public function getCheckoutForm(){
$arr = array(
'cmd' => '_cart',
'business' => 'some@mail',
'no_shipping' => '1',
'upload' => '1',
'return' => 'url',
'cancel_return' => 'url1',
'no_note' => '1',
'currency_code' => 'url2',
'bn' => 'PP-BuyNowBF');
$cpt=1;
foreach($this->items as $item){
$arr1[] = array(
'item_number_'.$cpt.'' => $item['item_id'],
'item_name_'.$cpt.'' => $item['item_name'],
'quantity_'.$cpt.'' => $item['item_q'],
'amount_'.$cpt.'' => $item['item_price']
);
$cpt++;
}
return array_merge($arr,$arr1[0],$arr1[1]);
}
这会返回如下数组:
Array
(
[cmd] => _cart
[business] => some@mail
[no_shipping] => 1
[upload] => 1
[return] => url1
[cancel_return] =>url2
[no_note] => 1
[currency_code] => EUR
[bn] => PP-BuyNowBF
[item_number_1] => 28
[item_name_1] => item_name_1
[quantity_1] => 1
[amount_1] => 5
[item_number_2] => 27
[item_name_2] => item_name_2
[quantity_2] => 1
[amount_2] => 30
)
问题是返回$ arr1 [0]和$ arr1 [1]是硬编码的。如果在循环中我有超过2个数组,让我们说0,1,2,3 ans等等,这段代码将无效。任何的想法?也许我的逻辑是完全错误的......
答案 0 :(得分:4)
答案 1 :(得分:3)
我可能会做类似
的事情$count = count($arr1);
for($i=0;$i<$count;$i++){
$arr = array_merge($arr,$arr1[$i]);
}
return $arr;
答案 2 :(得分:2)
我希望,我明白,你的意思^^
foreach ($i = 0, $n = count($arr1); $i < $n; $i++) {
$arr = array_merge($arr, $arr1[$i]);
}
return $arr;
答案 3 :(得分:0)
您可以在每次迭代中进行合并:
foreach($this->items as $item){
$temp_arr = array(
'item_number_'.$cpt.'' => $item['item_id'],
'item_name_'.$cpt.'' => $item['item_name'],
'quantity_'.$cpt.'' => $item['item_q'],
'amount_'.$cpt.'' => $item['item_price']
);
$arr = array_merge($arr,$temp_arr)
$cpt++;
}
,其优势在于您可以从函数中获取$temp_arr
,
或者只是将所有元素添加到一个数组中:
foreach($this->items as $item){
$arr['item_number_'.$cpt.''] => $item['item_id'];
$arr['item_name_'.$cpt.''] => $item['item_name'];
$arr['quantity_'.$cpt.''] => $item['item_q'];
$arr['amount_'.$cpt.''] => $item['item_price'];
$cpt++;
}
答案 4 :(得分:0)
这样做
$count = count($data);
$sum = 1;
$arr = [];
for($i=0;$i<$count;$i++){
$temp = $arr;
if($i == $count - 1){
$sum = 0;
}
$arr = array_merge($temp,$data[$i + $sum]);
}
return $arr;