Finding Matching Variable In Array

Question

I want to determine if $today is a known non-working day or holiday by comparing two arrays and report the result. I do not know what belongs in the if and elseif structures to compare if the variable within the array is a match to $today:

<?php
// Date & Time
$timeadjust = '-5 hours';
$today_long = date('l, d F Y', strtotime($timeadjust));
$today = date('m-d', strtotime($timeadjust));
$today_day_name = date('l', strtotime($timeadjust));
// Normal Non-Working Days
$saturday = 'Saturday';
$sunday = 'Sunday';
// 2015 Holidays
$new_years_day = '01-01';
$fourth_of_july = '07-04';
$thanksgiving = '11-26';
$thanksgiving_friday = '11-27';
$christmas_eve = '12-24';
$christmas = '12-25';
$new_years_eve = '12-31';
// Normal Non-Working Day Array
$no_work = array($saturday,$sunday);
// Holiday Array
$holiday = array($new_years_day,$fourth_of_july,$thanksgiving,$thanksgiving_friday,$christmas_eve,$christmas,$new_years_eve);
// Compare Today To Normal Non-Working Day & Holiday Arrays To Determine If Today Is Normal Non-Working Day Or Holiday
if ($today_day_name = $no_work) {$operating = 'CLOSED';}
elseif ($today = $holiday) {$operating = 'CLOSED';}
else {$operating = 'OPEN';}
// Display Result Of Comparison & Report Operating Status
echo '<h3>Today is&nbsp;'.$today_long.'.&nbsp;We are&nbsp;'.$operating.'&nbsp;today!</h3>';
?>

You want to take a look at in_array() for this

scrowler
– scrowler

2015-01-07 21:14:19 +00:00
Commented Jan 7, 2015 at 21:14 — scrowler
– scrowler, Commented Jan 7, 2015 at 21:14

Jay Blanchard · Accepted Answer · 2015-01-07 21:24:16Z

1

For what you're doing you can use in_array() which will test if the value is in a selected array -

// Compare Today To Normal Non-Working Day & Holiday Arrays To Determine If Today Is Normal Non-Working Day Or Holiday
if (in_array($today_day_name, $no_work)) {
    $operating = 'CLOSED';
} elseif (in_array($today, $holiday)) {
    $operating = 'CLOSED';
} else {
    $operating = 'OPEN';
}

EXAMPLE

In your original code you're also assigning instead of testing -

if($foo = $bar) // one equals sign assigns
if($foo == $bar) // two equals signs compares
if($foo === $bar) // three equals signs tests for equivalency, does it match value and type?

Because you were assigning your if statement would always evaluate to true.

answered Jan 7, 2015 at 21:24

Jay Blanchard

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