跪求算法大神！！列表中部分元素相同的行如何合并成一行！！

[时光瞄]

我的需求是把province,city,area,address,mobile,consiginee这几个字段相同的时候合并成一行，
并且order_sn拼接起来，number累加，price拼接，total_amount累加
其实就是收货人信息完全一致的时候，要合并成一行方便仓库发货
我下面的数据结构中，就希望王俊凯的两条数据能合并成一行
高分求解啊！！
数组结构如下：

[{
	"order_sn": "20191031030746713817",
	"consignee": "易烊千玺",
	"mobile": "15019252840",
	"province": "北京",
	"city": "北京市",
	"area": "朝阳区",
	"address": "牛皮",
	"number": 30,
	"price": 5500,
	"total_amount": 165000,
	"created_at": "2019-10-31 03:07:46"
}, {
	"order_sn": "20191031031248713766",
	"consignee": "王俊凯",
	"mobile": "18834876789",
	"province": "北京",
	"city": "北京市",
	"area": "朝阳区",
	"address": "嘻嘻",
	"number": 5,
	"price": 5500,
	"total_amount": 30250,
	"created_at": "2019-10-31 03:12:48"
}, {
	"order_sn": "20191031050229713346",
	"consignee": "王俊凯",
	"mobile": "18834876789",
	"province": "北京",
	"city": "北京市",
	"area": "朝阳区",
	"address": "嘻嘻",
	"number": 90,
	"price": 5500,
	"total_amount": 544500,
	"created_at": "2019-10-31 05:02:29"
}]

[时光瞄]

引用 1 楼 zoumin0323 的回复:

$data='[{
			"order_sn": "20191031030746713817",
			"consignee": "易烊千玺",
			"mobile": "15019252840",
			"province": "北京",
			"city": "北京市",
			"area": "朝阳区",
			"address": "牛皮",
			"number": 30,
			"price": 5500,
			"total_amount": 165000,
			"created_at": "2019-10-31 03:07:46"
		}, {
			"order_sn": "20191031031248713766",
			"consignee": "王俊凯",
			"mobile": "18834876789",
			"province": "北京",
			"city": "北京市",
			"area": "朝阳区",
			"address": "嘻嘻",
			"number": 5,
			"price": 5500,
			"total_amount": 30250,
			"created_at": "2019-10-31 03:12:48"
		}, {
			"order_sn": "20191031050229713346",
			"consignee": "王俊凯",
			"mobile": "18834876789",
			"province": "北京",
			"city": "北京市",
			"area": "朝阳区",
			"address": "嘻嘻",
			"number": 90,
			"price": 5500,
			"total_amount": 544500,
			"created_at": "2019-10-31 05:02:29"
		}]';
		$data=json_decode($data,true);
		$newData=[];
		foreach($data as $k=>$v){
			$key=md5($v['province'].$v['city'].$v['area'].$v['address'].$v['mobile'].$v['consignee']);
			if(!isset($newData[$key])){
				$newData[$key]=$data[$k];
			}else{
				$newData[$key]['order_sn'].=','.$v['order_sn'];
				$newData[$key]['number']+=$v['number'];
				$newData[$key]['total_amount']+=$v['total_amount'];
				$newData[$key]['price'].=','.$v['price'];
			}
		}
		print_r($newData);
		exit;

太感谢兄弟了，解决了我的问题

[乞丐921]

$data='[{
			"order_sn": "20191031030746713817",
			"consignee": "易烊千玺",
			"mobile": "15019252840",
			"province": "北京",
			"city": "北京市",
			"area": "朝阳区",
			"address": "牛皮",
			"number": 30,
			"price": 5500,
			"total_amount": 165000,
			"created_at": "2019-10-31 03:07:46"
		}, {
			"order_sn": "20191031031248713766",
			"consignee": "王俊凯",
			"mobile": "18834876789",
			"province": "北京",
			"city": "北京市",
			"area": "朝阳区",
			"address": "嘻嘻",
			"number": 5,
			"price": 5500,
			"total_amount": 30250,
			"created_at": "2019-10-31 03:12:48"
		}, {
			"order_sn": "20191031050229713346",
			"consignee": "王俊凯",
			"mobile": "18834876789",
			"province": "北京",
			"city": "北京市",
			"area": "朝阳区",
			"address": "嘻嘻",
			"number": 90,
			"price": 5500,
			"total_amount": 544500,
			"created_at": "2019-10-31 05:02:29"
		}]';
		$data=json_decode($data,true);
		$newData=[];
		foreach($data as $k=>$v){
			$key=md5($v['province'].$v['city'].$v['area'].$v['address'].$v['mobile'].$v['consignee']);
			if(!isset($newData[$key])){
				$newData[$key]=$data[$k];
			}else{
				$newData[$key]['order_sn'].=','.$v['order_sn'];
				$newData[$key]['number']+=$v['number'];
				$newData[$key]['total_amount']+=$v['total_amount'];
				$newData[$key]['price'].=','.$v['price'];
			}
		}
		print_r($newData);
		exit;