Duplicate values when populating select dropdown from mysql query

Question

I am trying to populate an html select dropdown with data from multiple columns in a mysql query.

When I retrieve the first column it populates the dropdown without issue. When I add the second or third columns I get issues. Originally I was getting blank fields, this was resolved by adding !empty().

However I am now getting values repeated in place of the blanks

 <?php
        $base = '';
        $base2 = '';
        $base3 = '';

 while($row = mysqli_fetch_assoc($resultpmb)) {
  $code=$row["id"];
  $name=$row["gw_name"];
  if (!empty(trim($row['vg_name']))){
  $vgname=$row["vg_name"];}
  if (!empty(trim($row['vm_name']))){
  $vmname=$row["vm_name"];}

$base .= "<option value=" .$code.">".$name."</option>";
$base2 .= "<option value=" .$code.">".$vgname."</option>";
$base3 .= "<option value=" .$code.">".$vmname."</option>";

}
?>
        <div class="form-group col-4 col-m-12">
         <div class="input-group">
            <span class="input-group-addon"><span class="glyphicon 
glyphicon-lock"></span></span>
<select name="base_colour1" class="form-control" id="base_colour1">
<option value="">Base...</option>
<?php 
echo "<option value=''>---- GW Golour ----</option></br>";
echo $base;
echo "<option value=''>---- Vallejo Game Golour ----</option>";
echo $base2; 
echo "<option value=''>---- Vallejo Model Golour ----</option>";
echo $base3;
?>
</select>
</div>
        </div>

That's because they are the different fields in the same row. — Foster
– Foster, Commented Dec 21, 2018 at 9:31

Nigel Ren · Accepted Answer · 2018-12-20 14:39:54Z

3

It may be easier to add the strings within your test of !empty() otherwise it will always add values in - even when there aren't any...

while($row = mysqli_fetch_assoc($resultpmb)) {
  $code=$row["id"];
  $name=$row["gw_name"];
  if (!empty(trim($row['vg_name']))){
      $base2 .= "<option value=" .$code.">".$row["vg_name"]."</option>";
  }
  if (!empty(trim($row['vm_name']))){
      $base3 .= "<option value=" .$code.">".$row["vm_name"]."</option>";
  }

  $base .= "<option value=" .$code.">".$name."</option>";


}

answered Dec 20, 2018 at 14:39

Nigel Ren

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