使用PHP while循環創建高級篩選器

Question

我有一個問題，那就是我想在網站上創建一個鏈接，就像人們可以單擊鏈接來顯示某些產品僅依賴於百分比。例如，我在我的數據庫中有折扣百分比的列，它會顯示最小折扣和最大折扣。假設我們有最低和最高折扣。 $ min = 12和$ max = 94;並且我想將它們放入鏈接中，以僅顯示具有某些折扣的產品，例如過濾。下面是鏈接的例子。

<a href="#">12% to 20%</a 
    <a href="#">21% to 30%</a> 
    <a href="#">31% to 40%</a> and so on until it reaches 
    <a href="#">81% to 90%</a> and the last will be 
    <a href="#">91% to 94%</a> 

最小和最大的數字將來自數據庫中的一列，它們可以頻繁更改。我想出瞭解決方案，它的工作正常，但我的代碼太長，就像我採取了很多步驟，可以在幾行代碼中完成。我在下面粘貼了我的工作代碼，但我相信這可以減少爲幾行代碼。

$catsql25 = "SELECT MAX(down_percentage) as largest FROM hot_deals"; 
$catquery25 = mysqli_query($conn, $catsql25); 
while ($row25 = mysqli_fetch_array($catquery25, MYSQLI_ASSOC)){ 
    $largest_number = $row25['largest']; 
} 

$catsql26 = "SELECT MIN(down_percentage) as smallest FROM hot_deals"; 
$catquery26 = mysqli_query($conn, $catsql26); 
while ($row26 = mysqli_fetch_array($catquery26, MYSQLI_ASSOC)){ 
    $smallest_number = $row26['smallest']; 
} 

$array_tens = array(10,20,30,40,50,60,70,80,90,100); 


foreach ($array_tens as $value){ 

    if(($value - $smallest_number <= 10) && ($value - $smallest_number > 0)){ 
     echo '<a href="/exp.php?fst='.$smallest_number.'&lst='.$value.'"><div class="lfmen2">'; 
     echo $smallest_number." to ".$value."</div></a>"; 
     $next_num = $value + 1; 
     $next_ten = 9; 
     $stop_num = floor($largest_number/10); 
     $stop_num2 = $stop_num * 10; 
     //echo $stop_num2.'<br>'; 
     $num_rounds = $stop_num2 - $value; 
     $num_rounds2 = $num_rounds/10; 
     //echo $num_rounds2; 
     for ($i = 1; $i <= $num_rounds2; $i++){ 
     $end_num = $next_num + $next_ten; 
     echo '<a href="/exp.php?fst='.$next_num.'&lst='.$end_num.'"><div class="lfmen2">'; 
     echo $next_num; 
     echo " to "; 
     echo $end_num; 
     echo "</div></a>"; 
     $next_num += 10; 
     $end_num += 10; 
     } 
    } 
} 


foreach ($array_tens as $value2){ 

    if(($largest_number - $value2 < 10) && ($largest_number - $value2 > 0)){ 
     $lsst = $value2 + 1; 
     if($lsst != $largest_number){ 
     echo '<a href="/exp.php?fst='.$lsst.'&lst='.$largest_number.'"><div class="lfmen2">'.$lsst." to ".$largest_number."</div></a>"; 
     } 
     elseif($lsst == $largest_number){ 
      echo '<a href="/exp.php?fnly='.$largest_number.'"><div class="lfmen2">'.$largest_number.'</div></a>'; 
     } 
    } 

} 

我知道它的所有爛攤子，但.. 謝謝。

Answer 1

你可以做的第一件事情就是隻有一個SQL查詢：

$catsql = "SELECT MAX(down_percentage) as largest, MIN(down_percentage) as smallest FROM hot_deals"; 

然後你只需要一個循環：

$catquery = mysqli_query($conn, $catsql); 
while ($row = mysqli_fetch_array($catquery, MYSQLI_ASSOC)){ 
    $largest_number = $row['largest']; 
    $smallest_number = $row['smalest']; 
} 

之後，你可以做只有一個foreach循環。這兩個「如果」條件可能是在同一個循環：

foreach ($array_tens as $value) { 

    if (($value - $smallest_number <= 10) && ($value - $smallest_number > 0)) { 
     echo '<a href="/exp.php?fst='.$smallest_number.'&lst='.$value.'"><div class="lfmen2">'; 
     echo $smallest_number." to ".$value."</div></a>"; 
     $next_num = $value + 1; 
     $next_ten = 9; 
     $stop_num = floor($largest_number/10); 
     $stop_num2 = $stop_num * 10; 
     //echo $stop_num2.'<br>'; 
     $num_rounds = $stop_num2 - $value; 
     $num_rounds2 = $num_rounds/10; 
     //echo $num_rounds2; 
     for ($i = 1; $i <= $num_rounds2; $i++) { 
      $end_num = $next_num + $next_ten; 
      echo '<a href="/exp.php?fst='.$next_num.'&lst='.$end_num.'"><div class="lfmen2">'; 
      echo $next_num; 
      echo " to "; 
      echo $end_num; 
      echo "</div></a>"; 
      $next_num += 10; 
      $end_num += 10; 
     } 
    } 

    if (($largest_number - $value < 10) && ($largest_number - $value > 0)) { 
     $lsst = $value + 1; 
     if ($lsst != $largest_number) { 
      echo '<a href="/exp.php?fst='.$lsst.'&lst='.$largest_number.'"><div class="lfmen2">'.$lsst." to ".$largest_number."</div></a>"; 
     } elseif ($lsst == $largest_number) { 
      echo '<a href="/exp.php?fnly='.$largest_number.'"><div class="lfmen2">'.$largest_number.'</div></a>'; 
     } 
    } 
} 

爲了使其更具可讀性，你也可以評論你的代碼知道什麼做什麼。

這是一個很好的縮進，你是對的。

希望它有幫助。