why multiple data doesn't insert in this code - php

Question

I have a tbl_employee table and tbl_time table, I want to insert multiple data insert for attendance but when I click submit button it's insert only single data . but where is the problem ,help me to find out this..this is insert code
require './db_connect.php';
class Time extends Db_connect {
protected $link;
public function __construct() {
$this->link = $this->database_connection();
}
public function attendance_insert($data) {
extract($data);
$cur_date = date('Y-m-d');
foreach ($time_attendance as $attn_key => $attn_value) {
if ($attn_value == 'P') {
$SQL = "INSERT INTO tbl_time(employee_id,time_date,time_attendance)VALUES('$attn_key','$cur_date','P')";
$atten_date=mysqli_query($this->link, $SQL);
} else if ($attn_value == 'A') {
$SQL = "INSERT INTO tbl_time(employee_id,time_date,time_attendance)VALUES('$attn_key','$cur_date','A')";
$atten_date=mysqli_query($this->link, $SQL);
}
if ($atten_date) {
$massage = "<div class='alert alert-success text-center'><h5>Attendance insert successfully</h5></div>";
return $massage;
} else {
die('Attendance insert query problem' . mysqli_error($this->link));
}
}
}
}
this is html code
<?php
require_once './time.php';
$obj_time = new Time()
$massage = '';
if (isset($_POST['btn'])) {
$massage = $obj_time->attendance_insert($_POST);
}
$employee_view = $obj_employee->employee_all_view();
?>
<div class="container-fluid">
<div class="row">
<center>
<span style="font-size:1.8em;">Attendance form</span>
</center>
</div>
</div>
<hr/>
<?php echo $massage; ?>
<div class="container-fluid">
<div class="row">
<div class="col-lg-12">
<div class="panel panel-body panel-default">
<div class="well text-center" style="font-size:15px;">
<strong>Date :</strong>
<?php $current_date = date('Y-M-d');
echo $current_date; ?>
</div>
<form class="form-horizontal" method="post">
<table class="table table-striped table-responsive text-center">
<tr>
<td><b>Serial</b></td>
<td><b>Name</b></td>
<td><b>ID</b></td>
<td><b>Attendance</b></td>
</tr>
<?php
$i = 0;
while ($employee_info = mysqli_fetch_assoc($employee_view)) {
extract($employee_info);
$i++;
?>
<tr>
<td><?php echo $i; ?></td>
<td><?php echo $employee_first_name . ' ' . $employee_last_name; ?></td>
<td><?php echo $employee_id; ?></td>
<td>
<input type="radio" name="time_attendance[<?php echo $employee_id; ?>]" value="P">P
<input type="radio" name="time_attendance[<?php echo $employee_id; ?>]" value="A">A
</td>
</tr>
<?php } ?>
<tr>
<td colspan="4">
<input type="submit" class="btn btn-primary btn-block" name="btn" value="submit"/>
</td>
</tr>
</table>
</form>
</div>
</div>
</div>
</div>

Answer 1

This :
foreach($time_attendance as $attn_key => $attn_value){
$SQL = "INSERT INTO tbl_time(employee_id,time_date,time_attendance)
VALUES('$attn_key','$cur_date','A')";
$atten_date = mysqli_query($this->link, $SQL);
if ($atten_date) {
$massage = "<div class=''><h5>Attendance insert successfully</h5></div>";
return $massage;
}
}
Literally ask to break the foreach with return $massage; if you query succeed... So yes, you'll only have one record.
Put your return outside the foreach.