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我有一套产品。根据页面的状态，我显示了一种产品，然后显示了最多 4 种其他产品。产品的结果集可以是大于 5 个产品的任何大小。我想总是展示 5 个产品。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。

例子：

如果有 10 个结果并且关键产品是 5。我想显示 3、4、5、6、7。

如果有 10 个结果并且关键产品是 9。我想显示 6、7、8、9、10。

如果有 10 个结果并且关键产品是 1。我想显示 1、2、3、4、5。

现在我正在使用 min() 和 max() 以及一些“IF”来解决它，它需要大量的代码行，当有一个优雅的解决方案时，我只是找不到它！下面的示例数组结果集

$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++;
}
4

2 回答 2

1

如果您愿意，您可以删除变量配置，将其压缩一点，并使其成为 1-liner ;-) 我不擅长这些东西，所以可能有更高效和/或更短的选项。

// 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);
于 2018-06-11T19:02:05.120 回答
0 
<?php
// example code
$productKey = 7;
$resultsShown = 5; 
$totalResults = 10;//num_rows()?
$limit = floor($resultsShown/2);
$min = $limit;
$max = $totalResults - $limit;
if($productKey<=$min){
for($i = 1;$i<=$resultsShown; $i++){
//Display result
echo $i;
}
}
else if($productKey>$max){
for( $i = $totalResults - $resultsShown+1; $i <=$totalResults;$i++){
//Display result
echo $i;
}}
else{
for( $i = $productKey - $limit; $i<=$productKey + $limit; $i++){
//Display result
echo $i;
}
}

因为我在移动设备上，所以没有机会进行测试，但我能想到的最简单的方法是解决问题，同时让你有机会在未来改变那些固定的限制，不优雅，但看不到你有什么比较的时刻！

于 2018-06-08T22:30:56.130 回答