如果元素匹配，合并数组-算法

我在不同的商店(简化商店1、2、3)有一系列设备(ibm、索尼、戴尔)。我要做的是对所有产品的商店进行比较。在下面的例子中，唯一符合条件的商店是shop_id: 2(商店1没有dell &3号店不卖ibm)。

的例子:

[
    [  {"id": "1", "name": "ibm", "price": "22", "shop_id": "1"}, {"id": "2", "name": "ibm", "price": "27", "shop_id": "1"}, {"id": "3", "name": "ibm", "price": "21", "shop_id":" 2"} ],
    [  {"id": "4", "name": "sony", "price": "19", "shop_id": "1"}, {"id": "5", "name": "sony", "price": "21", "shop_id": "2"}, {"id": "6", "name": "sony", "price": "28", "shop_id": "3"} ],
    [  {"id": "7", "name": "dell", "price": "22", "shop_id": "2"}, {"id": "8", "name": "dell", "price": "27", "shop_id": "2"}, {"id": "9", "name": "dell", "price": "21", "shop_id": "3"} ]
]

必须转换为:

[
    [ {"id": "3", "name": "ibm", "price": "21", "shop_id": "2"}, {"id": "5", "name": "sony", "price": "21", "shop_id": "2"}, {"id": "7", "name": "dell", "price": "22", "shop_id": "2"}],
    [ {"id": "3", "name": "ibm", "price": "21", "shop_id": "2"}, {"id": "5", "name": "sony", "price": "21", "shop_id": "2"}, {"id": "8", "name": "dell", "price": "27", "shop_id": "2"}]  
]

我必须在PHP中做，但我需要的是一个算法来解决这个问题。设备的数量是无限的——目前只有3家商店，但可能会更多，所以假设它也是无限的。

到目前为止，我所拥有的几乎是有效的…但是当一个设备在一个商店中有多个交易时(它只获得第一个交易)，它就失效了。

public function getDeals($prices){
    // define return array
    $multi_deals = array();
    // get number of devices 
    $no_devices = count($prices);
    // loop over each deal for first device
    foreach ($prices[0] as $dd){
        // reset other arrays
        for ($j=1; $j<$no_devices; $j++)
            reset($prices[$j]);
        // remember deal shop
        $shop = $dd['shop_id'];
        $success = true;
        $i=0;
        // remember deal
        $multi_deal = array();
        $multi_deal[$i] = $dd;
        // loop over rest of the devices 
        while ($i < ($no_devices-1)){
            $i++;
            // only continue if found a deal from the same shop
            if ( !($multi_deal[$i] = self::getDevice($shop, $prices[$i]))){
                $success = false;
                break;
            }    
            // THIS IS WHERE PRICES ARRAY WILL BE RESET IF THE SAME DEVICE IS TWICE IN ONE SHOP        
        }
        // we looped over all devices and find deals for each one of them
        if ($success)
            $multi_deals[] = $multi_deal;
    }
}
public function getDevice($current_shop, &$deals){
    while ($deal = next($deals)){
        if ($deal['shop_id'] == $current_shop)
            return $deal;
    }
    return false;
}

我已经想了好几个小时了，希望你能给我点提示。

假设你有供应商(商店)销售产品(ibm，索尼，戴尔)。作为一名顾客，我想买1*ibm + 1*索尼+ 1*戴尔，而且必须从一家商店购买。

作为结果，我需要显示可能交易的完整列表，按商店划分

在我给出的例子中，只有一个商店有所有免费产品- shop_id: 2。更重要的是，这家商店有两笔戴尔优惠。这就是为什么我们有两个集合，因为有两种可能的组合。

我尝试了不同的方法，我想我离成功越来越近了。这里有

[
    { "shop_id": "2", "deals": [ 
        { "ibm": [ 
            {"id": "3", "name": "ibm", "price": "21", "shop_id":" 2"} 
        ] },
        { "sony": [ 
            {"id": "5", "name": "sony", "price": "21", "shop_id": "2"} 
        ] },
        { "dell": [ 
            {"id": "7", "name": "dell", "price": "22", "shop_id": "2"}, 
            {"id": "8", "name": "dell", "price": "27", "shop_id": "2"}
        ] }
    ] }
]

必须再次转换为:

[
    [ {"id": "3", "name": "ibm", "price": "21", "shop_id": "2"}, {"id": "5", "name": "sony", "price": "21", "shop_id": "2"}, {"id": "7", "name": "dell", "price": "22", "shop_id": "2"}],
    [ {"id": "3", "name": "ibm", "price": "21", "shop_id": "2"}, {"id": "5", "name": "sony", "price": "21", "shop_id": "2"}, {"id": "8", "name": "dell", "price": "27", "shop_id": "2"}]  
]