implode value does not count as an array - php

Question

So this is my code for set the date from selected date to auto set date excluding weekend (sat&sun) and holiday ( i put it in database ) :
function number_of_working_dates($from, $days) {
/* get holiday from database */
include('includes/config.php');
$sss = "SELECT tanggal_awal FROM libur_nasional GROUP BY tanggal_awal ASC";
$qqq = mysqli_query($konek,$sss);
$arr = array();
while ( $tam = mysqli_fetch_array($qqq)) {
$date = $tam['tanggal_awal'];
$reformat_date = date("Y-m-d", strtotime($date));
$arr[] = $reformat_date;
}
$array_date_2 = '"'.implode('", "' , $arr) . '"';
The output of implode() is a list of dates:
"2018-08-09", "2018-08-10", "2018-08-15", "2018-08-17"
Then the continuation:
$holidayDays = [$array_date_2];
//$holidayDays = ["2018-08-09", "2018-08-10", "2018-08-15", "2018-08-17"]
$from = new DateTime($from);
$dates = [];
$dates[] = $from->format('Y-m-d');
**echo count($holidayDays);**
while ($days) {
$from->modify('+1 day');
if (!in_array($from->format('N'), $workingDays)) continue;
if (in_array($from->format('Y-m-d'), $holidayDays)) continue;
if (in_array($from->format('*-m-d'), $holidayDays)) continue;
$dates[] = $from->format('Y-m-d');
$days--;
}
return $dates;
}
When I count the "holidayDays" array using "count" it only shows 1 item, but when I hardcoded the date list like in the comment line it works and shows 4 items.
Can someone help me?

Answer 1

A variable containing a string contains that string and it will continue to be treated as a string unless you do something like evaluating it as source code (which is very dangerous).
So if you have:
$foo = ["1", "2", "3"];
Then you have an array with three things in it.
But if you have:
$string = '"1", "2", "3"';
$foo = [$string];
Then you have an array with one thing in it: a string.
It is the same as writing:
$foo = ['"1", "2", "3"'];
If you want to have an array of strings then look at this line of your code:
$array_date_2 = '"'.implode('", "' , $arr) . '"';
That converts the array of strings into a single string.
Do not do that because that is not what you want.
Just use the value you already have in $arr.

Answer 2

You have a variable named "$array_date_2" and are trying to assign a String to it with the following code :
$array_date_2 = '"'.implode('", "' , $arr) . '"';
This assignment , after the "implode" has been executed , would result in the following :
$array_date_2 = ""2018-08-09", "2018-08-10", "2018-08-15", "2018-08-17"";
Notice the quotes at the start and end of the String. Even though you do not see them when "echoing" the String is wrapped around those quotes. PHP does not permit a String without it being enclosed in quotes , it will throw a Parse Error.
So in the end you do not have a String like :
"2018-08-09", "2018-08-10", "2018-08-15", "2018-08-17"
But more like
""2018-08-09", "2018-08-10", "2018-08-15", "2018-08-17""
Now you are also trying to create an array by concatenating Strings and Special Characters together. This is not how PHP works though.
If you want to "create" a function out of a String , or generally evaluate a String as PHP Code , then you need to use "eval" but i would strongly suggest against it.
For reference only , if you went with eval your code should be like this :
eval("\$holidayDays = [".$array_date_2 ."];");