Display value in a dedicated field selected from a list - php

Question

in a php form I have the list of brands scrollable from a select field and I would modify the selected brand displaying it in a field beside.
The php form is this:
<?php
include '../sys/conn.php';
$brands = mysqli_query ($conn, "
*(query to select brands name and id)*
") or die ("Query not valid: " . mysqli_error($conn));
mysqli_close($conn);
?>
...
<form role="form" >
<label>Brands List</label>
<select class="form-control" name='brands list'>
<?php while ($listabrand=mysqli_fetch_array($brands)){
echo '<option>'.$listabrand['0'].' - '.$listabrand['1'].'</option>';
}?>
</select>
...
<label>Modify Brand</label>
<input type="text" name='brand-name' class="form-control" required placeholder="Brand Name to modify">
</form></html>
Basically I need to select the brand's name from the named "brands list" field and display it into the 'brand-name' in order to modify and save it.
Any help?

Answer 1

First of all, name attribute cannot contain space.
Correct this in <select class="form-control" name='brands list'>
And for the displaying selected brand in the value of text field, use javascript or jquery.
Showing example in JQuery:
$('select[name="brands-list"]').change(function(){
var selectedBrand = $(this).val();
$('input[name="brand-name"]').val(selectedBrand);
});

Answer 2

This is the right code.
<select class="form-control" name='brands-list'>
<?php
while ($listabrand=mysqli_fetch_array($brands)){
echo '<option value="'.$listabrand['1'].'">'.$listabrand['0'].' - '.$listabrand['1'].'</option>';
}?>
</select>
<div class="col-lg-8">
<form role="form">
<div class="form-group">
<label>Marca</label>
<input type="text" name='brand-name' class="form-control">
</div>
</form>
</div>
<script>
$('select[name="brands-list"]').change(function(){
var selectedBrand = $(this).val();
$('input[name="brand-name"]').val(selectedBrand);
});
</script>