5

我正在为我的目录设计一个包引擎。在这里，您可以将一定数量的产品添加到包装和折扣中。当您订购产品时，脚本必须检测哪些包裹交易适用于您的订单。

这是我的代码：

// packages
$packages["package1"] = array(1,1,2);
$packages["package2"] = array(1,2);

//orderlist
$orderlist = array(1,1,2,1,2,2);

// put the order list in a temp array
$temp_product_array = $orderlist;

foreach($packages as $pname => $package_array)
{
  $no_more_package = 0;
  do
  {
    // put the package products in a temp array
    $temp_package_array = $package_array;

    $is_packages_array = array_intersect($temp_package_array,$temp_product_array);

    // if all package values are present
    if(count($is_packages_array) == count($temp_package_array))
    {
      // add package name
      $packages_in_order[] =  $pname;

      // filter the package out of the product list but keep duplicate values
      foreach($temp_product_array as $key1 => $pid1)
      {
        foreach($temp_package_array as $key2 => $pid2)
        {
          if($pid1==$pid2)
          {
            unset($temp_product_array[$key1]);
            unset($temp_package_array[$key2]);
            break;  // after removing go to the next product to prevent double erasing
          }
        }
      }
    }
    else
    {
      $no_more_package = 1;
    }

  }
  while($no_more_package<1);
}

print_r($packages_in_order);
print_r($temp_product_array);

结果是：

Array ( [0] => package1 [1] => package1 ) Array ( [5] => 2 )

但我希望结果是：

Array ( [0] => package1 [1] => package2 ) Array ( [5] => 2 )

我试过了array_diff，array_intersect但它们都不能很好地处理重复值。

有没有人有更好/可行的方法来解决这个问题？
（PS 因为不同的来源我不能使用关联数组）

4

2 回答 2

0

我会解决这个问题。其中一部分是在列表中找到包。一个完全可以做到这一点的现有函数consecutive_values在一个可能相关的问题中被命名：在数组中搜索连续值。

可以按确切的顺序在另一个数组中定位一个数组。这可能就是你想要的。

剩下的部分是搜索包然后非常简单。如果您正确理解了您的问题，那么您也想返回剩菜：

list($found, $rest) = find_packages($packages, $orderlist);
var_dump($found, $rest);

function find_packages(array $packages, array $list)
{
    $found = array();
    foreach($packages as $name => $package) {
        # consecutive_values() is @link https://stackoverflow.com/a/6300893/367456
        $has = consecutive_values($package, $list);
        if ($has === -1) continue;
        $found[] = $name;
        array_splice($list, $has, count($package));
    }

    return array($found, $list);
}

输出：

array(2) {
  [0] =>
  string(8) "package1"
  [1] =>
  string(8) "package2"
}
array(1) {
  [0] =>
  int(2)
}

编辑：多次搜索同一个包需要稍作修改。这里创建了一个内部 while 循环，如果找不到当前包，则需要中断：

function find_packages(array $packages, array $list)
{
    $found = array();
    foreach($packages as $name => $package) {
        while (true) {
            # consecutive_values() is @link https://stackoverflow.com/a/6300893/367456
            $has = consecutive_values($package, $list);
            if ($has === -1) break;
            $found[] = $name;
            array_splice($list, $has, count($package));
        }
    }

    return array($found, $list);
}
于 2012-10-17T10:10:35.743 回答
0 
// packages
$packages["package1"] = array(1,1,2);
$packages["package2"] = array(1,2);

//orderlist
$orderlist = array(1,1,1,2,2,2);



// put the order list in a temp array
$temp_product_array = $orderlist;
$product_count_array = array_count_values($temp_product_array);

foreach($packages as $pname => $temp_package_array)
{
  $no_more_package = 0;
  do
  {
    $test_package_array = array();

    foreach($temp_package_array as $key => $pid)
    {
      // check if the product is still in the order totals 
      if(isset($product_count_array[$pid]) && $product_count_array[$pid]>0)
       {
         $product_count_array[$pid]--;
         $test_package_array[] = $pid;
       }
       else
       {
         $no_more_package = 1;
       }
    }
    // check if the found products match the package count
    if(count($temp_package_array)==count($test_package_array))
    {
      $packages_in_order[] = $pname;
    }
    else
    {
      // add the extracted products in case of incomplete package 
      foreach($test_package_array as $pid)
       {
         $product_count_array[$pid]++;
       }
    }


  }
  while($no_more_package<1);
}

print_r($packages_in_order);
print_r($product_count_array);
于 2012-10-17T12:05:29.643 回答