Question

I am trying to create 2 dropdown menus. One for displaying a list of buildings and then when user selects a building from the list, it will display the list of rooms in that building.

Problem is I have an error in my code. Below is the code:

      $sql="SELECT Building, Room FROM Room WHERE Building = '".$building."'";

      $sqlresult = mysql_query($sql);

      $sqldataArray = array();

      while($sqlrow = mysql_fetch_array($sqlresult))
   {
      $sqldataArray[$sqlrow['Building']]; 
      $sqldataArray[$sqlrow['Building']]['Rooms'][$sqlrow['Room']]; 
   }


       $buildingHTML = ""; 
       $buildingHTML .= '<select name="buildings" id="buildingssDrop">'.PHP_EOL;
       $buildingHTML .= '<option value="">Please Select</option>'.PHP_EOL; 

   foreach ($sqldataArray as $building => $buildingData) {      

            $buildingHTML .= "<option value='".$building."'>" . $building . "</option>".PHP_EOL;        

            }
            $buildingHTML .= '</select>';


       $roomHTML = ""; 
       $roomHTML .= '<select name="rooms" id="roomsDrop">'.PHP_EOL;
       $roomHTML .= '<option value="">Please Select</option>'.PHP_EOL;      
            foreach ($buildingData['Rooms'] as $roomId => $roomData) {        

            $roomHTML .= "<option value='".$roomId."'>" . $roomId . "</option>".PHP_EOL;        
  } 

            $roomHTML .= '</select>';

The error I am getting is this:

Undefined variable: buildingData in /web/stud/u0867587/Mobile_app/create_session.php on line 372

This is the line of code where the error is:

$buildingHTML .= "<option value='".$building"'>" . $building . "</option>".PHP_EOL;

Does anyone know how to fix this error. I believe it is because it is not in the other foreach loop but if I put that in, then does it affect the display of the dropdown menu?

Answer 1

Not entirely sure what you’re trying to get as a final output – there’s a few logical issues with your code

Take a look at this minor rewrite – this will produce a <select> for buildings, and a <select> for all the rooms in your first building.

<?php

$sqlresult = mysql_query($sql);

$buildings = array(); // easier if you don't use generic names for data

while($sqlrow = mysql_fetch_array($sqlresult))
{
    // you need to initialise your building array cells
    if (!isset($buildings[$sqlrow['Building']])) {
        $buildings[$sqlrow['Building']] = array('Rooms' => array());
    }

    // you can add the room to the building 'Rooms' array
    $buildings[$sqlrow['Building']]['Rooms'][] = $sqlrow['Room']]);
}


$buildingHTML = ""; 
$buildingHTML .= '<select name="buildings" id="buildingssDrop">'.PHP_EOL;
$buildingHTML .= '<option value="">Please Select</option>'.PHP_EOL; 

foreach ($buildings as $building => $buildingData) {      
    $buildingHTML .= "<option value='".$building."'>" . $building . "</option>".PHP_EOL;        
}
$buildingHTML .= '</select>';

$roomHTML = ""; 
$roomHTML .= '<select name="rooms" id="roomsDrop">'.PHP_EOL;
$roomHTML .= '<option value="">Please Select</option>'.PHP_EOL;      
foreach ($buildings['Building Number 1']['Rooms'] as $roomId => $roomData) {        
    $roomHTML .= "<option value='".$roomId."'>" . $roomId . "</option>".PHP_EOL;        
} 

$roomHTML .= '</select>';