合并共享相同id的数组
合并共享相同id的数组
提问于 2019-08-05 00:08:20
目前我的JSON是这样的：
[
{
"user_id": 6,
"room_id": 2,
"date_from": "2019-09-15 00:00:00",
"date_to": "2020-09-01 00:00:00",
"created_at": null,
"updated_at": null,
"users": [
{
"id": 6,
"name": "Makenna",
"surname": "Boyle",
"email": "nparker@yahoo.com",
"university_id": 3
}
],
"room": {
"id": 2,
"room_number": 327,
"floor": 3,
"slots": 2,
"dorm_id": 1,
"price": 0,
"features": "[{\"feature\":\"Nemokamas WiFi\",\"feature\":\"Erdvus dušas\"}]",
"photos": "[{\"img\":\"\\/images\\/rooms\\/9_1_5d3f6735884627.19832875.jpg\"},{\"img\":\"\\/images\\/rooms\\/9_1_5d3f673589a070.51161328.jpg\"},{\"img\":\"\\/images\\/rooms\\/9_1_5d3f67358b0419.09559739.jpg\"}]",
"created_at": null,
"updated_at": null,
"dorm": {
"id": 1,
"name": "Statybininkų prospekto bendrabutis",
"university_id": 2,
"photo": "https://www.umt.edu/housing/rh/descriptions/photos/Miller%20West%20Side%20View.jpg",
"university": {
"id": 2,
"name": "Klaipėdos Universitetas",
"city": "Klaipėda"
}
}
}
},
{
"user_id": 8,
"room_id": 2,
"date_from": "2019-09-15 00:00:00",
"date_to": "2020-09-01 00:00:00",
"created_at": null,
"updated_at": null,
"users": [
{
"id": 8,
"name": "Kassandra",
"surname": "Blick",
"email": "fschuppe@armstrong.com",
"university_id": 2
}
],
"room": {
"id": 2,
"room_number": 327,
"floor": 3,
"slots": 2,
"dorm_id": 1,
"price": 0,
"features": "[{\"feature\":\"Nemokamas WiFi\",\"feature\":\"Erdvus dušas\"}]",
"photos": "[{\"img\":\"\\/images\\/rooms\\/9_1_5d3f6735884627.19832875.jpg\"},{\"img\":\"\\/images\\/rooms\\/9_1_5d3f673589a070.51161328.jpg\"},{\"img\":\"\\/images\\/rooms\\/9_1_5d3f67358b0419.09559739.jpg\"}]",
"created_at": null,
"updated_at": null,
"dorm": {
"id": 1,
"name": "Statybininkų prospekto bendrabutis",
"university_id": 2,
"photo": "https://www.umt.edu/housing/rh/descriptions/photos/Miller%20West%20Side%20View.jpg",
"university": {
"id": 2,
"name": "Klaipėdos Universitetas",
"city": "Klaipėda"
}
}
}
}
]通过函数实现：
$rooms = OccupiedRoom::with([
'users' => function ($query) {
$query->select(['id', 'name', 'surname', 'email', 'university_id']);
},
'room.dorm' => function ($query) {
$query->select('dorms.id', 'dorms.name', 'dorms.university_id', 'dorms.photo');
},
'room.dorm.university' => function ($query) {
$query->select('universities.id', 'universities.name', 'universities.city');
},
])->get();我需要的是，如果数组共享相同的room_id，它将合并到一个具有多个users的数组中，如下所示：
[
{
"room_id": 2,
"date_from": "2019-09-15 00:00:00",
"date_to": "2020-09-01 00:00:00",
"created_at": null,
"updated_at": null,
"users": [
{
"id": 6,
"name": "Makenna",
"surname": "Boyle",
"email": "nparker@yahoo.com",
"university_id": 3
},
{
"id": 8,
"name": "Kassandra",
"surname": "Blick",
"email": "fschuppe@armstrong.com",
"university_id": 2
}
],
"room": {
"id": 2,
"room_number": 327,
"floor": 3,
"slots": 2,
"dorm_id": 1,
"price": 0,
"features": "[{\"feature\":\"Nemokamas WiFi\",\"feature\":\"Erdvus dušas\"}]",
"photos": "[{\"img\":\"\\/images\\/rooms\\/9_1_5d3f6735884627.19832875.jpg\"},{\"img\":\"\\/images\\/rooms\\/9_1_5d3f673589a070.51161328.jpg\"},{\"img\":\"\\/images\\/rooms\\/9_1_5d3f67358b0419.09559739.jpg\"}]",
"created_at": null,
"updated_at": null,
"dorm": {
"id": 1,
"name": "Statybininkų prospekto bendrabutis",
"university_id": 2,
"photo": "https://www.umt.edu/housing/rh/descriptions/photos/Miller%20West%20Side%20View.jpg",
"university": {
"id": 2,
"name": "Klaipėdos Universitetas",
"city": "Klaipėda"
}
}
}
}
]回答 1
Stack Overflow用户
回答已采纳
发布于 2019-08-05 00:31:35
这段代码将帮助你实现你的目标。
$data = json_decode($json);
$newFormat = [];
foreach ($data AS $each) {
if (!isset($newFormat[$each->room_id])) {
$newFormat[$each->room_id] = $each;
} else {
$newFormat[$each->room_id]->users[] = $each->users[0];
}
}
echo "<pre>" . var_export($newFormat, true) . "</pre>";这将显示与您期望的确切结果
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/57348139
复制相关文章
相似问题