用户请求产品类别并说明他们需要的数量,即糖7磅。PHP数据处理因模糊错误而失败
在搜索结果中,从数据库中,我有以下的项目(其各自不同的产品):
糖,8盎司0.75
糖,1磅1.50
糖,2磅4.00
糖,5磅7.00
糖,10磅11.00
当前进程:
- 获取搜索结果
- 检查然而,对于结果的一个结果(仅10磅就满足吧)或其倍数的任何一个(最便宜的选择许多1需要匹配的标准,所以7 X 1磅)
- 把单个产品ID放入一个数组
- 获得1:1排列,但在其中,使用一些代码添加多达3个重复(因为5磅+ 2 X 1磅是最便宜的选择N,不是5磅+ 2磅)
- 在此,检查不同的数量单位(盎司VS磅),能够转换和比较
- 的比较最便宜,最便宜的回报
- 我甚至不顾地方有在置换超过6元剔除掉不太可能的选项,并减少开销
除非有> 9个产品ID(对于一些我有> 25)这工作得很好,然后我获取日志中的这个错误(以及浏览器中完全不相关的错误):脚本标题的提前结束
这是很多代码。我很抱歉,只想彻底!有没有更高效/更有效的方法?
function processCombos($arr, $qty_search, $qty_unit_search){ //$arr is the dataset, $qty_search is 7, $qty_unit_search is 1 for lbs
$combo=array();
$pid_arr = arrayifyProductIDs($arr);
$count = count($pid_arr);
$members = pow(2,$count);
for ($i = 0; $i < $members; $i++) {
$b = sprintf("%0".$count."b",$i);
$out = array();
for ($j = 0; $j < $count; $j++) {
if ($b{$j} == '1'){
$out[] = $pid_arr[$j];
}
}
$minLength=2;
$out_max = count($out);
if ($out_max >= $minLength) {
// now add in different repeats of each of them
$repeat_max = 3;
$indiv = array();
for($k=0;$k<$out_max;$k++){
$tmp = array();
for ($r = 0; $r < $repeat_max; $r++) $tmp[$r] = array_fill(0, $r + 1, $out[$k]);
$indiv[] = $tmp;
}
$x_ct = count($indiv[0]);
$y_ct = count($indiv[1]);
$z_ct = count($indiv[2]) > 0 ? count($indiv[2]): 0;
$perm = array();
for($x=0;$x<$x_ct;$x++){
for($y=0;$y<$y_ct;$y++){
if($z_ct > 0){
for($z=0;$z<$z_ct;$z++){
$perm = array_merge($indiv[0][$x],$indiv[1][$y],$indiv[2][$z]);
}
}else{
$perm = array_merge($indiv[0][$x],$indiv[1][$y]);
}
$p=0;
$max_p=count($perm);
if($max_p >=7){
}else{
$product_ids = array();
$qty = 0;
$price = 0;
while($p < $max_p){
$product_id = $perm[$p];
$data = $arr[$product_id];
if(!$data['qty_unit_id'] OR !$data['qty']){continue;} // go to the next one if it doens't have qty or qty_unit
if($data['qty_unit_id'] == $qty_unit_search){
$product_ids[] = $product_id;
$qty += $data['qty'];
$price += $data['price'];
}else{
$unit_to_convert_data = getQtyUnitName($qty_unit_search);
$unit_to_convert = $unit_to_convert_data['abbr'];
$unit_to_convert_type = $unit_to_convert_data['conv_file'];
if($unit_to_convert_type == $data['conv_file']){
if($data['conv_file'] == "Mass"){
$product_conv = new PhpUnitsOfMeasure\PhysicalQuantity\Mass($data['qty'], $data['qty_unit']);
}else{
$product_conv = new PhpUnitsOfMeasure\PhysicalQuantity\Volume($data['qty'], $data['qty_unit']);
}
$data['qty_CONV'] = number_format($product_conv->toUnit($unit_to_convert),3,".",",");
$product_ids[] = $product_id;
$qty += $data['qty_CONV'];
$price += $data['price'];
}
}
$p++;
}
if(count($combo)==0 AND $qty >= $qty_search){
$combo = array('product_ids' => $product_ids, 'qty' => $qty, 'price' => $price);
}elseif(($qty >= $qty_search AND $price < $combo['price']) OR
($qty >= $qty_search AND $price == $combo['price'] AND $qty > $combo['qty'])){
$combo = array('product_ids' => $product_ids, 'qty' => $qty, 'price' => $price);
}else{
}
}/// i think it should go here
}
}
}
}
return $combo;
}
嗯,在那个列表中,我们可以从理论上[直接丢弃1lb&2lb,直接在我们的解决方案中](http://en.wikipedia.org/wiki/Knapsack_problem#Dominance_relations),因为0.75磅的0.75可以合并到2个相同或更少价钱。你想这样做,还是有一个优势(额外维度),应该更喜欢更大的包装?我们是否应该找到平等的“最佳”解决方案,然后选择最少数量的解决方案? – Wrikken
@Wrikken更大的包装是最佳的,所有其他因素相同。这是在包含代码的最终if/elseif中处理的。也许为了节省时间/资源,我可以开始排除过大的数量,即10磅的产品?!? –