Question

我有一套产品。根据页面状态，我显示一个产品，然后显示最多4个其他产品。产品的结果集可以是大于5种产品的任何尺寸。我想总是展示5种产品。如果可用，我想显示下面的2个产品（在结果集中）和上面的2个产品。

示例：

如果有10个结果且关键产品是5.我想显示3,4,5,6,7。

如果有10个结果且关键产品是9.我想显示6,7,8,9,10。

如果有10个结果且关键产品是1.我想显示1,2,3,4,5。

现在我正在使用min（）和max（）以及一些＆＃34; IF＆＃34; s来解决这个问题，当有一个优雅的解决方案时，它会占用大量的代码行在那里，我只是没找到它！示例数组结果集

$similar_products = array(
  array(
    "id" => 1,
    "title" => "Product One"
  ),
  array(
    "id" => 2,
    "title" => "Product Two"
  ),
  array(
    "id" => 3,
    "title" => "Product Three"
  ),
  array(
    "id" => 4,
    "title" => "Product Four"
  ),
  array(
    "id" => 5,
    "title" => "Product Five"
  ),
  array(
    "id" => 6,
    "title" => "Product Six"
  ),
  array(
    "id" => 7,
    "title" => "Product Seven"
  ),
  array(
    "id" => 8,
    "title" => "Product Eight"
  ),
  array(
    "id" => 9,
    "title" => "Product Nine"
  ),
  array(
    "id" => 10,
    "title" => "Product Ten"
  )
);
$i = 8; //change this value to test different key product array positions
$arrOut = array();
$floor = 0;
if($i <= 1) { //the key product is either in the first or second position in the array
  $floor = 0;
  $arrOut[] = $similar_products[0];
  $arrOut[] = $similar_products[1];
  $arrOut[] = $similar_products[2];
  $arrOut[] = $similar_products[3];
  $arrOut[] = $similar_products[4];
} elseif((count($similar_products)-1)-$i <= 1) {  //the key product is either in the last or second to last in the array
  $floor = count($similar_products)-5;
  $arrOut[] = $similar_products[count($similar_products)-5];
  $arrOut[] = $similar_products[count($similar_products)-4];
  $arrOut[] = $similar_products[count($similar_products)-3];
  $arrOut[] = $similar_products[count($similar_products)-2];
  $arrOut[] = $similar_products[count($similar_products)-1];
} else {  //otherwise, just grab two above and two below
  $floor = $i-2;
  $arrOut[] = $similar_products[$i-2];
  $arrOut[] = $similar_products[$i-1];
  $arrOut[] = $similar_products[$i];
  $arrOut[] = $similar_products[$i+1];
  $arrOut[] = $similar_products[$i+2];
}
$x = $floor;  //set x, our counter, to the floor (floor = the very first output postion)
foreach($arrOut as $ao) {
  if($x == $i) {  //current key product
    echo "<strong>" . $ao['id'] . ":" . $ao['title'] . "</strong><hr/>";
  } else {  //other NON key products
    echo $ao['id'] . ":" . $ao['title'] . "<hr/>";
  }
  $x++;
}

Answer 1

如果你想要你可以删除变量配置，将其压缩一点，并使其成为1-liner ;-) 我对这些东西并不擅长，所以可能有更高效和/或更短的选择。

// Set array index, starts with 0
// If you need to find with specific ID, just find the index by the ID
$primaryIndex = 4;// change this number to test

// How many extra items to show
// Must be divisible by 2
$extraToShow = 4;

// Find total items available - 1 to work with array indexes
$maxIndex = count($similar_products) - 1;

// Find the slice start
$low = min($maxIndex - $extraToShow, max(0, $primaryIndex - 1 - $extraToShow / 2));

// Slice to needed
$items = array_slice($similar_products, $low, $extraToShow + 1);

var_dump($items);

Answer 2

hash, err := contract.Send(transaction, "vote", value)

Haven没有机会在移动设备上进行测试，但这是我能想到解决问题的最简单方法，同时让你有机会在未来改变这些固定限制，而不是优雅，但可以＆＃39;看看你现在有什么比较！

以优雅的方式基于关键结果获取中间x结果

2 个答案: