鉴于以下两个数组,它们如何有效地合并以产生第三个数组? 的 productData
$productData =
[
{
"product_id": 4,
"type": "electronic",
"name": "monitor",
"specs": {
"HDMI": true,
"VGA": false
}
},
{
"product_id": 5,
"type": "electronic",
"name": "HDMI cable",
"specs": {
"length": "3ft"
}
},
{
"product_id": 6,
"type": "kitchen",
"name": "spoon"
}
]
产品
$products =
{
"products": 3,
"per_page": 10,
"current_page": 1,
"data": [
{
"id": 4,
"product_type": "electronic",
"product_id": 6
},
{
"id": 6,
"type": "electronic",
"product_id": 5
},
{
"id": 9,
"type": "kitchen",
"product_id": 4
}
]
}
productsFinal ($productData
合并到$products
- 基于product_id/product_id
和type/product_type
的匹配组合
$productsFinal =
{
"products": 3,
"per_page": 10,
"current_page": 1,
"data": [
{
"id": 4,
"product_type": "electronic",
"product_id": 6,
// How to merge product data and wrap with "data" key
"data": {
"product_id": 6,
"type": "kitchen",
"name": "spoon"
}
},
{
"id": 6,
"type": "electronic",
"product_id": 5,
// How to merge product data and wrap in "data" key
"data": {
"product_id": 5,
"type": "electronic",
"name": "HDMI cable",
"specs": {
"length": "3ft"
}
}
},
{
"id": 9,
"type": "kitchen",
"product_id": 4,
// How to merge product data and wrap in "data" key
"data": {
"product_id": 6,
"type": "kitchen",
"name": "spoon"
}
}
]
}
我在foreach循环中尝试了不同的结果,但仍无法按预期进行渲染:
foreach($productData as $productDataItem) {
// when $productDataItem.product_id == $product.product_id && $productDataItem.type == $product.product_type
// move the matching $productDataItem object into matching $product object, wrapped in a new "data" key
}
答案 0 :(得分:3)
我不太了解Laravel。但是,您可以非常轻松地加入数据对象:
<?php
$productData = json_decode('[
{
"product_id": 4,
"type": "electronic",
"name": "monitor",
"specs": {
"HDMI": true,
"VGA": false
}
},
{
"product_id": 5,
"type": "electronic",
"name": "HDMI cable",
"specs": {
"length": "3ft"
}
},
{
"product_id": 6,
"type": "kitchen",
"name": "spoon"
}
]');
$products = json_decode('{
"products": 3,
"per_page": 10,
"current_page": 1,
"data": [
{
"id": 4,
"type": "electronic",
"product_id": 6
},
{
"id": 6,
"type": "electronic",
"product_id": 5
},
{
"id": 9,
"type": "kitchen",
"product_id": 4
}
]
}');
// combine both data objects
foreach($products->data As &$p) {
foreach($productData As $d) {
if(property_exists($p, "product_id") && property_exists($d, "product_id") && property_exists($p, "type") && property_exists($d, "type")) {
if($p->product_id==$d->product_id && $p->type==$d->type) {
//$p = (object) array_merge((array) $p, (array) $d);
$p->data = $d; // updated answer
continue;
}
}
}
}
echo("<pre>");
echo json_encode($products, JSON_PRETTY_PRINT);
?>
您可以在此处测试代码:http://sandbox.onlinephpfunctions.com/code/98a50c35ee32c30f0d2be1661f7afb5895174cbe 更新:http://sandbox.onlinephpfunctions.com/code/aeebfdcf4f4db5e960260e931982570cfed19e0e
答案 1 :(得分:0)
我建议检查此包dingo/api。我假设你想要显示某种JSON响应。看看Transformers。你可以这样做:
<?php
namespace App\Http\Transformers;
use App\Http\Controllers\ProductData;
use League\Fractal\TransformerAbstract;
class ProductsDataTransformer extends TransformerAbstract
{
/**
* Turn this item object into a generic array
*
* @return array
*/
public function transform(ProductData $productdata)
{
return [
'id' => $productdata->id,
'product_type' => $productdata->product_type,
'product /*or data*/' => Product::find($productdata->product_id),
];
}
}
这可以通过它的ID找到产品,如下所示:
{
"id": 4,
"product_type": "electronic",
"product" {
"product_id": 6,
"type": "kitchen",
"name": "spoon"
},
},
然后,您还可以为Product创建一个转换器来处理您的specs属性以执行相同的操作。