Check if the year in line = is or is greater than 2017
I have a lot of posts and want to add a special to the ones I post this year.
offer name - July 20, 2015
offer name - July 20, 2016
offer name - July 20, 2017
...
I'm doing it:
$a = "offer name - July 20, 2000";
if (preg_match('/2017/',$a)) {
echo 'offer';
}
running for 2017, but I want to add suggestions for posts from 2017 and beyond ... 18, 19. any ideas?
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This option iterates over the inputs and filters by year:
<?php
$input = [
'offer name - July 20, 2015',
'offer name - July 20, 2016',
'offer name - July 20, 2017'
];
$output = [];
array_walk($input, function($entry) use (&$output){
if ( preg_match('| - \w+\s+\d+,\s+(\d+)$|', $entry, $tokens)
&& intval($tokens[1]) >= 2017 ) {
$output[] = $entry;
}
});
print_r($output);
This is a slightly more compact option:
<?php
$input = [
'offer name - July 20, 2015',
'offer name - July 20, 2016',
'offer name - July 20, 2017'
];
$output = array_filter($input, function($entry){
return
preg_match('| - \w+\s+\d+,\s+(\d+)$|', $entry, $tokens)
&& (intval($tokens[1]) >= 2017);
});
print_r($output);
The output is obviously equal to:
Array
(
[0] => offer name - July 20, 2017
)
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you can use something like this
function isYearAllowed($date, array $allowedYears)
{
try {
$date = new \DateTime($date);
} catch (\Exception $e) {
return false;
}
if(in_array($date->format("Y"), $allowedYears)){
return true;
}else{
return false;
}
}
Usage example
$allowedYears = ['2017','2016'];
if(isYearAllowed("2016-12-31", $allowedYears)){
echo "offer";
}else{
echo "don't offer";
}
exit;
live demo ( https://eval.in/835914 )
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You can use a date function to extract the year from a date string and then directly compare it to 2017
as the value it must be equal to or greater.
This function provides basic functionality whereas it returns TRUE
if the year is greater than or equal to 2017
, otherwise it returnsFALSE
function validate_year($in_string_date){
// Particular to the case in this question, format date for date() function
$in_string_date = str_replace(',','',$in_string_date);
$in_string_date = str_replace(' ','-',$in_string_date);
$valid_year = FALSE;
// Get the year component from the date passed into the function.
$year_in_date = date('Y',strtotime($in_string_date));
/* Check if the year component extracted from the date passed
to the function is greater or ewual to 2017 */
if($year_in_date >= 2017){
$valid_year = TRUE;
}
return $valid_year;
}
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There might be an easier way to do this, but until you find one, you can try doing a while loop.
1) Set the variable to the original year, you can do it statically or by the season function in php date("Y")
.
2)
while('your variable'){ //will always be true
if (preg_match('/'.'your variable'.'/')) //if there is an offer for this year
{
echo 'offer';
'your variable'++;
}
else
{
break;
}
}
This may not be what you are looking for, but it should work.
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You can use the functions substr
and intval
to achieve the same.
Below is the working code. You can see how it works here :
<?php
$val = array("offer name - July 20, 2000", "offer name - July 20, 2017","offer name - July 20, 2018");
foreach($val as $a)
{
$b = substr($a,strlen($a)-4, 4);
if(intval($b)>= 2017)
{
echo "Offer";
}
else
{
echo "No Offer";
}
echo "\n";
}
?>
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