I have a table called tasks (id, date, id_user ).
I have an other table called calendar (id, date ).
to find the number of occurrences of the date field for a user between 2 dates I use the query :
$test1 = date(Y-m-d', strtotime('-7'));
$test2 = date('Y-m-d');
$query= $connect->prepare("SELECT *, COUNT(*) AS counter FROM tasks WHERE matricule ='".$_SESSION['matricule']."' AND date < '".$test2."' AND date >= '".$test1."' GROUP BY date");
This query works.
I also want to display 0 for dates that are not in my table tasks but are in the table calendar
You want to count the daily tasks performed by a given user over a range of days, including the days without tasks.
If you have a calendar table - say calendar(dt), where dt is a date - you can do:
select c.dt, count(t.date) as counter
from calendar c
left join tasks t
on t.matricule = :matricule
and t.date = c.dt
where c.date >= :start_date and c.date < :end_date
group by c.dt
:matricule, :start_date and :end_date are parameters to the query; you should really be using prepared statements instead of concatenating POSTed values in the query string, which opens up your code to SQL injection.
If the time range always is the last week:
where c.date >= current_date - interval 7 day and c.date < current_date
Note how the aggregation differs from your original code: columns in the select clause are repeated in the group by clause (left apart those that are aggregated).
Related
I use MariaDB and have a table where each row has a date and a score.
I want to first show the rows where the date is 3 days old or newer, sorted by the score - then show the rest (more than 3 days old) sorted by date.
Since my date is stored in unix time, it's fairly easy to have php calculate 3 days from before now and use that as my $scoreTimeLimit variable in the below:
Here are my two queries:
SELECT * FROM myTable WHERE myDate > $scoreTimeLimit ORDER BY myPopularityScore DESC
SELECT * FROM myTable WHERE myDate < $scoreTimeLimit ORDER BY myDate DESC
However, I would VERY much like to have only 1 query instead of two. Can it be done...?
This is a job for UNION.
SELECT * FROM (
SELECT 0 ord1, NOW() as ord2, *
FROM myTable WHERE myDate > NOW() - INTERVAL 3 DAY
UNION ALL
SELECT 1 ord1, myDate as ord2, *
FROM myTable WHERE myDate <= NOW() - INTERVAL 3 DAY
) a
ORDER BY ord1, ord2 DESC, myPopularityScore
The inner query gives you a single result set with a couple of extra columns added on to help you manage your sorting.
Am trying to find the min value from past 30 days, in my table there is one entry for every day, am using this query
SELECT MIN(low), date, low
FROM historical_data
WHERE name = 'bitcoin'
ORDER BY STR_TO_DATE(date,'%d-%m-%Y') DESC
LIMIT 7
But this value not returing the correct value. The structure of my table is
Table structure
And table data which is store is like this
Table data style
Now what i need is to get the minimum low value. But my query not working it give me wrong value which even did not exist in table as well.
Updates:
Here is my updated Table Structure.
enter image description here
And here is my data in this table which look like this
enter image description here
Now if you look at the data, i want to check the name of token omisego and fatch the low value from past 7 days which will be from 2017-12-25 to 2017-12-19
and in this cast the low value is 9.67, but my current query and the query suggested by my some member did not brings the right answer.
Update 2:
http://rextester.com/TDBSV28042
Here it is, basically i have more then 1400 coins and token historical data, which means that there will me more then 1400 entries for same date like 2017-12-25 but having different name, total i have more then 650000 records. so every date have many entries with different names.
To get the lowest row per group you could use following
SELECT a.*
FROM historical_data a
LEFT JOIN historical_data b ON a.name = b.name
AND a.low > b.low
WHERE b.name IS NULL
AND DATE(a.date) >= '2017-12-19' AND DATE(a.date) <= '2017-12-25'
AND a.name = 'omisego'
or
SELECT a.*
FROM historical_data a
JOIN (
SELECT name,MIN(low) low
FROM historical_data
GROUP BY name
) b USING(name,low)
WHERE DATE(a.date) >= '2017-12-19' AND DATE(a.date) <= '2017-12-25'
AND a.name = 'omisego'
DEMO
For last 30 day of 7 days or n days you could write above query as
SELECT a.*, DATE(a.`date`)
FROM historical_data2 a
LEFT JOIN historical_data2 b ON a.name = b.name
AND DATE(b.`date`) >= CURRENT_DATE() - INTERVAL 30 DAY
AND DATE(b.`date`) <= CURRENT_DATE()
AND a.low > b.low
WHERE b.name IS NULL
AND a.name = 'omisego'
AND DATE(a.`date`) >= CURRENT_DATE() - INTERVAL 30 DAY
AND DATE(a.`date`) <= CURRENT_DATE()
;
DEMO
But note it may return more than one records where low value is same, to choose 1 row among these you have specify another criteria to on different attribute
Consider grouping the same and running the clauses
SELECT name, date, MIN(low)
FROM historical_data
GROUP BY name
HAVING name = 'bitcoin'
AND STR_TO_DATE(date, '%M %d,%Y') > DATE_SUB(NOW(), INTERVAL 30 DAY);
Given the structure, the above query should get you your results.
// Try this code ..
SELECT MIN(`date`) AS date1,low
FROM historical_data
WHERE `date` BETWEEN now() - interval 1 month
AND now() ORDER by low ASC;
I have a database with the rows: SearchTerm | userId | date | historyId
I need to get the amount of entries every hour in the last 24 hour period where the userId=userid.
So far I have as follows:
$stmt = $conn->prepare("SELECT historyId FROM webHistory WHERE date >= now() - INTERVAL 1 DAY GROUP BY HOUR(date) AND userId=?");
I'm now a little stuck, how would I go about getting the num_rows for each hour group? I though about using count(*), but would this be the right method, if so how would I go about doing this?
Lastly, for mobile displays I would need to group by every two hour period, is this possible as I can only seem to find documentation on HOUR(), possibly DATEPART()?
You just need count(*):
SELECT HOUR(date) as hr, historyId, COUNT(*) as num_rows
FROM webHistory
WHERE date >= now() - INTERVAL 1 DAY AND userId=?
GROUP BY HOUR(date);
The condition on userId goes in the where clause. It is good form to include the hour(date) in the select, so you know which hour a given count refers to.
EDIT:
To just get today's hours, hour can do:
SELECT HOUR(date) as hr, historyId, COUNT(*) as num_rows
FROM webHistory
WHERE date(date) = date(now()) AND userId=?
GROUP BY HOUR(date);
To list data by two-hour periods:
SELECT FLOOR(HOUR(date)/2) AS period,historyId FROM webHistory, COUNT(*) as num_rows
WHERE date >= now() - INTERVAL 1 DAY
GROUP BY date, period
We can use SQL BETWEEN operator
SELECT HOUR(date) as hr, historyId, COUNT(*) as num_rows
FROM webHistory
WHERE date BETWEEN SUBDATE(date(now()),1) AND date(now())
AND userId=?
GROUP BY HOUR(date);
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.
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