Display The Number Of Times A Location Appeared In The Respective Table After Corresponding Option From Dropdown List Was Chosen
I would like to ask help in displaying the number of times a location value appeared in a table that corresponds to an option user clicks from dropdown list. Here's what I did so f
Solution 1:
If I understood correctly, I believe your problem is exactly the same as last one. You're just missing a WHERE clause.
See if this works:
$con = new mysqli("localhost" ,"root" ,"" ,"user_databases");
//query buildings table for the dropdown
$bquery = mysqli_query($con, "SELECT building_ID, building_name FROM buildings");
$selectedbldg = null;
// if the form was submitted
if (!empty($_POST['bldg'])) {
// store selected building_ID
$selectedbldg = $_POST['bldg'];
// the subquery is used to count how many times each location appears
// for a particular building
$count = mysqli_query($con, "
SELECT lo.location_ID, lo.location_name, dt.num_visits
FROM location lo
JOIN (
SELECT location_ID, COUNT(location_ID) AS num_visits
FROM delivery_transaction
WHERE building_ID = {$selectedbldg}
GROUP BY location_ID
) AS dt ON lo.location_ID = dt.location_ID
");
// like before, better to use prepared statement
}
?>
<!-- ... --><sectionclass="row text-center placeholders"><divclass="table-responsive"><tableclass="table table-striped"><thead><tr><th>Location</th><th>Number of Visits</th></tr></thead><tbody><!-- PHP alternative syntax for control structures: easier to read (imo) --><!-- isset function is to ensure variable $count exist as it only gets declared in the IF condition (you would get an error otherwise) --><!-- mysqli_num_rows is to check if there are any results to loop over --><?phpif (isset($count) && mysqli_num_rows($count)) : ?><?phpwhile($row = mysqli_fetch_assoc($count)) : ?><tr><td><?=$row['location_ID'] ?></td><td><?=$row['num_visits'] ?></td></tr><?phpendwhile?><?phpelse : ?><tr><td>No results to display</td></tr><?phpendif?></tbody></table></div></section>
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