How do i make each movie appear on under their respective date - php

Question

i cant seem to make each movie appear under their respective date and have a correct time dropdown menu for the movie time selection
$sql_date = "SELECT DISTINCT sDate, sTitle FROM movieScreenings ";
$result_date = mysqli_query($db, $sql_date);
while($row = mysqli_fetch_array($result_date)) {
echo "<h2>" . $row['sDate'] . "</h2>";
$sql_movie = "SELECT * FROM movieList, movieScreenings WHERE title = '" . $row['sTitle'] . "'";
$result_movie = mysqli_query($db, $sql_movie);
while($row2 = mysqli_fetch_array($result_movie)) {
echo "<div class='box'>
<img class='poster' src='posters/" . $row2['poster'] . "'/>
<h2>" . $row2['title'] . "</h2>
<p>" . $row2['description'] . "</p>";
$sql_time = "SELECT DISTINCT sTime FROM movieScreenings WHERE sTitle = '" . $row2['title'] . "' AND sDate = '" . $row['sDate'] . "'";
$result_time = mysqli_query($db, $sql_time);
while($row3 = mysqli_fetch_array($result_time)) {
echo "<select name='sTime'>
<option value='" . $row3['sTime'] . "'>" . $row3['sTime'] . "</option>
</select>";
echo "</div>";
}
}
}
mysqli_free_result($result_date);
mysqli_free_result($result_movie);
mysqli_free_result($result_time);
// Close connection
mysqli_close($db);

Answer 1

You probably want to sort your first query by sDate with ORDER BY sDate.
In your second query you are again selecting from movieScreenings which is not necessary because I imagine all the data you are accessing (poster, title, description) is stored in movieList.
Thirdly, in your last while loop you are repeating the <select> tag for every screening time. This will result in multiple dropdown inputs. Also, you are including the closing </div> tag in this loop. So only loop the <option> tag.
With these changes your code would look like this:
$sql_date = "SELECT DISTINCT sDate, sTitle FROM movieScreenings ORDER BY sDate ASC";
$result_date = mysqli_query($db, $sql_date);
while($row = mysqli_fetch_array($result_date)) {
echo "<h2>" . $row['sDate'] . "</h2>";
$sql_movie = "SELECT * FROM movieList WHERE title = '" . $row['sTitle'] . "'";
$result_movie = mysqli_query($db, $sql_movie);
while($row2 = mysqli_fetch_array($result_movie)) {
echo "<div class='box'>
<img class='poster' src='posters/" . $row2['poster'] . "'/>
<h2>" . $row2['title'] . "</h2>
<p>" . $row2['description'] . "</p>";
$sql_time = "SELECT DISTINCT sTime FROM movieScreenings WHERE sTitle = '" . $row2['title'] . "' AND sDate = '" . $row['sDate'] . "'";
$result_time = mysqli_query($db, $sql_time);
echo "<select name='sTime'>";
while($row3 = mysqli_fetch_array($result_time)) {
echo "<option value='" . $row3['sTime'] . "'>" . $row3['sTime'] . "</option>";
}
echo "</select>";
echo "</div>";
}
}
...
And lastly, I would suggest using ID's as your foreign keys to prevent duplicates in the future. There might be multiple movies with the same title.