I have table temp with structure on sqlfiddle:
id(int 11 primary key)
name(varchar 100)
name2(varchar 100)
date(datetime)
I would like get record on this week, example if now 21.11.2013 i would like all rows on 18.11.2013 to 24.11.2013(on week)
Now I see next algorithm:
obtain weekday
calculate how many days ago was Monday
calculate the date Monday
calculate future date Sunday
make a request on date
Tell me please, is exist a shorter algorithm (preferably in the query MySQL)?
ADD Question is: Why this query select record on date 17.11.2013(Sunday) - 23.11.2013(Saturday) and how get records on date 18.11.2013(Monday) - 24.11.2013(Sunday) ?
query:
select * from temp
where yearweek(`date`) = yearweek(curdate())
Thanks!
Use YEARWEEK():
SELECT *
FROM your_table
WHERE YEARWEEK(`date`, 1) = YEARWEEK(CURDATE(), 1)
Use YEARWEEK. If you use WEEKOFYEAR you will get records of previous years also.
SELECT id, name, date
FROM table
WHERE YEARWEEK(date)=YEARWEEK(NOW());
For selecting records of day, week and month use this way:
function my_func($time, $your_date) {
if ($time == 'today') {
$timeSQL = ' Date($your_date)= CURDATE()';
}
if ($time == 'week') {
$timeSQL = ' YEARWEEK($your_date)= YEARWEEK(CURDATE())';
}
if ($time == 'month') {
$timeSQL = ' Year($your_date)=Year(CURDATE()) AND Month(`your_date`)= Month(CURDATE())';
}
$Sql = "SELECT * FROM your_table WHERE ".$timeSQL
return $Result = $this->db->query($Sql)->result_array();
}
You can do it by following method
SELECT DATE_FORMAT("2017-06-15", "%U");
Get number of week (From 00 to 53 )
Where (Sunday 1st day and Monday last day of week)
May be useful for you.
example:
SELECT DISTINCT DATE_FORMAT('dates', '%U') AS weekdays FROM table_name;
The short solution would be this:
SELECT * FROM my_table WHERE DATE_FORMAT("2021-08-19 15:40:33", "%U") = WEEK(CURDATE());
Related
This question already has answers here:
How to convert a string to date in MySQL?
(5 answers)
Closed 2 years ago.
I am trying to compare the following:
SELECT g_m.user_id
, g_m.group_id
FROM Group_Members g_m
WHERE g_m.gender_id = 2
AND g_m.partner_gender_id = 1
AND g_m.birthday >= '01-01-1955'
AND g_m.birthday <= '12-31-2002'
AND g_m.user_id != 12
g_m.birthday in this case is '02-15-1998' which should show up, but this returns an empty array, because the date comparison does not seem to be accurate?
Here is the entire function and the dates are being passed from age minimum and age maximums brought from user.
var today = new Date();
var minYear = "01-01-" + (today.getFullYear() - userPref.age_max); //min year to start but oldest age
var maxYear = "12-31-" + (today.getFullYear() - userPref.age_min); //max year to end but youngest age
var qSelect = "SELECT g_m.user_id, g_m.group_id" +
" FROM Group_Members g_m WHERE g_m.gender_id = ? AND g_m.partner_gender_id = ?" +
" AND g_m.birthday >= STR_TO_DATE(?, '%m/%d/%Y') AND g_m.birthday <= STR_TO_DATE(?, '%m/%d/%Y')" +
" AND g_m.user_id != ?";
var qValues = [userPref.partner_gender_id, userObj.gender_id, minYear, maxYear, userObj.id];
Anyone know how to compare dates in a mysql query?
What is the data type of your dates ? You should declare them as "date", otherwise you won't be able to compare them.
With strings, '02-15-1998' < '03-15-1990'
With dates, your mysql request should be :
SELECT g_m.user_id, g_m.group_id FROM Group_Members g_m WHERE g_m.gender_id = 2 AND g_m.partner_gender_id = 1 AND g_m.birthday >= '1955-01-01' AND g_m.birthday <= '2002-12-31' AND g_m.user_id != 12
Sorry for my english, I'm french...
As the comments have already pointed out, you appear to be storing your dates as actual text. For a short term workaround, you may use STR_TO_DATE to convert your text dates to bona fide MySQL dates. Then, compare them against valid MySQL date literals:
SELECT
g_m.user_id,
g_m.group_id
FROM Group_Members g_m
WHERE
g_m.gender_id = 2 AND
g_m.partner_gender_id = 1 AND
STR_TO_DATE(g_m.birthday, '%m-%d-%Y') >= '1955-01-01' AND
STR_TO_DATE(g_m.birthday, '%m-%d-%Y') < '2003-01-01' AND
g_m.user_id != 12;
Longer term, you should make the birthday column datetime or timestamp.
Side note: I have rewritten the date range to include those born in the calendar years from 1955 to 2002, inclusive on both ends.
I am using PHP with MySQL and would like to select rows that have a booking time within 2 hours from now. How do I compare what is in my database with the NOW() MySQL function?
I have columns pickupDate in the format yyyy-mm-dd and pickupTime in the format HH:mm (24-hour). I have tried creating a query with NOW() which returns the a 12-hour time as HH:mm:ss e.g. 2019-05-24 07:54:06 . I can't figure out how to format this to 19:54, or if I should use a different function instead.
For example, if the current date and time is 24/05/19 19:54:06, I would like to select rows between 19:54 and 21:54 on this date.
My table structure is:
referenceNo VARCHAR(100)
pickupDate DATE
pickupTime VARCHAR(100)
You need to create a DATETIME compatible value out of your pickupDate and pickupTime (which you can do by CONCATing them together), then you can compare that with a time range from NOW() to 2 hours later:
SELECT *
FROM yourtable
WHERE CONCAT(pickupDate, ' ', pickupTime) BETWEEN NOW() AND NOW() + INTERVAL 2 HOUR
Demo on dbfiddle
To add two hours in php
$hoursnow = date('H:i');
$timestamp = strtotime(date('H:i')) + 60*60*2;
$plusTwohours = date('H:i', $timestamp);
And $PlusTwohours using this variable frame the query like below
Sql Query:
$sqlQuery = 'select * from foodorder where pickupDate=DATE(NOW()) AND pickupTime>='.$hoursnow.' and pickupTime<='.$plusTwohours;
$result = mysql_query($sqlQuery);
variable $result will have the values of query
For Second Scenario: Adding hours to end of the day May 24 23:30:00
This should be handle by two different date for same column pickupDate
$d = new DateTime('2011-01-01 23:30:30');
$startDate = $d->format('Y-m-d H:i:s'); // For testing purpose assigned manually
$starttime = date('H:i');
// Here Process start, storing end date by adding two hours
$enddate1 = strtotime($startDate) + 60*60*2;
$enddate = date('Y-m-d', $enddate1); // Extracting date alone
$endtime = date('H:i', $enddate1); // Extracting time alone
Have to compare start and end date for column pickupDate, here is the query
$sqlQuery = "select * from foodorder where pickupDate>=DATE(".$startDate.") AND pickupDate<=DATE(".$enddate.") AND pickupTime>='".$starttime."' AND pickupTime<='".$endtime."'";
$result = mysql_query($sqlQuery);
I have table temp with structure on sqlfiddle:
id(int 11 primary key)
name(varchar 100)
name2(varchar 100)
date(datetime)
I would like get record on this week, example if now 21.11.2013 i would like all rows on 18.11.2013 to 24.11.2013(on week)
Now I see next algorithm:
obtain weekday
calculate how many days ago was Monday
calculate the date Monday
calculate future date Sunday
make a request on date
Tell me please, is exist a shorter algorithm (preferably in the query MySQL)?
ADD Question is: Why this query select record on date 17.11.2013(Sunday) - 23.11.2013(Saturday) and how get records on date 18.11.2013(Monday) - 24.11.2013(Sunday) ?
query:
select * from temp
where yearweek(`date`) = yearweek(curdate())
Thanks!
Use YEARWEEK():
SELECT *
FROM your_table
WHERE YEARWEEK(`date`, 1) = YEARWEEK(CURDATE(), 1)
Use YEARWEEK. If you use WEEKOFYEAR you will get records of previous years also.
SELECT id, name, date
FROM table
WHERE YEARWEEK(date)=YEARWEEK(NOW());
For selecting records of day, week and month use this way:
function my_func($time, $your_date) {
if ($time == 'today') {
$timeSQL = ' Date($your_date)= CURDATE()';
}
if ($time == 'week') {
$timeSQL = ' YEARWEEK($your_date)= YEARWEEK(CURDATE())';
}
if ($time == 'month') {
$timeSQL = ' Year($your_date)=Year(CURDATE()) AND Month(`your_date`)= Month(CURDATE())';
}
$Sql = "SELECT * FROM your_table WHERE ".$timeSQL
return $Result = $this->db->query($Sql)->result_array();
}
You can do it by following method
SELECT DATE_FORMAT("2017-06-15", "%U");
Get number of week (From 00 to 53 )
Where (Sunday 1st day and Monday last day of week)
May be useful for you.
example:
SELECT DISTINCT DATE_FORMAT('dates', '%U') AS weekdays FROM table_name;
The short solution would be this:
SELECT * FROM my_table WHERE DATE_FORMAT("2021-08-19 15:40:33", "%U") = WEEK(CURDATE());
How can I group a query result by a field that is not saved in the database.
For example I want to group the result by duration which is came from subtraction of start time and end time.
here is how i find out the duration
date1= $row_TicketRS['CloseDate'];
$date2 = $row_TicketRS['OpenDate'];
$diff = abs(strtotime($date2) - strtotime($date1));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));
if ( $days > 0)
{
$time1 = $row_TicketRS['OpenTime'];
$time2= $row_TicketRS['CloseTime'];
$t1=($time1);
$t2=($time2);
$end=('14:30');
$start=('07:30');
$n = $end- $t1;
$n2 = $t2- $start;
$Hours2 = floor(($n+$n2)+(($days-1)*7));
echo $Hours2.' Hours';
but know i do not know how to add it to the query
here is my query
$strQuery = "SELECT count(`ticket`.TicketID) as TotOutput, department.`DeptName` FROM `ticket`, `user`, department where ticket.OwnerID = user.EmpNo and user.`DepartmentID` = department.`DepartmentID` and OpenDate between'".$DateFrom."' And '".$DateTo."'"
It'd be better to have details, but a derived table/inline view would allow you to group by a computed value:
SELECT x.duration,
COUNT(*)
FROM (SELECT t.col,
t.end_time - t.start_time AS duration
FROM YOUR_TABLE t) x
GROUP BY x.duration
How about adding that computed value to the query with an alias like this:
SELECT some_fields, end - start AS duration FROM table ORDER BY duration
dont put alias for hidden column , use directly
exmaple:
SELECT id, FLOOR(value/100)
FROM tbl_name
GROUP BY id, FLOOR(value/100);
Reference
MySQL permits expressions in GROUP BY
clauses, so the alias is unnecessary:
I need to retrieve data from within a specific date range.Anybody can help me to create a query to getting the information within date range 12-12-2009 to 12-15-2009 from a mysql table.(including 12 and 15)
SELECT * FROM foo WHERE timestamp BETWEEN "2009-12-12" AND "2009-12-15"
Use this function in php first
function ChangeDateforDB($inputdate) {
if($inputdate>0) {
$month = substr($inputdate,0,2);
$date = substr($inputdate,3,2);
$year = substr($inputdate,6,4);
$show = $year."-".$month."-".$date;
return $show;
}
}
After that you can use this in query like this in the checking condition,
checkdate >= '".ChangeDateforDB($fromdate)."' and checkdate <= '".ChangeDateforDB($todate)."'
Check this one, you can get the correct answer.
SELECT ... WHERE DATEDIFF('2009-12-15',yourdatefield) <= 3 ...