拆分PHP和循環邏輯

你好，我有一個邏輯的問題，使我的測試完成這是我的數組拆分PHP和循環邏輯

Array ( 
[0] => Array (
    [id] => 1, 
    [p] => 150, 
    [w] => 120, 
), 
[1] => Array (
    [id] => 2, 
    [p] => 10, 
    [w] => 20, 
), 
[2] => Array (
    [id] => 3, 
    [p] => 70, 
    [w] => 10, 
), 
[3] => Array (
    [id] => 4, 
    [p] => 100, 
    [w] => 45, 
), 
[4] => Array (
    [id] => 5, 
    [p] => 110, 
    [w] => 500, 
) 
)

更新代碼

$pt = 0; 
foreach($data as $k => $plist){ 
    $pl[] = $plist['id']; 
    $pt += $plist['p']; 
    if($pt >= 250) break; 
} 
// Get filter max p 250 
foreach($data $k => $dat){ 
    foreach($pl as $paa => $pat){ 
     unset($data[$paa]); // Delete list key of Data 
    } 
    $jres[] = array("id" => $dat['id'],"p" => $dat['p'], "w" => $dat['w']); 
} 
// callback function and filter that will be printed only 
foreach($jres as $k => $dat){ 
    foreach($data as $paa => $pat){ 
     unset($jres[$paa]); // Delete list key of JRES 
    } 
} 
foreach($jres as $k => $dat){ 
    $subtotal += $dat['p']; 
} 
$subs = $subtotal + $pt; 
end($pl);   
$end = key($pl); 
if($subs > 250){ 
    unset($pl[$end]); // delete one key if total **p** sum total **price * w ** if above 250 will delete one key  
} 
$pack = '<table class="table"> 
    <thead> 
     <h3 style="background:#D8D8D8;">Package 1</h3> 
     <tr> 
      <td class="right">P</td> 
      <td class="right">W</td> 
     </tr> 
    </thead><tbody>'; 
foreach($jres as $k => $dat){ 
$subtotal += $dat['p']; 
$subw += $dat['w']; 
$list .='<tr> 
      <td class="right">$'.$dat['p'].'</td> 
      <td class="right">'.$dat['w'].$k.' gram</td> 
     </tr>'; 
    foreach($data as $dat2 => $pat){ 
     if($k == $dat2) unset($data[$k]); // Delete key from data 
    } 
} 
$gtotal = $subw * $sfee + $subtotal; 
$total = '<tr> 
      <th class="right" colspan="2"> Subtotal: </th> 
      <th class="right"> $'.$subtotal.' </th> 
     </tr> 
     <tr> 
      <th class="right" colspan="2">Shipping Cost ['.$subw.'g]: </th> 
      <th class="right"> $'.$subw * $sfee.' </th> 
     </tr> 
     <tr> 
      <th class="right" colspan="2"> Grandtotal: </th> 
      <th class="right"> $'.$gtotal.' </th> 
     </tr> 
     <tr> 
      <th class="right"></th> 
      <th class="right"></th> 
     </tr> 
    </tbody> 
</table>';

的目標是如何找到最小的「W 「在視圖分割中

如果總計爲250，則每個分割視圖的」p「的最大值

價格/W是0.25將總和與總P的

所以我要嘗試用循環一些邏輯，但它如此複雜，無法正常工作

你能不能幫我寫的邏輯這個的？

我上面的代碼只是顯示第一個分割視圖。

樣本輸出我希望它是這樣的：

First Split 
| id| w | p | 
| 2| 20| 10| 
| 3| 10| 70| 
| 4| 45|100| 
| stotal|180| 
| wtotal|18.75| // $w = 20 + 10 + 45 * 0.25; 
| gtotal|198.75| 

SecondSplit 
| id| w | p | 
| 1|120|150| 
| stotal|150| 
| wtotal| 30| // $w = 120 * 0.25; 
|Gtotal |180| 

Third Split 
| id| w | p | 
| 5|500|110| 
| stotal|110| 
| wtotal|125| // $w = 500 * 0.25; 
|Gtotal |235|

來源

2015-09-09 Alan El-nino Malmsteen

嗨，也許你可以告訴我們你嘗試過什麼使用循環，所以我們可以從那裏開始，看看別人能否爲您提供更好的解決方案。 – kdlcruz

如果你現在沒有發佈你正在使用的代碼，你至少應該發佈一個可用的PHP數組，供人們用來爲你粉碎一些東西 - print_r的輸出沒有多大用處，手動重寫數組 –

這將找到您的陣列中的最低值：

<?php 

$Arrays = Array ( 
'0' => Array (
    'p' => 150, 
    'w' => 120, 
), 
'1' => Array (
    'p' => 10, 
    'w' => 20, 
), 
'2' => Array (
    'p' => 70, 
    'w' => 10, 
), 
'3' => Array (
    'p' => 100, 
    'w' => 45, 
), 
'4' => Array (
    'p' => 110, 
    'w' => 500, 
) 
); 

$Encode = json_encode($Arrays); 

$Decoded = json_decode($Encode); 



foreach($Decoded as $Item){ 
$Num[] = $Item->w; 

} 
echo min($Num); //Will echo 10

來源

2015-09-09 01:38:26

拆分PHP和循環邏輯

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