I have table where there are columns username, visit and timestamp.
I select user by name and visit. But the result of query must contain: username, visit, datetime.
How do I go from here:
select username, visit, timestamp
from Table
where username like "Jane" and visit like "cafe"
to the result with date?
Thanks a lot
You can convert the timestamp to a date using the DATE() function in SQL. The following query should give you the desired result:
select username, visit, DATE(timestamp) as datetime
from Table
where username like "Jane" and visit like "cafe"
This query selects the username, visit, and timestamp columns from the Table, and also includes an additional column, "datetime", which is the result of applying the DATE() function to the timestamp column. This function extracts the date part of the timestamp and formats it as a date.
If you want to format the datetime in a specific format you could use DATE_FORMAT() function like this:
select username, visit, DATE_FORMAT(timestamp, '%Y-%m-%d') as datetime
from Table
where username like "Jane" and visit like "cafe"
This will format the datetime to YYYY-MM-DD format.
Related
I wish to sort my table with date order so that recently added data will be on the top of the table.
I have used query for sorting as:
select date from register_table order by date desc.
Currently table display data as:
date
02.04.2019
05.04.2019
09.04.2019
10.04.2019
06.02.2019
23.01.2019
11.01.2019
I expect my table to display as:
date
10.04.2019
09.04.2019
05.04.2019
02.04.2019
06.02.2019
23.01.2019
11.01.2019
How to display data in date order?
Your fundamental problem is not storing the date as a date. You should fix that.
For the query to work, use:
order by str_to_date(date, '%m.%d.%Y')
To fix the data, you can do:
update register_table
set date = str_to_date(date, '%m.%d.%Y');
alter table register_table
modify date date;
You can see how this works here.
I don't know why your date is stored like that but here, give this a try:
SELECT date FROM date ORDER BY STR_TO_DATE(REPLACE(date,'.','-'),'%d-%m-%Y') DESC;
If you want to see what exactly happen, run this query:
SELECT date,STR_TO_DATE(REPLACE(date,'.','-'),'%d-%m-%Y') FROM date;
In case you still don't quite understand, refer to MySQL STR_TO_DATE function and MySQL REPLACE function.
I have a column where the dates are type varchar. For example:
15-10-2018
16-10-2018
19-10-2018
23-10-2018
29-10-2018
8-11-2018
9-11-2018
10-11-2018
12-11-2018
when I consult with the following query
SELECT DISTINCT date FROM `test` WHERE date BETWEEN '15-10-2018' and '9-11-2018'.
I have the right result.
15-10-2018
16-10-2018
19-10-2018
23-10-2018
29-10-2018
8-11-2018
9-11-2018
but if the query is:
SELECT DISTINCT date FROM `test` WHERE date BETWEEN '15-10-2018' and '10-11-2018'.
or
SELECT DISTINCT date FROM `test` WHERE date BETWEEN '15-10-2018' and '12-11-2018'.
The answer I get is empty.
I think it's only validating the days in the sql.
I need to get the right dates.
I think the problem is the fact that the column is varchar, so it's comparing characters instead of a range of dates. I will recommend convert the column to date type and try again.
Alternative if you cannot change the type of the column you could cast it to date format like this:
SELECT DISTINCT `date` FROM `test` WHERE STR_TO_DATE(`date`,'%d-%m-%Y') BETWEEN '2018-10-15' AND '2018-11-10';
I tested with your data and it works. Of course this could put some extra effort on the database and will not use indexes.
You need to set the datatype to date and update your dates to be using date for a more reliable result. Once done you should be using the database format for the dates in your WHERE clause.
Try
SELECT DISTINCT date FROMtestWHERE date BETWEEN '2018-10-15' and '2018-11-10'
I have a table with a few columns. One of the column is named flight_date is varchar format.
It has a date stored in it in format like '08/12/2015' which is mm/dd/yyyy
I want to run a query which will return me a list of the records with date in 2016.
Can someone help me out please.
Thanks.
In this case, you can use SUBSTR function for just cut last 4 character.
SELECT * FROM yourtable where SUBSTR(flight_date, -4) = '2016'
Though, it is recommended to store date value in appropriate type column and not in varchar.
I have a mysql column where the data is stored as VARCHAR though the data values are of datetime in the format of yyyy-mm-dd hh:mm:ss.
Now my task is to group by the date part i.e yyyy-mm-dd by converting VARCHAR to date-time and then just taking date part out of it
QUERY
SELECT SUM(value)
FROM table
GROUP BY name , [date part of the varchar field]
Please let me know if this is at all possible and if yes, how?
Assuming that your data in this varchar field is properly formatted, you can work with the left function, like this:
SELECT LEFT(mydate, 10) AS myval,
SUM(myvalue)
FROM mytable
GROUP BY myval;
If this isn't a big issue; I'd advise converting your varchar column to datetime or timestamp. If not only for the possibly better data storage usage, it'll be way easier to do work with date and time related functions.
Just use the left function. You can leave the date as a string:
SELECT left(datecol, 10) as YYYYMMDD, SUM(value)
FROM table
GROUP BY left(datecol, 10);
I removed name from the group by because it doesn't seem relevant to the question. You can, of course, add it back in.
By the way, MySQL understands this format for dates, so if you really, really want a date:
SELECT date(left(datecol, 10)) as RealDate, SUM(value)
FROM table
GROUP BY RealDate;
My query is this. I have a bunch of entries and i want to group them by date. But instead of having date in my database, I have a datetime field. What do I do?
select * from follow_queue group by follow_date cast follow_date as date
That's not working.
Use DATE() function:
select * from follow_queue group by DATE(follow_date)