I have a table with next columns:
ID | Date | Hours |
1 20-05-2013 8:00
2 20-05-2013 2:00
I want to SUM the duplicate's hours if have the same date.
$sql2=mysql_query("SELECT count(distinct data_raport), SUM(ore) as TOTAL FROM sohy_works WHERE data_raport BETWEEN '".$_GET['data']."' AND '".$_GET['data-2']."' AND numes LIKE '%".$_GET['numes']."%' AND ore <= '8:00' GROUP BY data_raport ");
while ($row2=mysql_fetch_array($sql2)) {
$total = $row2[0];
echo "8 hours/day: " . $total;
}
I want to print total of days with 8 hours worked. conclusion it's must to be 0 worked days with 8 hours because the raport have the same date (same day)
Try this:
SELECT Date, SUM(Hours) as TOTAL
FROM your_table
GROUP BY Date;
Edit:
Hard to understand what you really want but this may help:
SELECT COUNT(*) as 8_Hours_day
FROM (
SELECT Date, SUM(Hours) as Total_Hours
FROM your_table
GROUP BY Date
HAVING Total_Hours = 8
) T;
SELECT SUM(Hours) from Table group by Date
Related
I have a question with mysql.
I have a table like records and records have a field registered_date which contains the dates. I want to select last n months but not before the current date. I must select the last date that registered_date contains.
registered_date
2015-05-30
2015-05-29
2015-05-28
2015-05-27
...
and we are in july. The last date is 2015-05-30. I want to select last 3 months before the 2015-05-30.
I tried to like these:
.... where registered_date > DATE_SUB(now(), INTERVAL 6 MONTH)
and
registered_Date between now() - interval 30 day and now()
Thank you.
You need to find the biggest date and then select three months before that:
select t.*
from table t cross join
(select max(registered_date) as maxrd from table t) m
where t.registered_date >= maxrd - interval 3 month;
Did you try like this
SELECT * FROM table WHERE registered_date >= now()-interval 3 month;
I have a mysql table with the following fields
Name | Email | Date | Status
I want to extract the records where date range is between 30 days
Assume today is 2014/12/9
ie. date values are
2014/11/25
2014/12/2
2014/12/1
2014/10/25
2014/11/9
I need the o/p as (the number of days should be with in 30 days from the db date to today date)
2014/11/25
2014/12/2
2014/12/1
2014/11/9
I want to extract records those have the interval of less than 30 days from the date in the db.
Yes. I want to fetch the record between 2 days. For this I used this query
SELECT * FROM tbl_jobboard WHERE dtDate <= ( dtDate +30 )
But it is not working.
How to write the select query?
USE DATE_SUB like this:
SELECT * FROM table1
WHERE `date` BETWEEN DATE_SUB(CURDATE(),INTERVAL 30 DAY) AND CURDATE()
Working Fiddle Demo: http://sqlfiddle.com/#!2/6344f2/1
use following query
select * from table_Name t where t.date<=now() and t.date>=DATE_SUB(now(),
INTERVAL 31 DAY)
This has me a stummped...
If I have a this MySQL table:
UserId | Commission | Date Of Commission
1 | 200.00 | 2014-02-12
1 | 50.00 | 2014-04-01
2 | 10.00 | 2014-04-05
and I would like to display the Total Commission for a specific user per week starting from his/her first record, and display 0 for that range if there's no record.
how would I go about it?
Sample Output
UserId | Date Range | Total Commission
1 | 02/10/14 - 02/16/14 | 200.00
1 | 02/17/14 - 02/23/14 | 0.00
...
1 | 03/31/14 - 04/06/14 | 50.00
I'm not a seasoned coder so any help will be much appreciated.
Thanks!
Edit:
I have tried this:
SELECT IFNULL(SUM(Commisssion),0) Total ,DATE_SUB(`DateOfCommission`,INTERVAL 7 DAY)
AS RangStart,DATE_SUB(`DateOfCommission`,INTERVAL 1 DAY) AS RangeEnd
FROM `comms` WHERE `UserId` = '$UserID' GROUP BY DATE(`DateOfCommission`) DESC
but it starts the week with whatever date the first record was entered..
This is very tricky to accomplish. Here is what I managed to do with small modifications it should work they way it needs to be. I have done it for userid = 1 and this could be done for other users as well.
In the query I have 2 lines
where a.Date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
and
WHERE date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
The query will try to generate the list of dates using the min() date of transaction for the user till today. Instead of now() this could be used as max() date of transaction for the user as well.
select
t1.date_range,
coalesce(SUM(t1.Commission+t2.Commission), 0) AS Commission
from
(
select
a.Date as date,
concat(
DATE_ADD(a.Date, INTERVAL(1-DAYOFWEEK(a.Date)) +1 DAY),
' - ',
DATE_ADD(a.Date, INTERVAL(7- DAYOFWEEK(a.Date)) +1 DAY)
) as date_range,
'0' as Commission
from (
select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as date
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 union all select 8 union all select 9) as a
cross join (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 union all select 8 union all select 9) as b
cross join (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 union all select 8 union all select 9) as c
) a
where a.Date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
)t1
left join
(
SELECT date ,
coalesce(SUM(Commission), 0) AS Commission
FROM transactions
WHERE date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
AND UserId = 1
GROUP BY date
)t2
on t2.date = t1.date
group by t1.date_range
order by t1.date_range asc
DEMO
So, this is sort of an algorithm you could use:
$Result = select distinct userid from table(this will fetch all userids from table)
while(There are rows in $Result)
{
$Userid = $Result['userid']
$StartDateRes = mysql_query(select Date, WEEKOFYEAR(Date) as week from table where userid = Userid order by date asc limit 1)
$StartDateRow = mysql_fetch_assoc($StartDateRes)
$StartDate = $StartDateRes['Date']
$StartWeekNumber = $StartDateRes['week']
$EndDateRes = mysql_query(select Date, WEEKOFYEAR(Date) as week from table where userid = Userid order by date desc limit 1)
$EndDateRow = mysql_fetch_assoc($EndDateRes)
$EndDate = $EndDateRes['Date']
$EndWeekNumber = $EndWeekRes['week']
for($i=$StartWeekNumber; $i<=$EndWeekNumber; $i++)
{
$StartDateOfWeek = FuncToFindStartDateOfWeek($i)
$EndDateOfWeek = FuncToFindEndDateOfWeek($i)
$Result2 = mysql_query(select sum(commission) as sum from table where date between StartDateOfWeek and EndDateOfWeek group by userid)
$Row2= mysql_fetch_assoc($Result2)
$Sum = $Row2['sum']
mysql_query("insert into OutputTable values($UserId, $StartDateOfWeek. '-'. $EndDateOfWeek ,$Sum");
}
}
SELECT UserId, COALESCE(SUM(Commission),0), YEARWEEK(DateOfCommission) AS TheWeek
GROUP BY UserId, TheWeek
ORDER BY UserId, TheWeek;
This will not print the nice date range, but should get you started in a SQL-only direction where the sum is broken down by the week of the year. I think you could take it from this point to add the nicer formatting of the Year/Week column. YEARWEEK() should give you pretty fast results.
The easiest way I can think of doing this as follows
Step 1: Get the date of the first record
"SELECT dateofcommission FROM comissionstable WHERE id='userid' ORDER BY dateofcommission ASC LIMIT 1"
The above query will return the first date of commission only
Step 2: Create a loop which starts from the date you got in Step 1 and continue the loop till the date is greater than or equal to today's date. Increment this date using PHP date function.
date('Y-m-d', strtotime($dateofcommission. ' + 7 days'));
Step 3: In this loop you can get the commission with-in the starting date and ending date. Starting date will be the date before adding 7 days and ending date will be the one after you have added 7 days.
SELECT SUM(commission) FROM commissiontable WHERE dateofcommission>= startingdate AND dateofcomission < endingdate AND id='userid'
The above logic should work. If you end up having some issues with this logic feel free to post in comments. I would be happy to help
The following is another solution
function getStartAndEndDate($week, $year) {
$time = strtotime("1 January $year", time());
$day = date('w', $time);
$time += ((7*$week)+1-$day)*24*3600;
$return[0] = date('Y-n-j', $time);
$time += 6*24*3600;
$return[1] = date('Y-n-j', $time);
return $return;
}
$query = mysqli_query($con, "SELECT userid, COALESCE( SUM( commission ) , 0 ) AS thecommission , YEARWEEK( doc ) AS TheWeek FROM commission GROUP BY userid, TheWeek ORDER BY userid, TheWeek");
while ($array = mysqli_fetch_array($query)) {
$test = $array['TheWeek'];
$store_array = getStartAndEndDate(substr($test,4,2), substr($test,0,4));
echo $array['userid']."-".$array['thecommission']."-".$store_array[0]."/".$store_array[1]."<br>";
}
I have a scenario where I need to pull up delivery dates based on a table below (Example)
job_id | delivery_date
1 | 2013-01-12
2 | 2013-01-25
3 | 2013-02-15
What I'm trying to do is show the user all the delivery dates that start with the earliest (in this case it would be 2013-01-12) and add an another 21 days to that. Basically, the output I would expect it to show of course, the earliest date being the starting date 2013-01-12 and 2013-01-25. The dates past the February date are of no importance since they're not in my 21 date range. If it were a 5 day range, for example, then of course 2013-01-25 would not be included and only the earliest date would appear.
Here is main SQL clause I have which only shows jobs starting this year forward:
SELECT date, delivery_date
FROM `job_sheet`
WHERE print_status IS NULL
AND job_sheet.date>'2013-01-01'
Is it possible to accomplish this with 1 SQL query, or must I go with a mix of PHP as well?
You can use the following:
select *
from job_sheet
where print_status IS NULL
and delivery_date >= (select min(delivery_date)
from job_sheet)
and delivery_date <= (select date_add(min(delivery_date), interval 21 day)
from job_sheet)
See SQL Fiddle with Demo
If you are worried about the dates not being correct, if you use a query then it might be best to pass in the start date to your query, then add 21 days to get the end date. Similar to this:
set #a='2013-01-01';
select *
from job_sheet
where delivery_date >= #a
and delivery_date <= date_add(#a, interval 21 day)
See SQL Fiddle with Demo
SELECT date,
delivery_date
FROM job_sheet
WHERE print_status IS NULL
AND job_sheet.date BETWEEN (SELECT MIN(date) FROM job_sheet) AND
(SELECT MIN(date) FROM job_sheet) + INTERVAL 21 DAY
SELECT j.job_id
, j.delivery_date
FROM `job_sheet` j
JOIN ( SELECT MIN(d.delivery_date) AS earliest_date
FROM `job_sheet` d
WHERE d.delivery_date >= '2013-01-01'
) e
ON j.delivery_date >= e.earliest_date
AND j.delivery_date < DATE_ADD(e.earliest_date, INTERVAL 22 DAY)
AND j.print_status IS NULL
ORDER BY j.delivery_date
(The original query has a predicate on job_sheet.date; the query above references the d.delivery_date... change that if it is supposed to be referencing the date column instaed.)
If the intent is to only show delivery_date values from today forward, then change the literal '2013-01-01' to an expression that returns the current date, e.g. DATE(NOW())
Let's say you have a simple table with:
ID IP CREATED_DATE
-------------------------------------------------------
1 111.111.111.111 2011-03-28 14:12:31
2 111.111.111.111 2011-03-29 03:38:12
3 222.222.222.222 2011-04-02 12:04:45
4 111.111.111.111 2011-04-02 22:13:23
5 333.333.333.333 2011-04-03 05:53:15
6 222.222.222.222 2011-04-05 02:13:51
7 111.111.111.111 2011-04-07 11:45:34
I need to query the last 30 days and get a count of unique ips and total rows per day even if there are no entries for that day. So days that are not entered would obviously just be 0 as the array value for both unique and total rows.
If you have an integers table, you can generate all the days in your desired timespan and join those with the tallies of IPs for each day appearing in your table:
SELECT timespan."day",
COALESCE(num_uniq_ips, 0) AS num_uniq_ips,
COALESCE(num_records, 0) AS num_records
FROM (SELECT DATE_SUB(CURRENT_DATE, INTERVAL i DAY) AS "day"
FROM integers
WHERE i < 30) timespan
LEFT JOIN (SELECT DATE(created_date) AS "day",
COUNT(DISTINCT ip) AS num_uniq_ips,
COUNT(1) AS num_records
FROM so6025149
GROUP BY 1) tallies
ON timespan."day" = tallies."day"
ORDER BY timespan."day" ASC;
You can create a temporary table, populate it with the past 30 days and use that in your query. From PHP code:
// create table with past 30 days
$buffer = "CREATE TEMPORARY TABLE IF NOT EXIST past_month (val DATETIME NOT NULL );";
$now = new DateTime(now);
for($i=30;$i>0;$i++)
{
$buffer += "INSERT INTO past_month VALUES(" . $now . ");";
$now->sub(new DateInterval('P1D'));
}
mysql_query($buffer);
// the query
$sql= "SELECT
val,
(SELECT COUNT(IP)
FROM table
WHERE DATE_FORMAT(CREATED_DATE,'%Y-%m-%d) = DATE_FORMAT(val,'%Y-%m-%d') As UNIQUE_IPS
FROM
past_month";
$result = mysql_query($sql);
SELECT COUNT(DISTINCT column_name)
FROM table_name
WHERE ( DateTime > (GetDate()-30) )
something along those lines