PHP返回已修改数据的JSON
PHP返回已修改数据的JSON
提问于 2017-06-23 03:06:36
我希望我的类别列表是很好的格式，同时返回给用户。我从数据库中得到的信息是：
[
{
"id": 1,
"name": "pet",
"parent_id": null
},
{
"id": 2,
"name": "page",
"parent_id": null
},
{
"id": 3,
"name": "dog",
"parent_id": 1
},
{
"id": 4,
"name": "cat",
"parent_id": 1
},
{
"id": 5,
"name": "rodent",
"parent_id": 1
},...我想让它保持树形结构，比如：
{
"id": 1,
"name": "pet",
"parent_id": null,
"children": [
{
"id": 3,
"name": "dog",
"parent_id": 1
},
{
"id": 4,
"name": "cat",
"parent_id": 1
},...等。
有没有简单的方法可以做到这一点，或者我必须遍历数据库结果并创建新的有组织的数组来返回？做到这一点的最佳方法是什么？问题是子类别也可以有子类别。或者，也许我应该保留从数据库中获得的结构，并将孩子的id添加为数组(因为我无论如何都可以引用它们)？
如果你能帮助我，我将不胜感激。
谢谢。
回答 2
Stack Overflow用户
发布于 2017-06-23 03:47:01
解决方案：
function normalize_db_animals(){
$values[] = ["id" => 1, "name" => "pet", "parent_id" => null];
$values[] = ["id" => 2, "name" => "dog", "parent_id" => 1];
$values[] = ["id" => 3, "name" => "cat", "parent_id" => 1];
$values[] = ["id" => 4, "name" => "rodent", "parent_id" => 1];
$values[] = ["id" => 5, "name" => "wild", "parent_id" => null];
$values[] = ["id" => 6, "name" => "tiger", "parent_id" => 5];
$values[] = ["id" => 7, "name" => "rhino", "parent_id" => 5];
$normalize = function () use ($values) {
$tree = [];
$i = 0;
do {
$pet = $values[$i];
if ($pet['parent_id']) {
if (array_key_exists($pet['parent_id'], $tree)) {
$tree[$pet['parent_id']]['children'][] = $pet;
}
} else {
$tree[$pet['id']] = $pet;
}
$i++;
} while ($i < count($values));
return $tree;
};
$tree = $normalize();
echo json_encode($tree);
}结果：
{"1":{"id":1,"name":"pet","parent_id":null,"children":[{"id":2,"name":"dog","parent_id":1},{"id":3,"name":"cat","parent_id":1},{"id":4,"name":"rodent","parent_id":1}]},"5":{"id":5,"name":"wild","parent_id":null,"children":[{"id":6,"name":"tiger","parent_id":5},{"id":7,"name":"rhino","parent_id":5}]}}Stack Overflow用户
发布于 2017-06-23 16:15:34
试试这个
$a = json_decode('[{
"id": 1,
"name": "pet",
"parent_id": null
},
{
"id": 2,
"name": "page",
"parent_id": null
},
{
"id": 3,
"name": "dog",
"parent_id": 1
},
{
"id": 4,
"name": "cat",
"parent_id": 1
},
{
"id": 5,
"name": "rodent",
"parent_id": 4
},
{
"id": 6,
"name": "rodent",
"parent_id": 2
}]');
$a = collect($a);
$filtered = $a;
foreach ($filtered as $key => $value) {
$children = $a->where('parent_id', $value->id);
if(!$children->isEmpty()){
$value->children = $children;
$filtered->forget(array_values(array_keys($children->toArray())));
}
}
dd($filtered);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/44707507
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