store all leave to array - php

Question

I am trying to fetch all employee leave data to an array. But first leave data of my employee is replacing by last leave data. I do not want that.
my data set:
id | emp_id | leave_from | leave_to
2894 886 2019-11-26 2019-11-27
2896 1970 2019-11-20 2019-11-24
2897 1016 2019-11-23 2019-11-23
2898 1870 2019-11-10 2019-11-10
2899 1895 2019-11-09 2019-11-11
2900 1895 2019-11-23 2019-11-23
2901 287 2019-11-27 2019-11-30
my sql query:
$leave_sql = "SELECT l.* FROM 0_employee e right join (SELECT * FROM 0_leave_ledger WHERE (leave_approved_from && leave_approved_to BETWEEN '".$fisrt_date."' AND '".$last_date."' )) l ON l.emp_id = e.emp_id where e.inactive=0";
$leave = get_tb($leave_sql);
$leave_date = array();
foreach ($leave as $leave_d) {
$leave_date[$leave_d["emp_id"]]['type'] = $leave_d["leave_type"];
$leave_date[$leave_d["emp_id"]]['lto'] = $leave_d["leave_approved_to"];
$leave_date[$leave_d["emp_id"]]['lfrom'] = $leave_d["leave_approved_from"];
}
let's look emp_id 1895.It has two rows in my data set. So I am expecting to insert those two rows in my array. But I am getting only last rows of data in my array.
my expecting output from $leave_date is:
[1895] => Array ( [type] => 4 [lto] => 2019-11-09 [lfrom] => 2019-11-11 )
[1895] => Array ( [type] => 4 [lto] => 2019-11-23 [lfrom] => 2019-11-23 )
but I am getting only
[1895] => Array ( [type] => 4 [lto] => 2019-11-23 [lfrom] => 2019-11-23 )

Answer 1

Your problem is that you have an employee who has taken multiple leaves, and the entries for each leave overwrite any previous ones in your array because they have the same array index. You need to change your data structure to allow for that. Try something like this:
$leave_date = array();
foreach ($leave as $leave_d) {
$leave_date[$leave_d["emp_id"]][] = array(
'type' => $leave_d["leave_type"],
'lto' => $leave_d["leave_approved_to"],
'lfrom' => $leave_d["leave_approved_from"]
);
}
For your sample data, this will give you something like
[1895] => Array (
[0] => Array ( [type] => 4 [lto] => 2019-11-09 [lfrom] => 2019-11-11 )
[1] => Array ( [type] => 4 [lto] => 2019-11-23 [lfrom] => 2019-11-23 )
)

Answer 2

The structure of your array $leave_date means it can only store one set of leave data per emp_id. When you iterate, it first fills up with the first leave record, and then the second iteration it gets replaced. If you were to put a print_r() at the end of your foreach you would see this happen for each iteration.
What you may want to do is try something like this:
foreach ($leave as $leave_d) {
$record = [
'type' => $leave_d["leave_type"],
'lto' => $leave_d["leave_approved_to"],
'lfrom' => $leave_d["leave_approved_from"],
]
if(array_key_exists($leave_d['emp_id'], $leave_date) {
$leave_date[$leave_d['emp_d']][] = $record;
} else {
$leave_date[$leave_d['emp_d'] = [$record];
}
This way, $leave_date[1234] will always be an array of leave records.