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
我的两个表要复杂得多,所以我从另一个用户的问题中获取了这段代码,这几乎是我想要的,但我不知道如何使用 PHP 正确获取输出格式
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
我想要得到的是
ford
mustang
focus
escort
我在下面发布了我的更新尝试,它有效,但我敢打赌,优秀的人可以让它变得更漂亮。
$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>';}}
}
结果:
福特
野马
焦点
护送
最佳答案
您可能需要一个以 key 作为汽车类型的数组。循环结果并根据需要附加到数组。
例如:
vehicle=[];
while($row = mysqli_fetch_assoc($result)) {
if (!array_key_exists($row['Car'], $vehicle)){
$vehicle[$row['Car']] = [];
}
$vehicle[$row['Car']][] = $row['Model'];
}
输出将如下所示:
array(1) {
["Ford"]=>
array(2) {
[0]=>
string(4) "focus",
[1]=>
string(6) "escort"
}
}
关于php - 从一个表中选择匹配值到另一个表中的多个值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48634947/