I need to only SELECT data where the date field is yesterdays date. The only problem I'm having is that the data in the date field looks like the following 20160412 062815.000
I don't really care about the time, I just want to search dynamically for anything with yesterdays date. I've tried a multitude of CURDATE() -1 but I'm unsure how to just search the first 8 digits of the field.
Assuming the date value is stored as a string, and the first 8 characters are always the date in YYYYMMDD format, then you can use a query like this:
select *
from your_table
where your_column like concat(date_format(current_date() - interval 1 day,'%Y%m%d'),'%')
One advantage of this query is that it can leverage an index on your date field, unlike the other answers so far.
Format yesterday's date to a number and convert the date string also to a number.
select * from your_table
where date_format(curdate() - interval 1 day, '%Y%m%d') * 1 = date_col * 1
SQLFiddle demo
*1 is a math operation that forces MySQL to convert strings to a number.
you can use subdate(currentDate, 1)
select * from your_table
where subdate(currentDate, 1) = DATE(your_date)
Related
I have a table that has dates stored as text and not as a date_time. I want to select rows that happened today. I can hard code the date, but obvious I want to do it dynamically.
SELECT * FROM view WHERE anniversary LIKE '%09-07'
SELECT * FROM view WHERE anniversary LIKE '%' + DATE_FORMAT(now(), "%m-%d")
First, you should be storing dates as dates, not as strings.
But, if they are stored as strings, you can use a comparison like this:
where right(anniversary_datestring, 5) = str_to_date(curdate(), '%m-%d')
But, I recommend fixing your data. If the values are in the format YYYY-MM-DD, simply do:
alter table t modify column anniversary date;
The month and day operators can be used:
SELECT * from view WHERE MONTH(anniversary) = MONTH(NOW()) AND DAY(anniversary) = DAY(NOW())
if the column is of type varchar the STR_TO_DATE function can be used to turn it into a date:
SELECT * from view WHERE MONTH(STR_TO_DATE(anniversary,'%Y-%m-%d')) = MONTH(NOW()) AND DAY(STR_TO_DATE(anniversary,'%Y-%m-%d')) = DAY(NOW())
This should work,
convert string to date and check in where clause
SELECT * FROM view WHERE STR_TO_DATE(anniversary, '%d-%m-%Y') = CURDATE()
I have a column where a date store in ddmmyy format (e.g. 151216). How can I convert it to yyyy-mm-dd format (e.g 2016-12-15) for calculating a date difference from the current date? I try using DATE_FORMAT function but its not appropriate for this.
If you want to get the date difference, you can use to_days() after converting the string to a date using str_to_date():
select to_days(curdate()) - to_days(str_to_date(col, '%d%m%y'))
or datediff():
select datediff(curdate(), str_to_date(col, '%d%m%y'))
or timestampdiff():
select timestampdiff(day, str_to_date(col, '%d%m%y'), curdate())
You can use the function, STR_TO_DATE() for this.
STR_TO_DATE('151216', '%d%m%y')
A query would look something like:
select
foo.bar
from
foo
where
STR_TO_DATE(foo.baz, '%d%m%y') < CURDATE()
Note: Since both STR_TO_DATE() and CURDATE() return date objects, there's no reason to change the actual display format of the date. For this function, we just need to format it. If you wanted to display it in your query, you could use something like
DATE_FORMAT(STR_TO_DATE(foo.baz, '%d%m%y'), '%Y-%m-%d')
To get the difference, we can simply subtract time
select
to_days(CURDATE() - STR_TO_DATE(foo.baz, '%d%m%y')) as diff
from
foo
If you wanted to only select rows that have a difference of a specified amount, you can put the whole to_days(...) bit in your where clause.
SELECT STR_TO_DATE('151216', '%d%m%y') FROM `table`
use this '%d%m%y'
I want to use a Datepicker to select specific entries in my Database.
The Problem i face is that the entries i get as a reposnse from my query are only sorted for the day part of the date.
I use the date format (varchar field in database):
*dd.mm.yyyy hh:mm ($date = date("Y-m-d") . ' ' . date("H:i");)*
For example:
01.04.2016
10.04.2016
11.04.2016
01.04.2017
10.04.2017
11.04.2017
The Query to get the entries:
SELECT *
FROM order_date
WHERE date >= '10.04.2017' AND date <= '10.04.2017'
Now i expected to get only the to matching entries but
the result of the Query are all four table entries which start with day 10 or 11. Even two of them are in the year 2016.
As i saw in other posts the between XX and YY work if the date is set with yyyy-mm-dd or yyyymmdd but the problem is that i got a database with around 20k entries on which i need a datepicker so i would prefer to use the givin format of the date without rewriting all entries.
Is there a possibilty to get a datepicker between with dd.mm.yyyy format aswell?
Any help highly appriciated
Try something like this,
STR_TO_DATE('10.04.2017', '%d.%m.%Y')
SELECT *
FROM order_date
WHERE date >= STR_TO_DATE('10.04.2017', '%d.%m.%Y') AND date <= STR_TO_DATE('10.04.2017', '%d.%m.%Y')
You can typecast the column as DATE, and use the ORDER BY clause.
Use STR_TO_DATE function may help.
STR_TO_DATE('10.04.2017','%d.%m.%Y')
How about something around this idea ?
SELECT *
FROM Mytable WHERE date
BETWEEN CONVERT(datetime,'10.04.2017',104 )
AND CONVERT(datetime,'10.04.2017',104 )
I have a table with year data in a year datatype field, now I want to insert new data with date data and I want to convert the year from the old data to date with default month and day 1.1.YearFromOldData.
I'm looking for something like the function STR_TO_DATE but for the datatype year NOT VARCHAR NOT VARCHAR and I fail to find it. How would I do this?
I want to do this
SELECT YEAR_TO_DATE(myYearField, '%1/%1/%Y')
FROM myTable
Assuming that you have a varchar field named year_dt with old years, use the following query to get a date with default day and month
SELECT DATE(CONCAT(table.year_dt, '-01-01')) as 'date' FROM table
this will return date in default format i.e. YYYY-MM-DD
If you want the first day of the year, I think the easiest way is with makedate():
select makedate(year, 1)
You can get the timestamp to insert to the database using
timestamp = (year - 1970) / 31557600
STR_TO_DATE() is exactly what you need. Try this if you have four-digit years, in a string, an integer, or a YEAR column. It works for them all.
select str_to_date(CONCAT(year_column,'-01-01'), '%Y-%m-%d')
If you have two-digit years try this (lower case %y)
select str_to_date(CONCAT(year_column,'-01-01'), '%y-%m-%d')
This is cool because you can do all sorts of date arithmetic: for example
select str_to_date(CONCAT(year_column,'-01-01'), '%Y-%m-%d')
+ INTERVAL 1 QUARTER
- INTERVAL 1 DAY
will give you the last day of the first quarter of your year.
I am trying to select records from a database only if they match today's date. The format for the date in the database is 2012-06-20 9:30:00 I am using the statement SELECT id FROMnewsreportsWHERE DATE(newsdate) = CURDATE() but it doesn't not return any records for today?
Screenshot of column with dates
http://img135.imageshack.us/img135/8399/2347f03df0394cd898c7fc5.png
DATE_FORMAT(NOW(),'%Y-%m-%d') = FORMAT_DATE(NOW(newsdate), '%Y-%m-%d')
Or better:
DATE(newsdate) = DATE(NOW())
The best way is to store the additional column with 2001-09-11 date format and compare this one
Example, thanks to #Conrad Frix
It looks like the curdate function is going to give you something different than the date format you've got in the database.
Take a look at the docs here
MySQL Date and Time Functions
It's supposed to return a date in this format:
2008-11-11
So, you could either search for a date range between curdate() and curdate() + INTERVAL 1 DAY (untested), or store the dates in the curdate() format.