从一个表中选择匹配值到另一个表中的多个值

Question

    cars                 models
car_id  car       car_id   model_id  model
   1    ford         1         1      mustang
   2    fiat         1         2      focus
   3    toyota       1         3      escort
                     2         4      500
                     2         5      spider
                     3         6      tacoma

The two tables I have are much more complicated so I took this code from another users question, It is almost what I want, but I don't know how to get the output to format with PHP correctly

SELECT c.Car, m.Model_id, m.Model
FROM models m
INNER JOIN car c ON c.Car_id = m.Car_id
WHERE m.Car_id = (SELECT Car_id FROM models WHERE Model = 'Escort');

$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {

$model = $row['Model'];
$vehicle = $row['Car'];

    ford mustang
    ford focus
    ford escort

What I am trying to get is

    ford
        mustang
        focus
        escort

I have posted my updated attempt below, which works, but I bet you fine people could make it prettier.

$sql = "
SELECT c.Car, m.Model_id, m.Model 
FROM models m 
INNER JOIN car c   
ON     c.Car_id = m.Car_id
WHERE m.Car_id = '1' ";

$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)) {
$vehicle[$row['Car']][] = $row['Model'];
}

echo "<table>";
foreach( $vehicle as $value => $key )
{echo '<tr><td>'.$value.'</td></tr>'; 
foreach( $key as $mod) 
{echo '<tr><td>'.$mod.'</td></tr>';}}

}

Results in:
ford
mustang
focus
escort

Answer 1

You probably want an array with the key as the car type. Loop you're results and append to an array as necessary.

Ex:

vehicle=[];
while($row = mysqli_fetch_assoc($result)) {
    if (!array_key_exists($row['Car'], $vehicle)){
        $vehicle[$row['Car']] = [];
    }
    $vehicle[$row['Car']][] = $row['Model'];
}

Output will look like this:

array(1) {
  ["Ford"]=>
  array(2) {
    [0]=>
    string(4) "focus",
    [1]=>
    string(6) "escort"
  }
}

Answer 2

<?php

$mo1 = "";

$sql = "SELECT c.Car, m.Model_id, m.Model FROM models m INNER JOIN car c ON c.Car_id = m.Car_id
WHERE m.Car_id = (SELECT Car_id FROM models WHERE Model = 'Escort')";

$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
    $model = $row['Model'];
    $vehicle = $row['Car'];

 if($mo1 != $mo){$data.="<tr>
                            <td>$model</td>
                        </tr>";
                }

$data.="<tr>
            <td>$vehicle</td>
           </tr>";

$mo1=$mo;

}

?>