MYSQL select data that is older then 2 years - php

So I want to purge some data from the system which is older then 2 years.
I have a datetime field where I store.
I know I can do something like
SELECT * FROM `table` where date_field < (??? what goes here)
My problem is i think how do I calculate the date, or maybe I am confused

For a specific answer in case the Manual isn't clear enough:
SELECT * FROM `table` where date_field < (now() - interval 2 year)

Use DATE_SUB
SELECT * FROM `table`
where date_field < DATE_SUB(CURDATE(), INTERVAL 2 YEAR)

may this can help :
SELECT EXTRACT(YEAR FROM `date`) AS year
from table
having year< (EXTRACT(YEAR from CURDATE())-2)

You can use INTERVAL:
SELECT * FROM `table` WHERE date_field < DATE_SUB(CURDATE(),INTERVAL 2 YEAR);
Referenced from https://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html.

SELECT * FROM table where date_field < DATE_SUB(NOW(),INTERVAL 2 YEAR);

Related

MySQL retrieve data for the past 24 hours when the column table is of type date NOT timestamp

I am trying to retrieve data for the passed 24 hours but I am getting nothing, but when I check on the table column (days) directly in db, I can find them. Here is my query:
select * from mytable
where
days between concat(date(date_sub(now(), interval 1 day))
between requires 2 arguments that specify the range:
select * from mytable
where days between date_sub(now(), interval 1 day) and now();
Or simply
select * from mytable
where days > date_sub(now(), interval 1 day);
You can use
select * from mytable where days > date_sub(CURDATE(),INTERVAL 1 DAY);

Get data from database inserted in latest 7 days

I am in problem in mysql database data retrieving. I have to get latest data inserted within a week or latest 7 days. I just know get data of specific date, but no within a span of days.
Please anyone help me. I am new in mysql.
You are looking for INTERVAL. For example, this will find all users whose created_time is in last 7 days and you have field created_time to tracked date of creation of record
SELECT * from users where created_time > (NOW()-INTERVAL 7 DAY)
You can do it using below query.
SELECT *
FROM `table`
WHERE `date` BETWEEN DATE_SUB(CURDATE(), INTERVAL 7 DAY) AND
CURDATE()
OR
SELECT * FROM `table` WHERE `date` >= DATE_SUB(CURDATE(), INTERVAL 7 DAY)
OR
SELECT * FROM table WHERE DATEDIFF(NOW(),dateField) < 7
we can use DATE_SUB Mysql default function
select * from yourtable where date_of_insertion >= DATE_SUB(NOW(),INTERVAL 7 DAY)
<?php
$date = date("your_date_format", strtotime(- 7 days));
$result = mysqli_query($link, "SELECT * FROM table WHERE `date` > '$date'");
?>
Simple as that.
try :
select * from tablename where add_time >= now() - interval 7 day

select from table where day difference is less than some number

I am trying to grab all SQL entries WHERE Day difference between current date and expiration date is < than 4 days
My first approach was following:
$sql_i_requested = "SELECT *, (To_days(date_return)-TO_DAYS(NOW())) as daydif FROM ".$tbl_name."
WHERE (status!='completed' AND status!='canceled')
AND owner_id=".$owner_id."
AND daydif < 4
ORDER BY date_created DESC";
My second aproach is (according to SQL DateDifference in a where clause):
$sql_i_requested = "SELECT * FROM ".$tbl_name."
WHERE (status!='completed' AND status!='canceled')
AND owner_id=".$owner_id."
AND date_return > DateAdd(day, -3, getdate())
ORDER BY date_created DESC";
Neither of them work, so how do I select FROM table WHERE day_difference between "date_return" and now() is less than 4 days?
EDIT:
changed
AND daydif < 4
to
AND (To_days(date_return)-TO_DAYS(NOW())) < 4
and now it's working. Anyway, maybe you guys could suggest other solutions.
Using the DATEDIFF
WHERE DATEDIFF(date_return, now()) < 4
try:
SELECT DATEDIFF(date_return, NOW()) AS dayDiff;
and
having dayDiff < 4
Try this:
select *
from table
where DATEDIFF(day, present, future) < 4

Get one day old record from database

How can i get one day old record from database from server . i have use datetime to insert record. below is how record is look like.
2013-01-15 23:44:02
i have use strtotime('-1 day') but it returns local system time.i want to get one day old record and do some stuff in condition..
Thanx in advance..
where date(date_column) = DATE_sub(NOW(), INTERVAL 1 DAY);
use DATE_ADD
SELECT *
FROM tablename
WHERE DATE(dateCol) = DATE_ADD(CURDATE(),INTERVAL -1 DAY)
DATE_ADD()
SELECT * from table WHERE datefield < DATE_SUB(NOW(), INTERVAL 1 DAY)
You will be archiving this with sql query.
SELECT * FROM tablename WHERE DATE(datefield)=DATE_ADD(CURDATE(),INTERVAL 1 DAY)
CURDATE() will return the current Mysql Server time.
You can use this function DATE_SUB(date,INTERVAL expr type)
http://www.w3schools.com/sql/func_date_sub.asp
Example provided in W3schools
Assume we have the following "Orders" table:
Columns: OrderId, ProductName, OrderDate
Values: 1, Jarlsberg Cheese, 2008-11-11 13:23:44.657
Now we want to subtract 5 days from the "OrderDate" date.
We use the following SELECT statement:
SELECT OrderId,DATE_SUB(OrderDate,INTERVAL 5 DAY) AS SubtractDate
FROM Orders
$yesterday = strtotime('Y-m-d h:i:s','yesterday');
SELECT * FROM tablename WHERE datefield = '$yesterday'

SQL time period query

I need to select from a table all rows that have date_added between CURDATE() and 6 weeks ago.
Help please.
SELECT *
FROM mytable
WHERE date_added BETWEEN CURDATE() - INTERVAL 6 WEEK AND CURDATE()
SELECT *
FROM a_table
WHERE date_added BETWEEN DATE_SUB(CURDATE(), INTERVAL 6 WEEK) AND CURDATE()
SELECT * FROM table_name WHERE TO_DAYS(NOW()) - TO_DAYS(date_added) <= 42
SELECT * FROM table_name WHERE DATEDIFF(NOW(),date_added)<=42
if you use dates in one year
select date_format(date, '%u') from tab
where (date_format(date, '%u')-date_format(now(), '%u'))>6
if you use dates with different years
you don't have to use dates with another year.
you can use
select
date_format(date, '%u') from tab
where (date_format(date, '%u')-date_format(now(), '%u'))>6
and
date_format(date, '%u') from tab
where (date_format(date, '%Y')-date_format(now(), '%Y'))=0
you can optimize query with join if you want. I think you know how to do it

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