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I'm using DateTime::createFromFormat in PHP like this:
$month = DateTime::createFromFormat("Y-m-d", '2014-08-01');
But echo $month is showing up as 08. How can I display it as a 3 letter name, i.e.: Aug?
Use capital M instead of m .
$getMonth = DateTime::createFromFormat("Y-m-d", '2014-08-01');
If you already have the date variable created then you can use directly the below code
$month = $getMonth->format('M');
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dob is date of birth and dob column in only dob's year
This is my code
$voters = DB::collection('users')->whereDate(Carbon::parse('dob')->age, '<', 25)->get();
This is my table
id, name, dob
1, Peter, 1974
2, John, 1988
3, Semi, 1995
Can you try this:
$now = Carbon::now();
$filterYear = $now->year - 25;
$voters = DB::collection('users')->where('dob', '<', $filterYear)->get();
Please dont forget to use Carbon on top of the class.
You can check the years difference:
$voters = DB::collection('users')->whereDate(Carbon::now()->year - 25, '>', 'dob')->get();
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I want to get a warning when a date is different like:
I have a while and wanted to when it reaches to a different date, it warns me
$hist=mysqli_query($db, "SELECT * from history");
while($his=mysqli_fetch_assoc($hist)){
If(certain_date != last_date){
echo certain_date;
}
}
Certain date and last date is the ones I wanna get
This is simply done like this
$hist=mysqli_query($db, "SELECT * from history");
$last_date = '';
while($his=mysqli_fetch_assoc($hist)){
If($his['ano'] != $last_date){
echo $his['ano'];
$last_date = $his['ano'];
}
}
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im new with the php and still studying about php and more. i have a question.
How to convert the age of human in years and then display the age in months,week and days.
(hint:1years=365days). i hope anyone could help me.thanks
You should look at the DateTime class
$previousDate = '1986-09-04 14:05:43';
$startDate = new DateTime($previousDate);
$endDate = new DateTime('now');
$interval = $endDate->diff($startDate);
echo $interval->format('%y years, %m months, %d days');
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<?php
$x=6;
$y=9;
$time = array(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0);
for ($i=$x;$i<=count($y);$i++)
{
If($x!=$y)
{
$time[$i]=1;
}
}
?>
Depending on the value of x and y,the values in array should change.
In this example... array[5] until array[8] should be value 1.
The value of x and y will not same.
Not a very good question, but I'm a little bored. So for fun:
array_splice($time, $x-1, $y-$x-1, array_fill(0, $y-$x+1, 1));
Not exactly sure of the logic using 6 and 9 and array[5] until array[8] is, but adjust the numbers to fit the range.
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I am currently trying to create a script/function that should set a value to all items (eshop goods) old 15 days or older.
This script will be used as cron.
Date is stored in database as integer, but I don't exactly know how to approach the 15 days gap.
Could somebody help me out please?
Thank you.
You can try something like this
$integer_format = strtotime("-15 days");
$dateTime_format = date("Y-m-d h:i:s A T",$integer_format);
echo 'In integer format: '.$integer_format.'</br>'; //1403953190
echo 'In date time format: '.$dateTime_format.'</br>'; //2014-06-28 12:59:50 PM CEST