Display results from an MDB file based on a selection using PHP - php

Question

I hope someone can help me with this?
I have a Joomla installation running and the website looks and works great.
Here's the problem, the website is for a car dealership, which means they need to display a list of their stock on the floor.
They are using a custom system to manage their stock and this system saves the data to a MS Access database.
I got it to work to a point where I can display a table from the database. (http://www.autodeal.co.za/newsite/used-car-sales-south-africa).
Now when someone clicks on the model, which is a link that will take them to another page which displays only the information relevant to the selected model.
That's what I can't figure out. The link works fine and it takes me to the page that I want, but it doesn't display the data like it's supposed to.
Please see the code below for connecting to the database and displaying the results:
<?php
$dbName = "F:/Domains/autodeal/autodeal.co.za/wwwroot/newsite/db/savvyautoweb.mdb";
// Throws an error if the database cannot be found
if (!file_exists($dbName)) {
die("Could not find database file.");
}
// Connects to the database
// Assumes there is no username or password
$conn = odbc_connect("Driver={Microsoft Access Driver (*.mdb)};Dbq=$dbName", '', '');
// This is the query
// You have to have each column you select in the format tableName.[ColumnName]
$sql = "SELECT Make, Model, Year, Price, SpecialPrice, Branch, StockNO FROM Vehicle ORDER BY Make";
// Runs the query above in the table
$rs = odbc_exec($conn, $sql);
echo "\t" . "<tr>\n";
echo "\t" . "<th>Make</th><th>Model</th><th>Year</th><th>Price</th><th>Special Price</th><th>Location</th><th>Stock Number</th>" . "\n";
while (odbc_fetch_row($rs))
{
$make = odbc_result($rs, Make);
$model = odbc_result($rs, Model);
$year = odbc_result($rs, Year);
$price = odbc_result($rs, Price);
$specialPrice = odbc_result($rs, SpecialPrice);
$branch = odbc_result($rs, Branch);
$stockNo = odbc_result($rs, StockNO);
echo "\t" . "<tr>\n";
echo "\t\t" . "<td>" . $make . "</td><td><a href=http://www.autodeal.co.za/newsite/selected-vehicles>" . $model . "</a></td><td>" . $year . "</td><td>" . $price . "</td><td>" . $specialPrice . "</td><td>" . $branch . "</td><td>" . $stockNo . "</td>\n";
echo "\t" . "</tr>\n";
}
odbc_free_result($rs);
odbc_close($conn);
// This message is displayed if the query has an error in it
if (!$rs) {
exit("There is an error in the SQL!");
}
?>
Please see the code below to display a specific vehicle information from the table based on a selection made from the above script.
<?php
$dbName = "F:/Domains/autodeal/autodeal.co.za/wwwroot/newsite/db/savvyautoweb.mdb";
// Throws an error if the database cannot be found
if (!file_exists($dbName)) {
die("Could not find database file.");
}
// Connects to the database
// Assumes there is no username or password
$conn = odbc_connect("Driver={Microsoft Access Driver (*.mdb)};Dbq=$dbName", '', '');
// This is the query
// You have to have each column you select in the format tableName.[ColumnName]
$selected_id = intval($_GET['Id']);
$sql = "SELECT Id, Make, Model, Year, Price, SpecialPrice, Branch, StockNO, MainPic FROM Vehicle WHERE Id = Id";
// Runs the query above in the table
$rs = odbc_exec($conn, $sql);
$id = odbc_result($rs, Id);
$make = odbc_result($rs, Make);
$model = odbc_result($rs, Model);
echo $make;
echo $model;
$image_path_main = "<img src=db/vehicleImages/" . $mainPic . "/>";
echo "this is a test";
odbc_free_result($rs);
odbc_close($conn);
// This message is displayed if the query has an error in it
if (!$rs) {
exit("There is an error in the SQL!");
}
?>
EDIT So I've updated the above code based an answer received, but the individual records aren't displayed. I printed a test line and that works fine, so that tells me that there's an issue with the query? The thing is, the query works fine to display all the records in a table, but I need to display a single record when that record has been clicked.
Furthermore, the $mainPic variable above is referencing the image name from the database. The actual image isn't saved in the database; it's in a different location. I assume I need to create a variable with the actual path of the image and use the reference variables above to display the image, but it's not working.
So to recap, I need some help displaying all information from the database based on a selection.
For example: in the table, I select 323i. On a new page, I need to display all the information that's in the database about the 323i on a new page.
Is that doable and if so, could anyone please assist me in this matter.
Thank you very much in advance.

Answer 1

You are not using given ID parameter in your query:
$sql = "SELECT ... FROM Vehicle WHERE Id = Id ORDER BY Make";
you need to get $ID from user and place it into the query like:
$id = intval($_GET['id']); // assuming it is an integer
$sql = "SELECT ... FROM Vehicle WHERE Id = $id; // no need to order