Getting a value of a field from an array

PHP
1

Hi,
First time poster -but this site has helped me out tremendously so far - so fingers crossed with this…
It may seem too simple but I have been struggling with this all day.

i am following the php/mysql book - the jokes database etc -but have hit a wall.

In my search form I have:

				<div>
				<label for="location">By location:</label>
				<select name="location" id="location">
					<option value="">Any location</option>
					<?php foreach ($locations as $location): ?>
						<option value="<?php htmlout($location['locationid']); ?>"><?php
								htmlout($location['location']); ?></option>
					<?php endforeach; ?>
				</select>
			</div>

That’s grand - I can select my locations no problem - all set up fine in my controller index.php (this is one out of 3 drop down boxes to search with)

Now when i want to return the results I cannot get the location, just locationid

index.php

if (isset($_GET['action']) and $_GET['action'] == 'search')
{
	include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php';

	// The basic SELECT statement - to do nothing - will have no effect on results unless added to
	$select = 'SELECT *';
	$from   = ' FROM tbljob';
	$where  = ' WHERE TRUE';

		$companyid =  mysqli_real_escape_string($link, $_GET['company']);

	if ($companyid != '') // A company is selected
	{
			$where .= " AND companyid='$companyid'";
	}
	
	$locationid =  mysqli_real_escape_string($link, $_GET['location']);

	if ($locationid != '') // A location is selected
	{

		$where .= " AND locationid='$locationid'";
	}
	
	$categoryid =  mysqli_real_escape_string($link, $_GET['category']);
	if ($categoryid != '') // A category is selected
	{

		$where .= " AND categoryid='$categoryid'"; 
	}
//matches categoryid in table with the categoryid in form
	
	$jobdesc = mysqli_real_escape_string($link, $_GET['jobdesc']);
	if ($jobdesc != '') // Some search text was specified
	{
		$where .= " AND jobdesc LIKE '%$jobdesc%'";
	}


$result = mysqli_query($link, $select . $from . $where);
	
	if (!$result)
	{
		$error = 'Error fetching jobs line109.';
		include 'error.html.php';
		exit();
	}
		while ($row = mysqli_fetch_array($result))
	{
		$jobs[] = array( 'jobtitle' => $row['jobtitle'], 'jobref' => $row['jobref'], 'jobsalshow' => $row['jobsalshow'], 'locationid' => $row['locationid'], 'location' => $row['location']);
	
	}


	include 'jobs.html.php';
	exit();
}

I have a problem when displaying the results. i can’t figure out how to link my locationid (tbljobs.locationid) to location (tblLocation.location).
How do I put ‘location’ => $row[‘location’] into the jobs array? I can’t figuere out how i can change the $result variable output or if that’s even what i do…
Any tips at all would be be brill - thanks a mill, Lynn