issues with php if statement - php

Question

ok so i am pretty new to php. i am making a site listing different attractions, i have a details page where i display information for each attraction stored in a mysql database. i'm trying to write an if statement so an icon will show if the relevant attraction has for example disabled facilities and so on. I have a tables attractions, type, facilities and faciliteslink in the database.
In the facilitieslink i have the columns:
AttractionID
FacilityID
in the Facility table i have:
FacilityID
FalilityName
in the attractions table i have:
AttractionID
name
summary ect...
also here is the if statement and queries i was trying to make
$myQuery = "SELECT Attraction.*, Type.TypeName ";
$myQuery .= "FROM Attraction ";
$myQuery .= "INNER JOIN Type ON Attraction.Type = Type.TypeID ";
$myQuery .= "WHERE AttractionID=" .$_GET['ID'];
//run query
$result = $con->query($myQuery);
if (!$result) die('Query error: ' . mysqli_error($result));
?>
<?php
$myQuery .= "INNER JOIN Type ON Attraction.AttractionID = facilitieslink.FacilityID ";
?>
<?php
//display attractions row by row
while($row = mysqli_fetch_array($result))
{
echo '<div class=" sixteen columns" id="attraction-details">';
echo '<h3>Attractions/<a href="details.php?ID=' . $row['AttractionID'] . '">' . $row['Name'] . '</a></h3>';
echo ' <div class="eight columns" id="big-a-image1">';
echo ' <img src="'. $row['ImageUrl'] . '"/>';
echo ' </div>';
echo ' <div class="eight columns" id="big-a-image2">';
echo ' <img src="'. $row['ImageUrl2'] . '"/>';
echo ' </div>';
echo ' <div class="eight columns" id="big-a-image3">';
echo ' <img src="'. $row['ImageUrl3'] . '"/>';
echo ' </div>';
echo ' <div class="eight columns" id="big-a-image4">';
echo ' <img src="'. $row['ImageUrl4'] . '"/>';
echo ' </div>';
echo' <div class="two columns" id="m-thumb1">';
echo' <img src="'. $row['ImageUrl'] . '"/>';
echo' </div>';
echo' <div class="two columns" id="m-thumb2">';
echo' <img src="'. $row['ImageUrl2'] . '"/>';
echo' </div>';
echo' <div class="two columns" id="m-thumb3">';
echo' <img src="'. $row['ImageUrl3'] . '"/>';
echo' </div>';
echo' <div class="two columns" id="m-thumb4">';
echo' <img src="'. $row['ImageUrl4'] . '"/>';
echo' </div>';
echo ' <div class="eight columns" id="a-description">';
echo' <h4> Attraction Description </h4>';
echo ' <p class="para"> ' . $row['Description'] . ' </p></a>';
echo ' </div>';
echo' <div class="two columns" id="thumb1">';
echo' <img src="'. $row['ImageUrl'] . '"/>';
echo' </div>';
echo' <div class="two columns" id="thumb2">';
echo' <img src="'. $row['ImageUrl2'] . '"/>';
echo' </div>';
echo' <div class="two columns" id="thumb3">';
echo' <img src="'. $row['ImageUrl3'] . '"/>';
echo' </div>';
echo' <div class="two columns" id="thumb4">';
echo' <img src="'. $row['ImageUrl4'] . '"/>';
echo' </div>';
// everything else works apart from this
$id = $row["AttractionID"];
if ($row['FacilityID'] == 1){
echo' <img src="'. $row['icon'] . '"/>';
}
else
echo'</div>';
echo' <div class="eight columns" id="directions">';
echo' <h3>Directions</h3>';
echo' <p class="para"> ' . $row['Address'] . ' </p>';
echo' </div>';
echo' <div class="eight columns" id="video">';
echo' <h3>Video</h3>';
echo' ' . $row['VideoEmbed'] . ' ';
echo' </div>';
echo' <div class="eight columns" id="contact">';
echo' <h3 id="con"> Contact </h3>';
echo' <p class="para"> ' . $row['Long'] . ' </p>';
echo' </div>';
echo' <div class="eight columns" id="opening-times">';
echo' <h3 id="open"> Opening times </h3>';
echo ' <p class="para"> ' . $row['OpeningHours'] . ' </p>';
echo' </div>';
}
?>

Answer 1

Regarding your main question:
"would i store the icon in this table or 1 of the others?"
Which table is "this table"? You already mentioned 3 Tables now (Attraction, Facility and Type)
What shall this Icon show? The answer to this question can tell you, where the data belongs to: Types or attraction. Can every attraction have another icon? or does the Type tell you, which icon you have to choose?
You try to join the Type by FacilityID and another time by TypeID. Are both unique in your table Type? Otherwise you're going to get multiple rows for the same Attraction. Due to your Edit i now know you want to join a third table: Facility. You have to write INNER JOIN facilitylink ON Attraction.FacilityId = facilitylink.FacilityID INNER JOIN Facility ON facilityLink.FacilityId = Facility.FacilityId to solve your problem with a join. But beware that you may get above mentioned multiple rows of attraction. Furthermore, due to the properties of an inner join, you will loose information on attraction which don't have a facility. Maybe you should switch the first INNER JOIN into a LEFT JOIN, what will produce NULL assignments to unknown Facilities. Don't forget to add the Information you additionally need (Icon, FacilityId) to the SELECT-Part. I suggest you to search further information on how to write select-statements and what effect which Statement parts will have.
The Syntax of your SQL-Statement is messed up because you append (PHP-Operator .=) a Join-Part after the Where-part. You don't execute the manipulated query, is the Manipulation needed in this case? Change the position of appending the Join-Part before the Query-Command and adjust the Order to get a correct SELECT-Statement.
You're appending a user generated input without proper escaping to the query (.$_GET['ID']). This is a very common beginners failure. Please Take a look into the topic "SQL Injection" and prevent this by using "prepared statements"
Your Appending the content of $row['icon'] directly into the generated HTML. Is the Content given by the users or do you generate/declare it? In the first case you should escape it with htmlspecialchars($row['icon'])
you're not starting the php file properly. I think you cut of some code, but just in case: all php Code need to be behind a <?php. the ?>will command the interpreter to stop interpreting php-code at all. A new <?php will start the interpreter again.
<p> HTML-Code here
<?php
echo "PHP-Code here";
echo "more PHP-Code here";
echo "much more PHP-Code here";
?> </p><p>some more html code </p><?php echo "and some PHP-Code again";
echo "but this time until the Ent of the File!";
This also means, that you have some redundant code in your example:
?>
<?php
Hopefully i have directed you in the right direction.
EDIT: Edited some of the Sentences above due to your Edit.

Answer 2

You should check if this query is good because i'm not sure about it.
The code whould go like this... Assuming the table is "attraction"
// Define query
$query = "SELECT * FROM `facilitieslinks` WHERE `AttractionID` = '".$_GET['ID']."'";
// Create connection
$con=mysqli_connect("localhost","&&&&&YOURUSERNAME&&&&&","&&&&&YOURPASSWORD&&&&&","&&&&&YOURDATABASE&&&&&");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//Execute the query to the DB
$result = mysqli_query($con,$query);
//Process data
while($row = mysqli_fetch_array($result))
{
$id = $row['AttractionID'];
if ($row['FacilityID']==1)
{
echo "<img src=\"".$row['icon'].\"">";
}
}
// Close the connection
mysqli_close($con);