I am retrieving data from a table and show the total SUM of entries. What I want to do is to show the total SUM of entries made on today's date, yesterday and this month. The table is using the unix timestamp format (e.g. 1351771856 for example).
Currently I am using this line to show todays results:
AND comment_date > UNIX_TIMESTAMP() - 24 * 3600";
but that gives me just the entries for the last 24 hours.
Example: So let's say its Friday, 17:00 PM - it gives me the count from Thursday 17:00 PM to Friday 17:00 PM
What I want is to get the results for
Thursday 00:00:00 - 23:59:59 (yesterday in this case)
the results for today (00:00:00 - 23:59:59)
and last week, results that start on Monday, 00:00:00 until "today" (in this case Friday).
I couldn't find a way in the MySQL documentation to achieve this.
This mysql code should work for you:
// Today
AND DATE(from_unixtime(comment_date)) = CURRENT_DATE
// Yesterday
AND DATE(from_unixtime(comment_date)) = DATE_SUB(CURRENT_DATE,INTERVAL 1 DAY)
// This week
AND YEARWEEK(from_unixtime(comment_date), 1) = YEARWEEK(CURRENT_DATE, 1)
// This month
AND YEAR(from_unixtime(comment_date)) = YEAR(CURRENT_DATE)
AND MONTH(from_unixtime(comment_date)) = MONTH(CURRENT_DATE)
Simply use this:
AND comment_date > date_sub(current_date, interval 1 day)
See my answer here, I think it's quite related.
Pull records from orders table for the current week
Consider getting intimate with MySQL's GROUP BY. You will most likely need to know this if you use MySQL.
Related
I'm building a date specific PHP script.
This script will perform a specific task when its Sunday. I've written my PHP code and I'm trying to test it. Since it's not Sunday today, I set my system date to last Sunday. The thing is that my MySQL query doesn't return the expected results
SELECT date FROM schedules WHERE name='someName' AND date >= (CURDATE() - INTERVAL 3 DAY )
The query above will return the records of the current date plus and minus 3 days.
Here is an example of the output I get:
2015-11-26
2015-11-29
2015-11-30
2015-12-04
2015-12-02
The result that I need is:
2015-11-26
2015-11-29
Another query:
SELECT date FROM schedules WHERE name='someName' AND date >= (CURDATE() - 3 )
The query above will return all the dates that are 3 days older than the current date not including the last 3 days of the current date.
I only want it to return the last 3 days of the current date.
I'm trying to understand the output you want. So you want to get dates that are no more than 3 days prior to the current date and nothing in the future.
So on '2015-12-04' the result would be:
2015-12-04
2015-12-03
2015-12-02
2015-12-01
So maybe the BETWEEN Clause will do the trick
SELECT date FROM schedules
WHERE name='someName'
AND (date BETWEEN (CURDATE() - INTERVAL 3 DAY ) AND CURDATE() );
I am trying to display only records two weeks in advance from the current date onwards. starttime is stored as a datetime data type in the database.
I have this,
SELECT id, date_format(starttime, '%d-%m-%Y %H:%i') AS formatted_start, date_format(starttime, '%Y-%m-%d') AS formatted_date,
date_format(endtime, ' %H:%i') AS formatted_end
FROM timedates WHERE user_id = 1 AND `status`='' AND YEARWEEK(formatted_date, 0) IN (YEARWEEK(NOW(), 0),
YEARWEEK(DATE_ADD(NOW(), INTERVAL 2 WEEK), 0))
But I am getting a syntax error YEARWEEK(formatted_date, 0) IN (YEARWEEK(NOW(), 0) AND YEARWEEK(DATE_ADD(NOW()
Could anyone tell me what's wrong with it?
It seems MySQL does not support calculated column in the where clause:
try
SELECT id, date_format(starttime, '%d-%m-%Y %H:%i') AS formatted_start,
date_format(starttime, '%Y-%m-%d') AS formatted_date,
date_format(endtime, ' %H:%i') AS formatted_end
FROM timedates
WHERE user_id = 1 AND `status`='' AND
YEARWEEK(starttime, 0) IN (
YEARWEEK(DATE_ADD(NOW(), INTERVAL 1 WEEK), 0),
YEARWEEK(DATE_ADD(NOW(), INTERVAL 2 WEEK), 0))
use starttime instead of formatted_date
As I said, I see no reason to use YEARWEEK function. You need start date set to today which is CURDATE() and plus 2 weeks period which is DATE_ADD(CURDATE(), INTERVAL 2 WEEK) and then we just check if starttime between those 2 dates.
SELECT id, date_format(starttime, '%d-%m-%Y %H:%i') AS formatted_start,
date_format(starttime, '%Y-%m-%d') AS formatted_date,
date_format(endtime, ' %H:%i') AS formatted_end
FROM timedates
WHERE user_id = 1
AND `status`=''
AND DATE(starttime) BETWEEN CURDATE() AND DATE_ADD(CURDATE(), INTERVAL 2 WEEK)
How do you define "two weeks in advance"? Our normal inclination, is to think "within two weeks", which would be to simply before 14 days from now.
WHERE starttime >= NOW()
AND starttime < NOW() + INTERVAL 14 DAY
But the original query is using YEARWEEK function, and the behavior with that is a bit different.
If today is Thursday, May 21, 1015. What is the range of dates that starttime should fall into?
For datetimes right now or in the future, we can just do:
WHERE starttime >= NOW()
(We can do just a > rather than >= if we want to exclude startime values that are an exact match for NOW().
The question is, the bit about "two weeks in advance". Adding 14 days, or 2 weeks, would get us up to Thursday, June 4, 2015. It looks like you might also want to include Friday and Saturday, June 5th and 6th. (Up until Sunday, June th.)
We can use an expression to return "Sunday on or following 14 days from today"
If it's a Sunday, we just use that date. (For convenience, we think of that as adding 0 days). If it's Saturday, we need to add 1 day. If it's a Friday, add 2 days, ... Monday, add 6 days.
Conveniently, the expression 6 - WEEKDAY(NOW()) gives us the number of days we need to add to today's date to get to the next Sunday.
Reference: https://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_weekday
We probably want to lop the time off at midnight, we can use the DATE() function to do that.
-- startime values on or after now and before midnight
-- of the first sunday on or after the date two weeks from now
WHERE starttime >= NOW()
AND starttime < DATE(NOW()) + INTERVAL 14+6-WEEKDAY(NOW()) DAY
It's much easier to read and understand what the query is doing when we reference bare columns from the table on one side, and do the gyration and manipulation on the other side. It's also easier to test. And, maybe more importantly, it allows MySQL to make effective use of an "range scan" operation on a suitable index, rather than having to evaluate a function on every row in the table.
If today is '2015-05-21', and what you meant by "two weeks in advance" was up until the second Sunday from now (Sunday, May 31, 2015... 1 week and 3 days from now) just replace the 14 with 7.
Again, it really depends on how you define "two weeks in advance", how you determine what that ending date boundary is. Once you can define that, you can write an expression that returns the value, and just compare to value stored in the starttime column.
I'm having troubles selecting future sessions were the date is 21 days from now. So not between now and 21 days but ONLY sessions that will take place 21 days from this day.
In my table dx_sessions_dates I have a field timestart of type BIGINT where a timestamp is saved (why BIGINT and not TIMESTAMP? -> Not my DB, but can't change it ... ).
My SQL Query is :
SELECT timestart, timefinish, sessionid FROM `dx_sessions_dates` WHERE timestart = UNIX_TIMESTAMP(DATE_ADD(NOW(), INTERVAL 21 DAY))
As you can see I want to select all the sessions where timestart is 21 days from now. 21 days from now should be 15 april 2015.
The query always returns 0 rows ... . While in my table I have a timestart with value = 1429081200 . And when you calculate the date with this you see it's 15 april 2015. Why don't I get any rows back?
The unixtimestamp you calculate is never exactly the moment of the value you stored. You just need the same day.
Try
SELECT timestart, timefinish, sessionid
FROM `dx_sessions_dates`
WHERE date(FROM_UNIXTIME(timestart)) = curdate() + interval 21 DAY
Try using BETWEEN to find any rows that have timestart values anywhere within the 24-hour period that is 21 days from today:
SELECT timestart, timefinish, sessionid
FROM dx_sessions_dates
WHERE timestart BETWEEN UNIX_TIMESTAMP(adddate( curdate(), 21)) AND UNIX_TIMESTAMP(adddate(curdate(), 22))
curdate() returns midnight at the start of today, so 21 days from today is up to 22 days from midnight.
I am currently trying to write a little program to track time-off requests for employees. I'm fairly new to MYSQL and PHP, so it's a learning project for me as well. I've run into this problem which I do not seem to be able to figure out.
I want to display time off requests for a given week (Mon-Fri). I've got the requests in a table 'requests', with a 'Starttime' and 'Endtime' in separate fields, both Date/Time.
I can currently easily search and retrieve requests that have either (or both) Starttime or Endtime values that fall within the given ISO week I am looking at (WEEKOFYEAR() ).
What I need to be able to do is search for requests that may include days in the ISO week I am displaying, but not have a Starttime or Endtime during that week.
Example:
Employee takes off Tuesday of Week 24 through Friday of Week 24.
Currently, I would correctly display that the employee was off starting Tuesday and show a return on that Friday, but on Wed and Thursday nothing would be entered.
Employee takes off Friday of Week 30 through Monday of Week 32.
Currently, I would show that employee as not being 'off' during Week 31 because the search would not show a Starttime or Endtime during that week even though they are actually off the entire week. Though the Starttime and Endtime would be noted on the correct days.
Right now, what I do to work around this is run 5 additional queries to check if the date for each day Mon-Fri during Week 31 is contained BETWEEN the Starttime and Endtime of each request in the db.
I hate to run a total of 6 queries to get this information. Is there an easier way to get that information?
I just wrote a calendar app for events and ran into this- how bout something like this:
SELECT * FROM Requests WHERE
Start_Date BETWEEN <first_day_of_week> AND <last_day_of_week>
OR
End_Date BETWEEN <first_day_of_week> AND <last_day_of_week>
OR
<day_of_week_monday> BETWEEN Start_Date AND End_Date
OR
<day_of_week_tuesday> BETWEEN Start_Date AND End_Date
OR
<day_of_week_wedenesday> BETWEEN Start_Date AND End_Date
OR
<day_of_week_thursday> BETWEEN Start_Date AND End_Date
OR
<day_of_week_friday> BETWEEN Start_Date AND End_Date
GROUP BY ID ORDER BY Start_Date ASC, Date ASC
While < value_names > are generated with php via the currently viewed week requested. Should cover your bases.
What format are these dates stored in the DB? Assuming they are using MySQL's DATE format, its pretty easy to do. You can just use comparison operators on the fields and MySQL will do the work for you in one query.
$sql = "SELECT * FROM table WHERE startdate <= $someday AND $someday <= enddate";
Sticking to your week of year way of doing it, you can run a check to see if the current week falls within the range of the starting week off and the ending week off.
SELECT * FROM table WHERE WEEKOFYEAR(Starttime) <= N AND WEEKOFYEAR(Endtime) >= N;
(where N is the week you're displaying)
What you'd want to do, once you get the rows, is parse each day in PHP to see if that day falls between the starttime and endtime.
A good method for that is using:
$start_timestamp = strtotime($row['Starttime']);
$end_timestamp = strtotime($row['Endtime']);
You can use a similar method to get a timestamp of the day you are displaying, and see if it falls between $start_timestamp and $end_timestamp to determine if that day is off.
since you mentioned PHP as the display:
$start=30; //Friday of Week 30
$end=32; //Monday of Week 32
foreach (range($start, $end) as $number) {
echo $number;
}
You would expect to get 30,31,32.
You will have to verify that $start is < $end though as well for December to January weeks.
The range for WEEKOFYEAR() is 1-53. In this case, add 53 to the $end and display mod of $end if greater than 53:
$start=52; //Friday of Week 52, 2009
$end=2; //Monday of Week 2, 2010
If($start>$end) $end+=53;
foreach (range($start, $end) as $number) {
if($number>53){
echo $number%53;}
else{
echo $number;
}
}
You would expect to get 52,53,1,2
Probably need the query to read either like the big one with all the ORs or something like this
WHERE
Starttime <= <value for last day/time of the week>
AND
Endtime >= <value for the first day/time of the week>
That should get all the dates you need that could fall in that week, then you parse the records you get with PHP to find the ones for that actual week... though there may be an easier way to do that.
I need to calculate a difference between a starting date/time and an ending date/time. But, I only want to do this for the 5-day work week (exclude Sat/Sun as days). What is the best way to do this? My thought is that from the date, I'll have to get the day of the week and if it is a weekday, then I add to the accumulator. If it's not, then I don't add anything.
I'm sure someone has done this before, but I couldn't seem to find anything searching. Any links or other help would be very useful.
Many thanks,
Bruce
DAYOFWEEK returns 1 for Sunday and 7 for Saturday. I'm not sure how your schema is set up, but this will perform a TIMEDIFF of two dates that are on a Monday - Friday work week.
select TIMEDIFF(date1,date2) from table
where DAYOFWEEK(date1) not in (1,7) and DAYOFWEEK(date2) not in (1,7)
MySQL DATE/TIME functions
EDIT: From Bruce's comment about holidays. If you have a table full of holiday dates, something like this would work to exclude processing those days:
select TIMEDIFF(date1,date2) from table
where date1 not in (select holiday from holiday_table) and
date2 not in (select holiday from holiday_table) and
DAYOFWEEK(date1) not in (1,7) and DAYOFWEEK(date2) not in (1,7)
NETWORKDAYS() "Returns the number of whole working days between start_date and end_date. Working days exclude weekends and any dates identified in holidays. Use NETWORKDAYS to calculate employee benefits that accrue based on the number of days worked during a specific term." according to the Excel 2007 help file.
The "between" description is a bit inaccurate because it includes the start and end dates, i.e. networkdays(21-01-2010. 22-01-2010) = 2. It also takes no account of times.
Here's a function in PHP that will give the same results. It doesn't work properly if the end date is less than the start date, nor does do anything about holidays (see below the function).
function networkdays($startdate, $enddate)
{
$start_array = getdate(strtotime($startdate));
$end_array = getdate(strtotime($enddate));
// Make appropriate Sundays
$start_sunday = mktime(0, 0, 0, $start_array[mon], $start_array[mday]+(7-$start_array[wday]),$start_array[year]);
$end_sunday = mktime(0, 0, 0, $end_array[mon], $end_array[mday]- $end_array[wday],$end_array[year]);
// Calculate days in the whole weeks
$week_diff = $end_sunday - $start_sunday;
$number_of_weeks = round($week_diff /604800); // 60 seconds * 60 minutes * 24 hours * 7 days = 1 week in seconds
$days_in_whole_weeks = $number_of_weeks * 5;
//Calculate extra days at start and end
//[wday] is 0 (Sunday) to 7 (Saturday)
$days_at_start = 6 - $start_array[wday];
$days_at_end = $end_array[wday];
$total_days = $days_in_whole_weeks + $days_at_start + $days_at_end;
return $total_days;
}
To take holidays into account, you'd have to work out the number of days using this function, then use a query like
Select count (holiday_date) from holidays
where holiday_date between start_date and end_date
and DAYOFWEEK(holiday_date) not in (1,7)
Be careful that there isn't a problem with the end_date being treated as 00:00 (i.e. first thing in the morning) - you may have to condition it to be 23:59:59 so that it works properly. It all depends on how your holidays are stored.
To return the holidays in the same time period and subtract that from the number you first thought of.