I wonder if its possible to use my $today in WHERE query.
If $today is Monday, i want my WHERE to be p650Wdat = 1 (1 stands for monday)
And and tuesday will be 2 and so on.
$today = date("l");
<?php
$today = date("l");
//Establish connection to database
$host = "randomtext";
$conn = myacc");
$query = " Select *
from P660F
where p660WDAT = $today
ORDER BY P660PRIO DESC";
?>
OK so you want to match Monday to 1 and Tuesday to 2 etc. Really you should have same values in table as variables for simplicity and clarity butyou can return the day name as an integer like so
<?php
$day_name = date("l");
$today = date('N', strtotime($day_name));
echo 'Today name is ' . $day_name .'<br>';
echo 'Today number is ' . $today;
?>
Then your query should work
Use this in the where clause:
SQL:
WHERE DATENAME(dw, p660WDAT) = $today
MySQL:
WHERE DAYNAME(p660WDAT)= $today
And your function Date(l) will also return day name as Sunday, Monday... instead of 1,2,3...
Depends on your RDBMS you can use different ways:
MySQL:
WHERE WEEKDAY(p660WDAT)+1 = $today;
PostgreSQL:
WHERE EXTRACT(DOW FROM TIMESTAMP p660WDAT) = $today;
P.S. Of course, you should think about escaping values and etc. I didn't mention to give cleaner answer.
P.P.S. If I understood you properly, just use date('w') to get day of week.
<?php
$day_name = date("l");
$today = date('N', strtotime($day_name));
echo 'Today name is ' . $day_name .'<br>';
echo 'Today number is ' . $today;
$today = date("l");
//Establish connection to database
$host = "randomtext";
$conn = myacc");
$query = " Select *
from P660F
where p660WDAT = $today
ORDER BY P660PRIO DESC";
?>
Related
I'm trying to use one's date of birth to calculate when he'll be 50, if he's not 50 already.
If person is not 50, add a year to his age then check if it'll be 50. If not, iterate until it's true. Then get the date he turned 50. in PHP
Here's the code, not complete.
$rAge = 50;
$retir = date('j F Y ', strtotime("+30 days"));
$oneMonthAdded = strtotime(date("d-m-Y", strtotime($DOB)). "+1 year");
$re = date("d-m-Y", $oneMonthAdded);
$futDate = date("d-m-Y", strtotime(date("d-m-Y", strtotime($re))));
$date_diff = strtotime($futDate)-strtotime($DOB);
$future_age = floor(($date_diff)/(60*60*24*365));
Help please.
try this code bro!
// your date of birth
$dateOfBirth = '1950-11-26';
// date when he'll turn 50
$dateToFifty = date('Y-m-d', strtotime($dateOfBirth . '+50 Years'));
// current date
$currentDate = date('Y-m-d');
$result = 'retired';
// checks if already fifty
if($currentDate <= $dateToFifty) {
$result = $dateToFifty;
}
echo $result;
I use simplest php code to find out retire date. y
<?php
$dob = '1970-02-01';
$dob_ex = explode("-",$dob);
$age_diff = date_diff(date_create($dob), date_create('today'))->y;
$year_of_retire = 50 - $age_diff;
$end = date('Y', strtotime('+'.$year_of_retire.'years'));
$date_of_retire = $end."-".$dob_ex[1]."-".$dob_ex[2];
echo $date_of_retire;
?>
you can use if...else condition to echo values according to you.
like
if($year_of_retire > 0){
echo $date_of_retire;
} else if($year_of_retire < 0){
echo "retired";
}
I have a PHP script which records things based on the day. So it will have a weekly set of inputs you would enter.
I get the data correctly, but when i do $day ++; it will increment the day, going passed the end of the month without ticking the month.
example:
//12/29
//12/30
//12/31
//12/32
//12/33
Where it should look like
//12/29
//12/30
//12/31
//01/01
//01/02
My script is as follows:
$week = date ("Y-m-d", strtotime("last sunday"));
$day = $week;
$run = array(7); //this is actually defined in the data posted to the script, which is pretty much just getting the value of the array index for the query string.
foreach( $run as $key=>$value)
{
$num = $key + 1;
$items[] = "($num, $user, $value, 'run', '$day')";
echo "".$day;
$day ++;
}
Should I be manipulating the datetime differently for day incrementations?
You can use
$day = date("Y-m-d", strtotime($day . " +1 day"));
instead of
$day++;
See live demo in ideone
You refer to $day as a "datetime" but it is just a string - that is what date() returns. So when you do $day++ you are adding 1 to "2015-12-02". PHP will do everything it can to make "2015-12-02" into a number and then add 1 to it, which is not date math. Here is a simple example:
<?php
$name = "Fallenreaper1";
$name++;
echo $name
?>
This will output:
Fallenreaper2
This is how I would do it, using an appropriate data type (DateTime):
<?php
$day = new DateTime('last sunday');
$run = array(7);
foreach ($run as $key => $value) {
$num = $key + 1;
$dayStr = $day->format('Y-m-d');
$items[] = "($num, $user, $value, 'run', '$dayStr')";
echo $dayStr;
$day->modify('+1 day');
}
To increase time you should use strtotime("+1 day");
here is simple example of using it
<?php
$now_time = time();
for($i=1;$i<8;$i++) {
$now_time = strtotime("+1 day", $now_time);
echo date("Y-m-d", $now_time) . "<br>";
}
?>
I'm working on a PHP page that calculates the percentage of completion of a project. For example, if you had a start date of January 1st, 2015, and an end date of March 3rd, 2015, and today was February 2nd, 2015, the project would be estimated to be about 50% done. So far I've attempted using the DateTime class and the date_diff function, but I couldn't divide the two, so I'm back at square one. Obviously I need to take Daylight Saving and leap years into account, so that adds an additional layer of complexity to the matter. Any ideas? Here the current block.
try {
$dbh = new PDO('mysql:host=localhost; dbname=jkaufman_hartmanbaldwin', $username, $password, array(
PDO::MYSQL_ATTR_SSL_KEY => '../php_include/codekaufman_com.key',
PDO::MYSQL_ATTR_SSL_CERT => '../php_include/codekaufman_com.crt',
PDO::MYSQL_ATTR_SSL_CA => '../php_include/codekaufman_com.ca_bundle'
));
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$projectName = $_GET['project'];
$sth = $dbh->prepare('SELECT start, end FROM projects WHERE name = ?');
$sth->execute([$projectName]);
if($sth->rowCount()) {
$row = $sth->fetchAll(PDO::FETCH_ASSOC);
date_default_timezone_set('America/Los_Angeles');
$date = strtotime($row[0]['start']);
$start = date('m/d/Y', $date);
echo $start;
$date = strtotime($row[0]['end']);
$end = date('m/d/Y', $date);
echo " " . $end;
$today = date('m/d/y');
echo $end - $start;
}
} catch(PDOException $e) {
echo $e->getMessage();
}
With reference to How to Minus two dates in php:
$start = new DateTime($row[0]['start']);
$end = new DateTime($row[0]['end']);
$today = new DateTime();
$total = $start->diff($end);
$current = $start->diff($today);
$completion = $current->days / $total->days;
MySQL has some pretty easy to use functions for this sort of thing, you can just gather the info you need in your query:
SELECT start, end, DATEDIFF(end, start) as total_days, DATEDIFF(end, NOW()) as days_remaining
FROM projects WHERE name = ?
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff
From there you just need to divide days_remaining by total_days to get the percentage. Datediff should take leap years and DST into account.
You can use TIMEDIFF in place of DATEDIFF if you need to be more precise, just make sure to convert the sql timestamps to integers with strtotime.
You may also need to set the timezone:
SET time_zone = 'America/Los_Angeles';
The formula is:
percentage = (date - start) / (end - start) * 100
As a PHP function:
function date_progress($start, $end, $date = null) {
$date = $date ?: time();
return (($date - $start) / ($end - $start)) * 100;
}
Example:
$start = strtotime("January 1st 2015");
$end = strtotime("March 3rd 2015");
$date = strtotime("February 2nd 2015");
$percent = date_progress($start, $end, $date);
// "You are 52.46% there!"
echo 'You are ', round($percent, 2), '% there!';
Get your SQL output in the right format (YYYY-MM-DD) and then shove it into the code below:
<?php
$startDate = date_create('2015-01-01');
$endDate = date_create('2015-01-30');
$currentDate = date_create('2015-01-08');
$totalTime = date_diff($endDate, $startDate);
$elapsedTime = date_diff($currentDate, $startDate);
$totalTimeDays = $totalTime->format("%d");
$elapsedTimeDays = $elapsedTime->format("%d");
echo "Total project time = " . $totalTimeDays . "<br/>";
echo "Elapsed project time = " . $elapsedTimeDays . "<br/>";
echo "Percent of project complete = " . ($elapsedTimeDays / $totalTimeDays) * 100.0;
?>
$start = new DateTime("<YOUR START DATE>"); // example input "2014/06/30"
$end= new DateTime("<YOUR END DATE>");
$now = new DateTime();
$intervalOBJ = $start->diff($end);
$totalDaysOfProject = $intervalOBJ->format('%a');
$intervalOBJ_2 = $now->diff($end);
$daysRemaining = $intervalOBJ_2->format('%a');
$completedPercentage = round(($daysRemaining/$totalDaysOfProject)*100);
echo $completedPercentage . "% of this project has been completed!";
Description: This calculates the percentage of days remaining. interval = $start to $end. Calculated percentage is in relation to $now.
I am trying to create a dropbox that will display the last year, the current year and the next year using the php DateTime object.
In my current code I create three objects and have to call a method on 2 of them. This seems a bit heavy on the resources.
$today = new DateTime();
$last_year=new DateTime();
$last_year->sub(new DateInterval('P1Y'));
$next_year = new DateTime();
$next_year->add(new DateInterval('P1Y'));
echo date_format($last_year, 'Y').' '.date_format($today, 'Y').' '.date_format($next_year, 'Y');
another way I found to only use 1 object is
$today = new DateTime();
echo date_format($today->sub(new DateInterval('P1Y')), 'Y').' '.date_format($today->add(new DateInterval('P1Y')), 'Y').' '.date_format($today->add(new DateInterval('P1Y')), 'Y');
but that will become very confusing.
Can someone tell me a better way to do this using DateTime()? As I will need something similar for months ?
Depending upon your version of PHP (>= 5.4), you could tidy it up a bit like this:-
$today = new DateTime();
$last_year=(new DateTime())->sub(new DateInterval('P1Y'));
$next_year = (new DateTime())->add(new DateInterval('P1Y'));
echo $last_year->format('Y').' '.$today->format('Y').' '.$next_year->format('Y');
See it working.
A more readable and concise option may be to use \DateTimeImmutable.
$today = new DateTimeImmutable();
$one_year = new DateInterval('P1Y');
$last_year = $today->sub($one_year);
$next_year = $today->add($one_year);
echo $last_year->format('Y').' '.$today->format('Y').' '.$next_year->format('Y');
See it working.
Other than that, this all looks fine. Worry about optimisation when it is needed.
Try this. This quite efficient.
echo date("Y");
echo date("Y",strtotime("-1 year"));
echo date("Y",strtotime("+1 year"))
May be you can also limit the call of new DateInterval('P1Y') by creating one object and using it for all three calculations?
$interval = new DateInterval('P1Y');
$dateTime = new DateTime();
$lastYear = $dateTime->sub($interval)->format('Y');
$dateTime = new DateTime();
$nextYear = $dateTime->add($interval)->format('Y');
$dateTime = new DateTime();
$thisYear = $dateTime->format('Y');
echo $lastYear . ' ' . $thisYear . ' ' . $nextYear;
and by breaking the single string into multiple commands always helps me in reducing confusions.
<?php
$d = new DateTime('now');
$cy = $d->format('Y');
// Get previous year
$d->modify('-1 year');
$py = $d->format('Y');
//Next year : Since object has previous year, so +2 to get next year
$d->modify('+2 year');
$ny = $d->format('Y');
echo "Previous Year: ".$py."<br>";
echo "Current Year : ".$cy."<br>";
echo "Next Year : ".$ny."<br>";
$d = new DateTime('now');
$cm = $d->format('m');
$d->modify('-1 month');
$pm = $d->format('m');
$d->modify('+2 month');
$nm = $d->format('m');
echo "Previous Month: ".$pm."<br>";
echo "Current Month : ".$cm."<br>";
echo "Next Month : ".$nm."<br>";
?>
Output
Previous Year: 2013
Current Year : 2014
Next Year : 2015
Previous Month: 12
Current Month : 01
Next Month : 02
Just use basic math:
$current = date('Y');
$prev = $current - 1;
$next = $current + 1;
after reading all the answers I came up with this function to create a dropdown box for year.
$name is the name of the
$year_number is the number of years you want to display
this function starts with one year before the current year but you can easily modify it to start displaying from earlier years.
function drop_down_box_date_year(&$dbconn, $name, $year_number){
$today = new DateTime();
$interval = new DateInterval('P1Y');
$startyear = (new DateTime())->sub($interval);
echo '<select name="'.$name.'_year">';
for($i=0;$i<$year_number;$i++){
if($startyear->format('Y')==$today->format('Y')){
echo '<option value="'.$startyear->format('Y').'" selected>'.$startyear->format('Y').'</option>';
}else{
echo '<option value="'.$startyear->format('Y').'">'.$startyear->format('Y').'</option>';
}
$startyear->add($interval);
}
echo'</select>';
}
I am trying to calculate the nearest Age based on DOB, but i cant wrap my head around how to do it. I have tried some methods which estimates but this is not good enough. We need to calculate the days from today and the next birthday, whether it is in the current year or next year. and also calculate the days from today and the last birthday again whether it is in the current year or last year.
Any suggestions?
I think this is what you want.... of course, you could just get a persons age accurate to the day and round it up or down to the closest year..... which is probably what I should have done.
It's quite brute force, so I'm sure you can do it better, but what it does is check the number of days until this year's, next year's, and last year's birthday (I checked each of those three separately instead of subtracting from 365, since date() takes care of leap years, and I don't want to). Then it calculates age from whichever one of those birthdays is closest.
Working example
<?php
$bday = "September 3, 1990";
// Output is 21 on 2011-08-27 for 1990-09-03
// Check the times until this, next, and last year's bdays
$time_until = strtotime(date('M j', strtotime($bday))) - time();
$this_year = abs($time_until);
$time_until = strtotime(date('M j', strtotime($bday)).' +1 year') - time();
$next_year = abs($time_until);
$time_until = strtotime(date('M j', strtotime($bday)).' -1 year') - time();
$last_year = abs($time_until);
$years = array($this_year, $next_year, $last_year);
// Calculate age based on closest bday
if (min($years) == $this_year) {
$age = date('Y', time()) - date('Y', strtotime($bday));
}
if (min($years) == $next_year) {
$age = date('Y', strtotime('+1 year')) - date('Y', strtotime($bday));
}
if (min($years) == $last_year) {
$age = date('Y', strtotime('-1 year')) - date('Y', strtotime($bday));
}
echo "You are $age years old.";
?>
Edit: Removed unnecessary date()s in the $time_until calcs.
If I understand correctly you want to "round" the age? Then how about something along these lines:
$dob = new DateTime($birthday);
$diff = $dob->diff(new DateTime);
if ($diff->format('%m') > 6) {
echo 'Age: ' . ($diff->format('%y') + 1);
} else {
echo 'Age: ' . $diff->format('%y');
}
Edit: rewrote to use DateInterval
This should do the trick for you...
$birthday = new DateTime('1990-09-03');
$today = new DateTime();
$diff = $birthday->diff($today, TRUE);
$age = $diff->format('%Y');
$next_birthday = $birthday->modify('+'. $age + 1 . ' years');
$halfway_to_bday = $next_birthday->sub(DateInterval::createFromDateString('182 days 12 hours'));
if($today >= $halfway_to_bday)
{
$age++;
}
echo $age;