How I can convert INT to date in php and than find this date in mysql table ?
Lets say I want to convert
$month = (int)$_GET['month'];
$month = 42014 ;//This means 04th month in 2014 year
$month = 04-2014; // I want this
And than find that in mysql table
$query=mysql_query("SELECT id FROM matches WHERE date=$month");// This need to select
// all data that is
// created in 04th month
// of 2014 year .
echo "Thanks :)";
Don't pass $month parameter like you do.
Send and pass both $month and $year and handle them after:
$month = (int)$_GET['month'];
$year = (int)$_GET['year'];
if($month < 10) { // add this check for months lesser then october (they containt 1 digit, which is wrong for t-sql)
$month = "0".$month;
}
$query=mysql_query("SELECT id FROM matches WHERE DATE_FORMAT(datefield, '%m-%Y') = '$month-$year'");
source
If you can't send both month and year in different variables, do this, like M Khalid Junaid suggested:
$query=mysql_query("SELECT id FROM matches WHERE DATE_FORMAT(datefield,'%m%Y')='$month'");
Try as below using DateTime::createFromFormat
$month = $_GET['month'];
$date = DateTime::createFromFormat('mY',$month);
echo $date->format('m-Y');
Try this
$input=42014;
$year = strlen(substr($input,-4));
$month = substr(0,(strlen($input)-strlen($year)));
$final = $month."-".$year
Related
I have a user in a database with a creation_date. This user can run a job in my app UI, but he is limited by a number of job to run in one year.
This user has been created in 2014. I would like to do something like :
function runJob($user){
$nbRemainingJob = findReminingJobs($user);
if ($nbRemainingJob > 0){
runJob($user);
}
else {
die("no more credits";)
}
}
findReminingJobs($user){
$dateRangeStart = ?; //start date to use
$endRangeStart = ?; //end date to use
$sql = "SELECT count(*) FROM jobs WHERE user_id=?";
$sql .= "AND job_created_at BETWEEN ($dateRangeStart AND $endRangeStart)";
$res = $pdo->execute($sql, [$user->id]);
$done = $res->fetchOne();
return ($user->max_jobs - $done);
}
Every user's creation birthday, the $user->max_jobs is reset.
The question is how to find starting/ending date ? in other words, I would like to get a range of date starting from the user's creation date.
For example, if the user was created on 2014-04-12, my start_date should be 2018-04-12 and my end_date = 2019-04-11.
Any idea ?
First get the user register date from db and split it into Year, Month and Day like
$register= explode('-', $userCridate);
$month = $register[0];
$day = $register[1];
$year = $register[2];
Then get the current year like
$year = date("Y");
$dateRangeStart = $year."-".$month."-".$day; //start date to use
Now, check if this date is greater then today date, then use last year as starting date
$previousyear = $year -1;
$dateRangeStart = $previousyear ."-".$month."-".$day; //start date to use
$endRangeStart = date("Y-m-d", strtotime(date("Y-m-d", strtotime($dateRangeStart))
. " + 365 day"));
It is a idea, check if it work for you.
function getRange($registrationDate) {
$range = array();
// Split registration date components
list($registrationYear, $registrationMonth, $registrationDay) = explode('-', $registrationDate);
// Define range start year
$currentYear = date('Y');
$startYear = $registrationYear < $currentYear ? $currentYear : $registrationYear;
// Define range boudaries
$range['start'] = "$startYear-$registrationMonth-$registrationDay";
$range['end'] = date("Y-m-d", strtotime($range['start'] . ' + 364 day'));
return $range;
}
And for your example:
print_r(getRange('2014-04-12'));
Array
(
[start] => 2018-04-12
[end] => 2019-04-11
)
print_r(getRange('2014-09-13'));
Array
(
[start] => 2018-09-13
[end] => 2019-09-12
)
$created='2025-04-12';
$date=explode('-',$created);
if($date[0]<date("Y")){
$newDate=date('Y').'-'.$date[1].'-'.$date[2];
$dateEnding = strtotime($newDate);
$dateEnding = date('Y-m-d',strtotime("+1 year",$dateEnding));
}
else{
$newDate=$created;
$dateEnding = strtotime($newDate);
$dateEnding = date('Y-m-d',strtotime("+1 year",$dateEnding));
}
echo 'starting date is: '.$newDate;
echo '</br>';
echo 'ending date is: '.$dateEnding;
This code will get the date you have and match it with the current year. If the year of the date you provided is equal or above the current year the start date will be your date and end date will be current date +1 year. Otherwise if the year is below our current year (2014) it will replace it with the current year and add 1 year for the end date. Some example outputs:
For input
$created='2014-04-12';
The output is :
starting date is: 2018-04-12
ending date is: 2019-04-12
But for input
$created='2025-04-12';
The outpus is :
starting date is: 2025-04-12
ending date is: 2026-04-12
The solution that match my need :
$now = new DateTime();
$created_user = date_create($created);
$diff = $now->diff($created_user)->format('%R%a');
$diff = abs(intval($diff));
$year = intval($diff / 365);
if ($year == 0){
$startDate=$created_user->format("Y-m-d");
}else{
$startDate=$created_user->add(new DateInterval("P".$year."Y"))->format("Y-m-d");
}
The problem was to define the starting date that is comprised in the one year range max from the current date and starting from the user's creation date.
So if the user's creation_date is older than one year, than I do +1 year, if not, take this date. the starting date must not be greater than the current date_time
thanks to all for your help
I need to compare different parts of dattime in php to see when to present a certain text to the user based on both time, day and season
I found an example that gives me every part of datetime I need
but when I compare it with a string value, I get a blank screen
then I tried to compare it with a number, same result
when I call the method without the comparing it does get executed
how can I compare these values and with what ?
here is the code
$dateFormat = 'Y-m-d';
$stringDate = date($dateFormat);
$date = DateTime::createFromFormat($dateFormat, $stringDate);
$dateValue = strtotime($q);
$year = $date->format('Y');
$month = $date->format('m');
$days = $date->format('d');
$hour = $date->format('H');
$minute = $date->format('i');
$second = $date->format('s');
I have problem getting the year
I have here in my database year of 11/11/2016 - 11/13/2015 - 11/13/2014
the problem is I want to get that year just deducting current year
so if the current year is 2017 I want to get 2016,2015,2014 , and the next year
2018 the year will get is 2017,2016,2015
$selectmemberpaid = mysql_query("SELECT
memberid,StartDate,EndDate,tblpayments.activityname,amount_paid FROM
tblattend_activity LEFT JOIN tblpayments ON tblattend_activity.memberid =
.profile_id WHERE memberid='201400001'");
That is my query..
and this is my while loop
while ($rows = mysql_fetch_assoc($selectmemberpaid)) {
# code...
$dateTimestamp = strtotime($rows['StartDate']);
$year = date("Y", $dateTimestamp);
}
You can get the year from a date string by doing something like this:
$dateTimestamp = strtotime('2015-12-12');
$year = date("Y", $dateTimestamp);
echo $year;
Once you get it you can add or subtract to get the previous/next year (after converting the value into an integer).
I'm not sure I understand what you're asking but:
Query!
$rest = substr(date('d-m-Y'), 6);
$rep = $idz->query("SELECT * FROM stack1 where SUBSTR(date_depart, 6) < $rest
order by SUBSTR(date_depart,6) desc limit 3");
while ($row = $rep->fetch())
{
echo $row['date_depart'];
//echo "<br>";
}
$rep->closeCursor();
Result
I have save year as input into database but it save the duration like if I get input 1990 as input in database save 26.Here 26 is difference of current date to selected date 2016-1990=26.I am using php Date function but it's not work proper.Using this get fixed year 1970
<?php
$dr = [];
for($i = date("Y"); $i > date("Y") - 100; $i--) {
$dr[] = $i;
}
echo $form->field($model, 'ven_established_date')->dropDownList($dr);//Here Select year
?>
$date = $model->ven_established_date;
echo '$date'.$date;
$year = date("Y", strtotime($date));
echo '$year'.$year;
As your question is not much clear so what i understood from your question is you have a year count(ex:-26) and you want exact year.
Solution:-
$yearCount = 26; //input you have, getted from DB
$currentYear = date("Y");
$pastYear = $currentYear - $yearCount;
echo $pastYear;
I have a database table that has two columns eventName|eventDate. I have created a function that takes in a startDate and endDate, I want to display the list of events in a ListView with each date as the header.
In my brief example below, I know I can retrieve the full event listings with SQL. How do I then slot the event headers in so that I can return them in a properly formatted array?
function retrieveEvents($startDate, $endDate) {
// run SQL query
//
if($stmt->rowCount() > 0) {
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
// how do I write this part such that I can output event headers in my array
$events = $row;
}
}
}
So my intended output is
1st July 2013 ($startDate)
- Tea with President - 1300h
- Mow the lawn - 1330h
- Shave the cat - 1440h
2nd July 2013
- Shave my head - 0800h
3rd July 2013
4th July 2013 ($endDate)
- Polish the car - 1000h
In your MYSQL query:
SELECT * FROM `yourTableName` WHERE `eventDate` >= $startDate AND `eventDate` <= $endDate
PS: I'm not sure about the quotes arount the variables in your query.
PPS: never use * to select your columns, always only select the columns you need. Here I'm using it because I don't know the names of your columns
I ended up doing my checking in PHP and print a new row only when a different date is detected.
Codes below in case it serves someone's needs in future.
<?php
$currentPrintDay = 0;
$currentPrintMonth = 0;
$currentPrintYear = 0;
echo "<table>"
foreach($reservationsToShow as $row):
// get day, month, year of this entry
$timestamp = strtotime($row['timestamp']);
$day = date('d', $timestamp);
$month = date('m', $timestamp);
$year = date('Y', $timestamp);
// if it does not match the current printing date, assign it to the current printing date,
// assign it, print a new row as the header before continuing
if($day != $currentPrintDay || $month != $currentPrintMonth || $year != $currentPrintYear) {
$currentPrintDay = $day;
$currentPrintMonth = $month;
$currentPrintYear = $year;
echo
"<tr>" .
"<td colspan='100%'>". date('d-m-Y', $timestamp) . "</td>" .
"</tr>";
}
// continue to print event details from here on...
?>