with them by executing this query over and over (using the returned start and end values from the results of the above query):
SELECT `employeenum` FROM `schedule` WHERE `overridden` =0 AND `date` >= '$startdate' AND `date` <= '$enddate' AND (`status` = '1' OR `status` = '7') AND ((('$start' > `start`) AND ('$start' < `end`)) OR ((`end` > '$start') AND (`end` < '$end'))) AND `employeenum` = '$employee';"
If there is a result, it pushes the employee number to an array of employees with conflicts. This then prevents the loop from checking for that employee again.
At any given time there could be 10,000 records, so it's executing 10,000+ queries. These records represent only 100-200 employees, so I am looking for a way to query one time to see if there are any overlapping (start and end overlap with another start or end) records between two date values for one employeenum without having to query the database 10,000 times.
This Query will give you the shift id, the date, the employee number, the conflicting employee numbers and a count of conflicting shifts. You have a ton of shifts that conflict in your dataset!!!
SELECT `schedule_test`.`id`, `schedule_test`.`date`, `schedule_test`.`employeenum`, GROUP_CONCAT(DISTINCT`join_tbl`.`employeenum`), COUNT(`join_tbl`.`employeenum`)
FROM `schedule_test`
INNER JOIN `schedule_test` AS `join_tbl` ON
`schedule_test`.`date` = `join_tbl`.`date`
AND (`join_tbl`.`status` = 1 OR `join_tbl`.`status` = 7)
AND (`join_tbl`.`start` BETWEEN `schedule_test`.`start` AND `schedule_test`.`end`
OR `join_tbl`.`end` BETWEEN `schedule_test`.`start` AND `schedule_test`.`end`)
AND `schedule_test`.`id` != `join_tbl`.`id`
WHERE (`schedule_test`.`status` = 1 OR `schedule_test`.`status` = 7)
GROUP BY `schedule_test`.`id`
ORDER BY `schedule_test`.`date`
Adapted from #cmorrissey 's answer. THANK YOU!!
SELECT `schedule_test`.`id`, `schedule_test`.`date`,
`schedule_test`.`employeenum`,
GROUP_CONCAT(DISTINCT`join_tbl`.`employeenum`),
COUNT(`join_tbl`.`employeenum`)
FROM `schedule_test`
INNER JOIN `schedule_test` AS `join_tbl` ON
`schedule_test`.`date` = `join_tbl`.`date`
AND (`join_tbl`.`status` = 1 OR `join_tbl`.`status` = 7)
AND (`join_tbl`.`employeenum` = `schedule_test`.`employeenum`)
AND (`join_tbl`.`start` BETWEEN `schedule_test`.`start` AND `schedule_test`.`end`
OR `join_tbl`.`end` BETWEEN `schedule_test`.`start` AND `schedule_test`.`end`)
AND `schedule_test`.`id` != `join_tbl`.`id`
WHERE (`schedule_test`.`status` = 1 OR `schedule_test`.`status` = 7)
GROUP BY `schedule_test`.`id`
ORDER BY `schedule_test`.`date`
Using PHP/MySQL
I'm trying to create a select statement that gets the data from the least day of the current week (I'm using it to show data on a certain player 'this week'). The week starts on Sunday. Sundays's data may not always exist therefore if the Sunday data isn't found then it would use the next earliest day found, Monday, Tuesday, etc.
My date column is named 'theDate' and the datatype is 'DATE'
The query would need to be something like:
SELECT *
FROM table_name
WHERE name = '$username'
AND [...theDate = earliest day of data found for the current week week]
LIMIT 1
It would return a single row of data.
This is a query I tried for getting the 'this week' data, It doesn't seem to work correctly on Sunday's it shows nothing:
SELECT *
FROM table_name
WHERE playerName = '$username'
AND YEARWEEK(theDate) = YEARWEEK(CURRENT_DATE)
ORDER BY theDate;
This is the query that I'm using to get 'this months' data and it works even if the first day of the months data is not found, it will use the earliest date of data found in the current month/year (this query works perfect for me):
SELECT *
FROM table_name
WHERE playerName = '$username'
AND theDate >= CAST( DATE_FORMAT( NOW(),'%Y-%m-01') AS DATE)
ORDER BY theDate
LIMIT 1
Without trying this, you probably need an inner query:
select *
from table_name tn
where tn.the_date =
(select min(the_date)
from table_name
where WEEKOFYEAR(the_date) = WEEKOFYEAR(CURDATE())
and YEAR(the_date) = YEAR(CURDATE()))
viz, give me the row(s) in the table with a date equal to the earliest date in the table in the current week and year.
Try this
SELECT * FROM table_name WHERE name = '$username'
AND your_data IS NOT NULL
AND WEEK(the_date,0 = WEEK(NOW(),0))
ORDER BY DATE_FORMAT(the_date,'%w') ASC
Try the following, replace YOUR_DATE with the date from the column you want (theDate):
SELECT ADDDATE(YOUR_DATE, INTERVAL 1-DAYOFWEEK(YOUR_DATE) DAY)
FirstDay from dual
Did you try:
SELECT ADDDATE(theDate , INTERVAL 1-DAYOFWEEK(theDate ) DAY) FirstDay
FROM table_name
WHERE playerName = '$username'
ORDER BY theDate DESC
LIMIT 1
I've been searching and I know there are similar questions, but none of them seems to answer this particular question.
I am trying to get a count of the total number of days an employee has worked on a given schedule. To do this, I am counting the total number of rows the employee appears on the "schedules" table. Only we run into a problem if the employee is scheduled twice on the same day.
To solve this, I want to count total number of rows and sort by DATE in a DATETIME field.
Current query:
$days = mysql_query("SELECT emp_id FROM schedules
WHERE sch_id = '$sch_id'
AND emp_id = '$emp_data[emp_id]'");
$tot_days = mysql_num_rows($days);
I would like to change it to:
$days = mysql_query("SELECT emp_id FROM schedules
WHERE sch_id = '$sch_id'
AND emp_id = '$emp_data[emp_id]'
GROUP BY start_date");
// "start_date" is a datetime field. Need to sort by date only YYYY-MM-DD
$tot_days = mysql_num_rows($days);
Any thoughts?
If your start_date column is a MySQL datetime type, you could use the following:
$days = mysql_query("SELECT start_date, count(*) FROM schedules
WHERE sch_id = '$sch_id'
AND emp_id = '$emp_data[emp_id]'
GROUP BY DATE(start_date)
HAVING count(*) > 1
ORDER BY DATE(start_date)");
The DATE function "Extracts the date part of the date or datetime expression"
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_date
This will give you only those rows where the emp_id being considered is used more than once in a given date. Remove the HAVING line if you want to see all.
I have a field on my table which is called X, and I store unixtime(php -> time()).
My question is:
How can I list all the months from my DB, something like: 6.2009 8.2009 etc..
And my second question is:
How can I make a query to list all the informations based on a month and year on the same field X(i store unixtime), I know this doesn't work, but mabe you can understand it better: where date = '06.2009'
Any solutions?
List all months:
SELECT DISTINCT EXTRACT(YEAR_MONTH FROM FROM_UNIXTIME(X)) FROM MyTable;
Return all rows with a given year & month:
SELECT * FROM MyTable
WHERE EXTRACT(YEAR_MONTH FROM FROM_UNIXTIME(X)) = 200906;
Try this
SELECT FROM_UNIXTIME(X,'%m.%Y') FROM TABLE GROUP BY FROM_UNIXTIME(X,'%m.%Y')
SELECT * FROM TABLE WHERE FROM_UNIXTIME(X,'%m.%Y')='06.2009'
Bye
To answer your first question, you would do this with grouping:
SELECT FROM_UNIXTIME(timestamp_column, '%c.%Y') AS month, some_other_column FROM table_name GROUP BY month;
As for your second question, this depends on what you're trying to do. For example:
SELECT AVG(payment), SUM(*), FROM_UNIXTIME(timestamp_column, '%c.%Y') AS month, some_other_column FROM table_name WHERE timestamp_column BETWEEN UNIX_TIMESTAMP(200906) AND UNIX_TIMESTAMP(200907) - 1 GROUP BY month;
Would return the average of the payments and the number (sum) of rows for each group.
To get information ungrouped from a specific timeframe, reduce the query like so:
SELECT payment, some_other_column FROM table_name WHERE timestamp_column BETWEEN UNIX_TIMESTAMP(200906) AND UNIX_TIMESTAMP(200907) - 1;
I'm looking for a best practice advice how to speed up queries and at the same time to minimize the overhead needed to invoke date/mktime functions. To trivialize the problem I'm dealing with the following table layout:
CREATE TABLE my_table(
id INTEGER PRIMARY KEY NOT NULL AUTO_INCREMENT,
important_data INTEGER,
date INTEGER);
The user can choose to show 1) all entries between two dates:
SELECT * FROM my_table
WHERE date >= ? AND date <= ?
ORDER BY date DESC;
Output:
10-21-2009 12:12:12, 10002
10-21-2009 14:12:12, 15002
10-22-2009 14:05:01, 20030
10-23-2009 15:23:35, 300
....
I don't think there is much to improve in this case.
2) Summarize/group the output by day, week, month, year:
SELECT COUNT(*) AS count, SUM(important_data) AS important_data
FROM my_table
WHERE date >= ? AND date <= ?
ORDER BY date DESC;
Example output by month:
10-2009, 100002
11-2009, 200030
12-2009, 3000
01-2010, 0 /* <- very important to show empty dates, with no entries in the table! */
....
To accomplish option 2) I'm currently running a very costly for-loop with mktime/date like the following:
for(...){ /* example for group by day */
$span_from = (int)mktime(0, 0, 0, date("m", $time_min), date("d", $time_min)+$i, date("Y", $time_min));
$span_to = (int)mktime(0, 0, 0, date("m", $time_min), date("d", $time_min)+$i+1, date("Y", $time_min));
$query = "..";
$output = date("m-d-y", ..);
}
What are my ideas so far? Add additional/ redundant columns (INTEGER) for day (20091212), month (200912), week (200942) and year (2009). This way I can get rid of all the unnecessary queries in the for loop. However I'm still facing the problem to very fastly calculate all dates that doesn't have any equivalent in database. One way to simply move the problem could be to let MySQL do the job and simply use one big query (calculate all the dates/use MySQL date functions) with a left join (the data). Would it be wise to let MySQL take the extra load? Anyway I'm reluctant to use all these mktime/date in the for loop. Since I have complete control over the table layout and code even suggestions with major changes are welcome!
Update
Thanks to Greg I came up with the following SQL query. However it still bugs me to use 50 lines of sql statements - build up with php - that maybe could be done faster and more elegantly otherwise:
SELECT * FROM (
SELECT DATE_ADD('2009-01-30', INTERVAL 0 DAY) AS day UNION ALL
SELECT DATE_ADD('2009-01-30', INTERVAL 1 DAY) AS day UNION ALL
SELECT DATE_ADD('2009-01-30', INTERVAL 2 DAY) AS day UNION ALL
SELECT DATE_ADD('2009-01-30', INTERVAL 3 DAY) AS day UNION ALL
......
SELECT DATE_ADD('2009-01-30', INTERVAL 50 DAY) AS day ) AS dates
LEFT JOIN (
SELECT DATE_FORMAT(date, '%Y-%m-%d') AS date, SUM(data) AS data
FROM test
GROUP BY date
) AS results
ON DATE_FORMAT(dates.day, '%Y-%m-%d') = results.date;
You definitely shouldn't be doing a query inside a loop.
You can group like this:
SELECT COUNT(*) AS count, SUM(important_data) AS important_data, DATE_FORMAT('%Y-%m', date) AS month
FROM my_table
WHERE date BETWEEN ? AND ? -- This should be the min and max of the whole range
GROUP BY DATE_FORMAT('%Y-%m', date)
ORDER BY date DESC;
Then pull these into an array keyed by date and loop over your data range as you are doing (that loop should be pretty light on CPU).
Another idea is not to use string inside the query. Transform the string parameter to datetime, on mysql.
STR_TO_DATE(str,format)
http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html