$today = CURDATE();
$result = mysqli_query($con,"SELECT * FROM Persons WHERE Day ='"$today"'");
The columns are:
Name
Day
Time
Reg
And a select * from Persons works fine.
You should be able to just specify your current date in the query (i.e. no need to calculate in PHP). This would also give you more consistant time handling in case web server and MySQL server have different timezones.
If Day is datetime or timestamp field use this:
SELECT * FROM Persons WHERE Day LIKE CONCAT(CURRENT_DATE(),'%')
If Day is date field use this:
SELECT * FROM Persons WHERE Day = CURRENT_DATE()
If day is a datetime, then use:
SELECT *
FROM Person
WHERE Day >= CURRENT_DATE and Day < CURRENT_DATE + interval 1 day;
The use of like for dates is bad practice, because it requires converting the date to a string. This prevents an index from being used.
If day has no time component, the above will work, but you can simplify it to:
SELECT *
FROM Person
WHERE Day = CURRENT_DATE;
You can also write:
SELECT *
FROM Person
WHERE date(Day) = CURRENT_DATE ;
(The parentheses on CURRENT_DATE are optional.)
$today = CURDATE();
$result = mysqli_query($con,"SELECT * FROM Persons WHERE Day LIKE '$today'");
Related
I have a varchar field contain a timestamp value, i want to get the tomorrow records.
This is my code as you can see:
$tomorrow = strtotime("1+ day");
SELECT * FROM tasks WHERE task_accomplish_date > $tomorrow
but the query is incorrect
thanks for the help in advance
Should be like this :
$tomorrow = strtotime("+1 day");
/* this will select all record before tomorrow*/
SELECT * FROM tasks WHERE task_accomplish_date < $tomorrow;
/* this will select all record after tomorrow*/
SELECT * FROM tasks WHERE task_accomplish_date > $tomorrow;
http://php.net/manual/en/function.strtotime.php
SELECT * FROM tasks WHERE task_accomplish_date > NOW() + INTERVAL 1 DAY
You should be able to do this all in the MySQL query
SELECT *
FROM `tasks`
WHERE DATE(`task_accomplish_date`) = DATE(NOW() + INTERVAL 1 DAY)
That converts your task_accomplish_date into a date value and looks for records where that date is equal to today + 1 day (tomorrow). That does not give you records from 2 days from today or beyond, just tomorrow. If you want tomorrow and all records beyond tomorrow you would use
SELECT *
FROM `tasks`
WHERE DATE(`task_accomplish_date`) >= DATE(NOW() + INTERVAL 1 DAY)
If you are storing a timestamp, these only care about the date part of the timestamp, not about the time part of the timestamp. If you care about the time part as well you can remove DATE() from both sides of the = in the WHERE clause
I do not understand how to get output from my sql table .
the table date is stored in int . for example date output from one row is like 1362157869 .
i want to show today orders in query :
php : $today = date("y-m-d", time());
query : SELECT * FROM test WHERE date = '$today'
but it didn't work . i also try this :
SELECT * FROM test WHERE date LIKE '$today'
Query without PHP var:
SELECT * FROM test WHERE date = DATE(NOW());
You are storing your date as a UNIXTIMESTAMP, so you have to convert it to DATETIME using FROM_UNIXTIME, and then you have to extract only the date part, ignoring the time, using the DATE() function:
SELECT * FROM test WHERE DATE(FROM_UNIXTIME(date)) = CURDATE()
Please see fiddle here. You can then extract yesterday records with something like this:
SELECT * FROM test WHERE DATE(FROM_UNIXTIME(date)) = CURDATE() - INTERVAL 1 DAY
Or a specific date with this:
SELECT * FROM test WHERE DATE(FROM_UNIXTIME(date)) = '2013-03-01'
To make use of an index
Assiming that your date column could contain not only the date part but also the time, you could also use this that will return all today records:
SELECT * FROM test WHERE date >= UNIX_TIMESTAMP(CURDATE())
and this for yesterday:
SELECT * FROM test WHERE date >= UNIX_TIMESTAMP(CURDATE() - INTERVAL 1 DAY)
AND date < UNIX_TIMESTAMP(CURDATE())
or this for a specific date:
SELECT * FROM test WHERE date >= UNIX_TIMESTAMP('2013-03-01')
AND date < UNIX_TIMESTAMP('2013-03-01' + INTERVAL 1 DAY)
Using PHP/MySQL
I'm trying to create a select statement that gets the data from the least day of the current week (I'm using it to show data on a certain player 'this week'). The week starts on Sunday. Sundays's data may not always exist therefore if the Sunday data isn't found then it would use the next earliest day found, Monday, Tuesday, etc.
My date column is named 'theDate' and the datatype is 'DATE'
The query would need to be something like:
SELECT *
FROM table_name
WHERE name = '$username'
AND [...theDate = earliest day of data found for the current week week]
LIMIT 1
It would return a single row of data.
This is a query I tried for getting the 'this week' data, It doesn't seem to work correctly on Sunday's it shows nothing:
SELECT *
FROM table_name
WHERE playerName = '$username'
AND YEARWEEK(theDate) = YEARWEEK(CURRENT_DATE)
ORDER BY theDate;
This is the query that I'm using to get 'this months' data and it works even if the first day of the months data is not found, it will use the earliest date of data found in the current month/year (this query works perfect for me):
SELECT *
FROM table_name
WHERE playerName = '$username'
AND theDate >= CAST( DATE_FORMAT( NOW(),'%Y-%m-01') AS DATE)
ORDER BY theDate
LIMIT 1
Without trying this, you probably need an inner query:
select *
from table_name tn
where tn.the_date =
(select min(the_date)
from table_name
where WEEKOFYEAR(the_date) = WEEKOFYEAR(CURDATE())
and YEAR(the_date) = YEAR(CURDATE()))
viz, give me the row(s) in the table with a date equal to the earliest date in the table in the current week and year.
Try this
SELECT * FROM table_name WHERE name = '$username'
AND your_data IS NOT NULL
AND WEEK(the_date,0 = WEEK(NOW(),0))
ORDER BY DATE_FORMAT(the_date,'%w') ASC
Try the following, replace YOUR_DATE with the date from the column you want (theDate):
SELECT ADDDATE(YOUR_DATE, INTERVAL 1-DAYOFWEEK(YOUR_DATE) DAY)
FirstDay from dual
Did you try:
SELECT ADDDATE(theDate , INTERVAL 1-DAYOFWEEK(theDate ) DAY) FirstDay
FROM table_name
WHERE playerName = '$username'
ORDER BY theDate DESC
LIMIT 1
I have the following relation in my schema:
Entries:
entryId(PK) auto_inc
date date
In order to count the total entries in the relation I use a query in my php like this:
$sql = mysql_query("SELECT COUNT(*) as Frequency FROM Entries WHERE date = '$date'");
My question is how can I count the number of entries for the CURRENT month..
You want a between query based on your date column.
WHERE date BETWEEN startdate AND enddate.
Between is equivalent to date >= startdate AND date <= enddate. It would of course be also possible to just use >= AND < explicitly which would simplify it a bit because you don't need to find the last day of the month, but just the first day of the following month using only DATE_ADD(..., INTERVAL 1 MONTH).
However startdate and enddate in this case would be derived from CURDATE().
You can use CURDATE(), MONTH(), DATE_ADD and STR_TO_DATE to derive the dates you need (1st day of current month, last day of current month). This article solves a similar problem and all the techniques needed are shown in examples that you should be able to adapt:
http://www.gizmola.com/blog/archives/107-Calculate-a-persons-age-in-a-MySQL-query.html
The first day of the current month is obvious YEAR-MONTH(CURDATE())-01. The last day you can calculate by using DATE_ADD to add 1 Month to the first day of the current month, then DATE_ADD -1 Days.
update-
Ok, I went and formulated the full query. Don't think str_to_date is really needed to get the index efficiency but didn't actually check.
SELECT count(*)
FROM entries
WHERE `date` BETWEEN
CONCAT(YEAR(CURDATE()), '-', MONTH(CURDATE()), '-', '01')
AND
DATE_ADD(DATE_ADD(CONCAT(YEAR(CURDATE()), '-', MONTH(CURDATE()), '-', '01'), INTERVAL 1 MONTH), INTERVAL -1 DAY);
Try this
SELECT COUNT(1) AS `Frequency`
FROM `Entries`
WHERE EXTRACT(YEAR_MONTH FROM `date`) = EXTRACT(YEAR_MONTH FROM CURDATE())
See EXTRACT() and CURDATE()
Edit: Changed NOW() to CURDATE() as it is more appropriate here
Try
$sql = mysql_query("SELECT COUNT(*) as Frequency FROM Entries WHERE MONTH(date) = MONTH(NOW()) );
I have a list of unix timestamps in a database, and I wanting to select the ones that are from today.
i.e If today is Tueday, I want to get all the timestamps that were made today? Is it possible? Is there such a things as strtotime("Today")?
Any help would be great
you can use mktime() to generate the timestamp for the start of the day and then find the database entries with a timestamp greater than that.
$start = strtotime(date('Y-m-d 00:00:00')); // Current date, at midnight
$end = strtotime(date('Y-m-d 23:59:59')); // Current date, at 11:59:59 PM
then, you can just select where the timestamp is between the above 2 timestamps:
"SELECT FROM `foo` WHERE `timestamp` BETWEEN '{$start}' and '{$end}'"
You can convert the unix timestamps to sql dates in the SQL using FROM_UNIXTIME(), then compare those to NOW()
SELECT * FROM `tablename` WHERE DATE(FROM_UNIXTIME(`dateFld`)) = DATE(NOW());
Check if DAY(NOW()) and MONTH(NOW()) and YEAR(NOW()) is equal to appropriate value of DAY(timestamp) and MONTH(timestamp) and YEAR(timestamp).
select timestamp from table where DAY(NOW()) = DAY(timestamp) AND MONTH(NOW()) = MONTH(timestamp) AND YEAR(NOW()) = YEAR(timestamp)
If you're using mysql:
SELECT * FROM table WHERE DATE(NOW()) = DATE(FROM_UNIXTIME(timestampcol))
FROM_UNIXTIME(somefield) can be compared to CURDATE() assuming you're using MySQL
SELECT * FROM mytable WHERE FROM_UNIXTIME(datefield,'%Y-%m-%d') = CURDATE();
ETA:
Okay, I was assailed by doubt when this answer was marked down. So I went and did a couple of tests. Given MySQL it definitely works. So why the downmod?
Consider this test which outputs 2 identical fields for every row in a table:
SELECT FROM_UNIXTIME(UNIX_TIMESTAMP(CURDATE()),'%Y-%m-%d') a , CURDATE() b
FROM tablewithsomerows
WHERE FROM_UNIXTIME(UNIX_TIMESTAMP(CURDATE()),'%Y-%m-%d') = CURDATE();