PHP - Display Yes No on the feature name - php

Question

Here is my database:
Here is the code that I tried:
<?php
//---DB Connection---
//---FIND ALL AMBANK RECORD---
$q = "SELECT *
FROM mtid
WHERE provider = Ambank";
$r = mysqli_query($dbc, $q);
//---CREATE TABLE---
echo '<table border="1" style="border-collapse:collapse">';
echo '<tr>';
echo '<th>Serial No</th>';
echo '<th>Feat A?(Y/N)</th>';
echo '<th>Feat B?(Y/N)</th>';
echo '<th>Feat C?(Y/N)</th>';
echo '<th>Feat D?(Y/N)</th>';
echo '</tr>';
while($row = mysqli_fetch_array($r, MYSQLI_ASSOC))
{
//---FIND FEATURE ID---
$q1 = "SELECT feat_ID
FROM functionfeatures
WHERE sNo = '".$row['sNo']."' && mmID = '".$row['mmID']."' &&
ttID = '".$row['ttID']."'";
$r1 = mysqli_query($dbc, $q1);
//---[HERE]---
echo '<tr>';
echo '<td>'.$row['sNo'].'</td>';
...
echo '</tr>';
}
echo '</table>';
?>
I want to make a table like below:
This table is consists of 5 fields, which is serial number, feature A, feature B, feature C, feature D. From my database, serial number 11-11-11-11 have feature B, feature C, feature D. Thus, I want to display No on feature A, Yes on the feature B, feature C and feature D in table.
However, I am stuck on how to compare the feature ID and display Yes/No on the feature name.
How should I continue from the [HERE] section?
Can someone help me please?

Answer 1

You can use only one query for that:
$query = "SELECT
m.sNo,
f.feat_name,
CASE
WHEN ff.sNo IS NULL THEN 'No'
ELSE 'Yes'
END AS yesno
FROM
mtid m
JOIN features f
LEFT JOIN functionfeatures ff ON ff.feat_ID = f.ID AND ff.sNo = m.sNo
WHERE provider = 'Ambank'
ORDER BY
m.sNo,
f.feat_name";
$result = mysqli_query($link, $query) or die(mysqli_error($link));
$sNo = '';
echo '<table border="1">
<tr>
<td>sNo</td>
<td>Feat A</td>
<td>Feat B</td>
<td>Feat C</td>
<td>Feat D</td>
</tr>
<tr>';
while($row = mysqli_fetch_array($result)){
if($sNo != $row['sNo']){
if($sNo != '') echo '</tr><tr>';
echo '<td>'.$row['sNo'].'</td>';
$sNo = $row['sNo'];
}
echo '<td>'.$row['yesno'].'</td>';
}
echo '</tr>
</table>';
Output is:

Answer 2

Try this:
while($row = mysqli_fetch_array($r, MYSQLI_ASSOC))
{
// set default values
$array = array(
"1" => "NO",
"2" => "NO",
"3" => "NO",
"4" => "NO"
);
$q1 = "SELECT feat_ID
FROM functionfeatures
WHERE sNo = '".$row['sNo']."' && mmID = '".$row['mmID']."' &&
ttID = '".$row['ttID']."'";
$r1 = mysqli_query($dbc, $q1);
// if we have a feat_ID match - set its value in the array to YES
while($row1 = mysqli_fetch_array($r1, MYSQLI_ASSOC))
{
$array[''.$row1['feat_ID'].''] = "YES";
}
echo '<tr>';
echo '<td>'.$row['sNo'].'</td>';
// output the values for each function ID
foreach ($array as $yesno) {
echo '<td>'.$yesno.'</td>';
}
echo '</tr>';
}

Answer 3

I would approach it a bit different. Using JOINs as mentioned but also separating the code to separate your front-end from back-end (Read up on The MVC Pattern):
Your back-end handles the data, obtains the SQL results and builds the array needed for the front-end:
<?php
// $dbc is set in your ---DB Connection---
$sql = "SELECT m.sNo, f.feat_name, ff.feat_ID
FROM mtid m
JOIN features f
LEFT JOIN functionfeatures ff ON ff.feat_ID = f.ID AND ff.sNo = m.sNo
WHERE provider = 'Ambank'
ORDER BY m.sNo, f.feat_name";
$result = mysqli_query($dbc, $sql) or die(mysqli_error($dbc));
$yes_no_feats = [];
while ($row = mysqli_fetch_assoc($result)) {
$yes_no_feats[$row['sNo']][] = $row;
}
mysqli_close($dbc);
?>
As you can see the variable $yes_no_feats holds the result which is then passed to the front-end so that it can then be parsed with HTML:
<table border="1">
<tr>
<th>Serial No</th>
<th>Feat A</th>
<th>Feat B</th>
<th>Feat C</th>
<th>Feat D</th>
</tr>
<?php
foreach ($yes_no_feats as $sno => $features) {
?>
<tr>
<td><?= $sno ?></td>
<?php foreach ($features as $f) { ?>
<td><?= (!empty($f['feat_ID'])) ? "Yes" : "No" ?></td>
<?php } ?>
</tr>
<?php
}
?>
</table>
Hope this helps.