Php 数组和元素的值相同吗?
如何对数组中元素的相同值求和Php 数组和元素的值相同吗?,php,Php,如何对数组中元素的相同值求和 $arr = [ ['id' => 1, 'qty' => 100, 'name' => 'a'], ['id' => 1, 'qty' => 100, 'name' => 'a'], ['id' => 2, 'qty' => 100, 'name' => 'b'] ]; 成为: $arr = [ ['id' => 1, 'qty' => 200, 'name' =>
$arr = [
['id' => 1, 'qty' => 100, 'name' => 'a'],
['id' => 1, 'qty' => 100, 'name' => 'a'],
['id' => 2, 'qty' => 100, 'name' => 'b']
];
成为:
$arr = [
['id' => 1, 'qty' => 200, 'name' => 'a'],
['id' => 2, 'qty' => 100, 'name' => 'b']
];
我很努力,但还是回来了
for($i=0; $i<count($cok);$i++){
$item_id = $cok[$i]['id'];
$quantity = $cok[$i]['quantity'];
if (isset($new_items[$item_id])) {
$new_items[$item_id] = ['quantity' => $new_items[$item_id]['quantity'] + $quantity];
} else {
$new_items[$item_id] = ['quantity' => $quantity];
}
}
[1=>['qty'=>200],2=>['qty'=>100]]
我很努力,但还是回来了
for($i=0; $i<count($cok);$i++){
$item_id = $cok[$i]['id'];
$quantity = $cok[$i]['quantity'];
if (isset($new_items[$item_id])) {
$new_items[$item_id] = ['quantity' => $new_items[$item_id]['quantity'] + $quantity];
} else {
$new_items[$item_id] = ['quantity' => $quantity];
}
}
潜入我的
和inplace snippet,返回您需要的准确信息:
<?
$arr = array(
array('id' => 1, 'qty' => 100, 'name' => 'a'),
array('id' => 1, 'qty' => 100, 'name' => 'a'),
array('id' => 2, 'qty' => 100, 'name' => 'b')
);
$ids = array();
foreach ($arr as $i => $subarray) {
if (!($remove_from_array = array_key_exists($subarray['id'], $ids))) {
$ids[$subarray['id']] = 0;
}
$ids[$subarray['id']] += $subarray['qty'];
if ($remove_from_array) {
unset($arr[$i]);
}
}
foreach ($arr as &$subarray) {
$subarray['qty'] = $ids[$subarray['id']];
}
print_r($arr);
非常简单:
<?php
$arr = array(
array('id' => 1, 'qty' => 100, 'name' => 'a'),
array('id' => 1, 'qty' => 100, 'name' => 'a'),
array('id' => 2, 'qty' => 100, 'name' => 'b')
);
$new_arr = array();
foreach($arr AS $item) {
if(isset($new_arr[$item['id']])) {
$new_arr[$item['id']]['qty'] += $item['qty'];
continue;
}
$new_arr[$item['id']] = $item;
}
$arr = array_values($new_arr);
var_dump($arr);
你尝试过什么吗?foreach循环+基本数学你可以使用一个映射,它存储唯一的键,可能重复的键与我尝试的相同,但与我想要的不同