Undefined offset: 1 after workin offset o - php

Question

i want to add same values to different id number than i make this code but it say " Undefined offset: 1 "after adding only one value to data base
her is may code
$d = 0;
$sql="SELECT * FROM emplyeelist WHERE StatuseUnder = '$EmpIdSE'";
$result = mysqli_query($db,$sql);
while($row = mysqli_fetch_assoc($result)){
echo "$row['fullName']"."<input type = 'Checkbox' name = 'EmpName[$d]' value = '".$row['fullName']."'>";
++$d;
}
$size = count($_POST['EmpName']);
$d = 0;
while ($d < $size){
$EmpName = $_POST['EmpName'][$d];
$sql = "INSERT INTO ...";
$result = mysqli_query($db,$sql);
++$d;
}

Answer 1

i finally change my cod to this
<?php
$d = 0;
$sql="SELECT * FROM emplyeelist WHERE StatuseUnder = '$EmpIdSE'";
$result = mysqli_query($db,$sql);
while($row = mysqli_fetch_assoc($result)){
echo "<input type = 'text' readonly name = 'EmpName[$d]' value ='".$row['fullName']."'>".
"<input type = 'checkbox' readonly name = 'CheckBoxEmpName[$d]' value = '".$row['fullName']."'>"."</br>";
++$d;}
?>
$d = 0;
while ($d < $size){
$EmpName = $_POST['EmpName'][$d];
$CheckBoxEmpName = isset( $_POST['CheckBoxEmpName'][$d]) $_POST['CheckBoxEmpName'][$d] : "";
if(!empty($CheckBoxEmpName)){
$sql = "INSERT INTO table_name (table_row_name)VALUES ('$EmpName')";
$result = mysqli_query($db,$sql);
}++$d;}
than All done all work wall