如果创建一个按位置id索引的数组，则可以使用索引为指定位置添加子项：

$locations_by_id = [];

foreach($locations as $location) {
    $location['children'] = []; //initialize children to an empty array
    $locations_by_id[$location['id']] = $location;
}

foreach($rooms as $room) {    
    //add room to location
    $locations_by_id[$room['locationId']]['children'][] = $room;
}

$locations = array_values($locations_by_id); //removes the id index

print json_encode($locations);

我注意到在你的一条评论中你正在使用一个数据库，所以我决定继续添加一个答案。另一个选择是通过在查询中连接位置和房间来避免一开始就有两个数组。我对表/列的名称做了一些假设，但是这样的查询应该可以工作：

SELECT l.id AS l_id, l.name AS l_name, r.id AS r_id, r.name AS r_name
FROM locations l LEFT JOIN rooms r ON l.id = r.locationId
ORDER BY l.id, r.id

然后，在从结果中提取行时，可以使用位置ID作为键来构造数组。

while ($room = $query->fetch()) { // fetch method depends on what db extension you're using
    $rooms[$room['l_id']]['name'] = $room['l_name'];
    $rooms[$room['l_id']]['id'] = $room['l_id'];
    $rooms[$room['l_id']]['children'][] = array('name' => $room['r_name'], 
                                                'id'   => $room['r_id']);
}

echo json_encode(array_values($rooms));

您可以使用locations数组的所有id创建一个数组，然后使用array_search将每个rooms数组直接添加到locations数组：

$index = array_column( $locations, 'id' );
foreach( $rooms as $key => $val )
{
    $found = array_search( $val['locationId'], $index );
    $locations[$found]['children'][] = array( 'name' => $val['name'], 'id' => $val['id'] );
}

$json = json_encode( $locations );

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