add option value dynamically - php

Question

I have a select with MySQL option values. My problem is with that if I have more then 1 value in the table, it's alway prints one option. How can I add dynamically more option value if an Employee has more then 1 company?
For example: modal
It has 2 company and I want to print all of the company he has, not just 1. How can I do this?
I print the current MySQL datas in the fetch.php.
index.php
<label>Company:</label>
<select name="company" id="company" class="form-control">
<?php
$query2 = "SELECT * FROM company GROUP BY company_id";
$result2 = mysqli_query($connect, $query2);
while($row2= mysqli_fetch_array($result2)){
?>
<option value="<?php echo $row2['company_id'];?>"><?php echo $row2['company_name'];?></option>
<?php
}
?>
</select>
<script>
$(document).on('click', '.edit_data', function(){
var employee_id = $(this).attr("id");
$.ajax({
url:"fetch.php",
method:"POST",
data:{'employee_id':employee_id},
dataType:"json",
success:function(data){
$('#company').val(data.company);
}
});
});
</script>
fetch.php
if(isset($_POST["employee_id"]))
{
$query = "SELECT employee.id AS id, employee.name AS name, employee.address AS address, company.company_id AS company
FROM employee
LEFT JOIN employee_company ON employee.id = employee_company.employee_id
LEFT JOIN company ON company.company_id = employee_company.company_id
WHERE employee.id = '".$_POST["employee_id"]."'";
$result = mysqli_query($connect, $query);
$row = mysqli_fetch_array($result);
echo json_encode($row);
}

Answer 1

First of all, you can't select multiple options from a dropdown menu. You need to use a multi-select.
<select name="company" id="company" class="form-control" multiple>
<?php
$query2 = "SELECT * FROM company GROUP BY company_id";
$result2 = mysqli_query($connect, $query2);
while($row2= mysqli_fetch_array($result2)){
?>
<option value="<?php echo $row2['company_id'];?>"><?php echo $row2['company_name'];?></option>
<?php
}
?>
</select>
Then you need to change fetch.php return an array of all the company IDs:
if(isset($_POST["employee_id"]))
{
$query = "SELECT employee.id AS id, employee.name AS name, employee.address AS address, company.company_id AS company
FROM szerepek
LEFT JOIN employee_company ON employee.id = employee_company.employee_id
LEFT JOIN company ON company.company_id = employee_company.company_id
WHERE employee.id = '".$_POST["employee_id"]."'";
$result = mysqli_query($connect, $query);
if (mysqli_num_rows($result) == 0) {
echo json_encode(null);
} else {
$companies = array();
while ($row = mysqli_fetch_assoc($result)) {
$companies[] = $row['company'];
}
$row['companies'] = $companies;
echo json_encode($row);
}
}
and change the Javascript to loop through them and select all of them.
$(document).on('click', '.edit_data', function(){
var employee_id = $(this).attr("id");
$.ajax({
url:"fetch.php",
method:"POST",
data:{'employee_id':employee_id},
dataType:"json",
success:function(data){
if (data) {
$("#company").val(data.companies);
}
}
});
});