Rebuilding array for dynamic dropdowns

Question

I'm trying to take a query result, build into an associative array where I use the high level for a general dropdown and that dropdown selection will load cooresponding children of the array.

My current function:

$getDisplays = "
SELECT *
FROM locations l
inner join displays d
ON d.location_id = l.id;";

$displayResult = $mysqlConn->query($getDisplays);

This returns:

Array
(
[ID] => 1
[location_name] => Office 1
[display_name] => lobby
[location_id] => 1
)
Array
(
[ID] => 2
[location_name] => Office 1
[display_name] => break room
[location_id] => 1
)
Array
(
[ID] => 3
[location_name] => Office 2
[display_name] => lobby
[location_id] => 2
)
Array
(
[ID] => 4
[location_name] => Office 2
[display_name] => break room
[location_id] => 2
)
Array
(
[ID] => 10
[location_name] => Office 5
[display_name] => Break Room
[location_id] => 5
)
Array
(
[ID] => 11
[location_name] => Office 5
[display_name] => Lobby
[location_id] => 5
)
Array
(
[ID] => 12
[location_name] => Office 5
[display_name] => Conference Room
[location_id] => 5
)

I'm trying to create dynamic dropdowns so that the first one shows each distinct location_name, and depending on the selection, the second dropdown contains the corresponding display_name.

So for this array, the first dropdown would have:

<option>Office 1</option>
<option>Office 2</option>
<option>Office 5</option>

and if I select Office 1 the next dropdown would contain:

<option>lobby</option>
<option>break room</option>

The way I'm building them now:

<label for="plantSelect">Select A Location</label>
<select class="form-control" id="plantSelect">
    <?php foreach($displayResult as $area=>$display):?>
    <option><?php echo $display['location_name']?></option>
    <?php endforeach;?>
</select>

<label for="areaSelect">Select An Area</label>
<select class="form-control" id="areaSelect">

    <option>Please Choose a Location</option>

</select>

The problem is, the first dropdown is showing each entry so it's showing multiple instances of Office 1, Office 2 and Office 5. I want it to only show one of each location name and then load the next dropdown with the children(display_name) of the selected location. I will also need the ID from display in order to save that to a database, so I'm thinking once I fix this issue I should just be able to include that later in a form.

What am I doing wrong with the array though?

Wololo · Accepted Answer · 2018-07-19 10:13:05Z

1

You can rebuild your array like this:

$displayNames = array();
foreach($displayResult as $subArray) {
  if(!array_key_exists($subArray['location_name'], $displayNames)) {
    $displayNames[$subArray['location_name']] = array();

  }
    $displayNames[$subArray['location_name']][] = $subArray['display_name'];
}

Now, you can do something like:

<?php foreach($displayNames as $displayName):?>
<option><?php echo key($displayName['location_name'])?></option>
<?php endforeach;?>

Things won't repeat. Plus, now you can easily create other dropdowns.

edited Jul 19, 2018 at 10:13

answered Jul 17, 2018 at 17:56

Wololo

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1 Comment

Geoff_S Over a year ago

So, I tried this and it looks like my first dropdown now only has the correct number of entries but they are all 0's

Amitesh Bharti · Accepted Answer · 2018-07-17 19:21:17Z

0

$displayNames = array();
foreach($displayResult as $subArray) {
  if(!array_key_exists($subArray['location_name'], $displayNames)) {
      $displayNames[$subArray['location_id']]['location'] = $subArray['location_name'];
      $displayNames[$subArray['location_id']]['display_name'] = array();
  }

  array_push( $displayNames[$subArray['location_id']]['display_name'], array($subArray['ID'] => $subArray['display_name']));
}

print_r($displayNames);

edited Jul 17, 2018 at 19:21

answered Jul 17, 2018 at 18:35

Amitesh Bharti

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