我正在尝试检查数组元素是否已经存在，如果不存在，我需要创建一个数组元素，只填充一个值，第二个值设置为null。增加的复杂性是我需要在检查数组时忽略第二级，而不必再次遍历数组，因为它可能是一个非常大的数组。
我的数组看起来像这样：
Array
(
[2016-05-28] => Array
(
[0] => Array
(
[store] => 1
[price] => 12
)
[1] => Array
(
[store] => 7
[price] => 18
)
[2] => Array
(
[store] => 9
[price] =>
)
)
)
我正在尝试检查是否存在具有商店值x的现有元素，如果它不存在，我创建一个新元素，如果它存在，我会忽略它并继续前进。
对于这个例子，我对$day和$store变量进行了硬编码，但这通常会在for循环中填充，然后下面的代码片段将在for循环中运行。
我的代码：
$day = '2016-05-28';
$store = 8;
if (!$history[$day][][$store]) {
$history[$day][] = array(
"store" => $store
, "price" => null
);
}
问题在于检查元素是否存在if (!$history[$day][][$store]) {，是否可以忽略$day元素和$store元素之间的第二级，以便它检查{{1} 1}}元素，看看它是否存在，我可以使用外卡还是会store工作？
这是我目前正在使用的完整代码。
in_array答案 0 :(得分：2)
我会遍历所有日期。对于每一天，循环遍历您希望找到的所有商店编号。使用array_filter查找所需的商店。如果您找不到所需的商店，请添加它。
$required_stores = [1,2,3,4]; // stores you wish to add if missing
$source = [
'2016-06-15'=>[
['store'=>1,'price'=>10],['store'=>2,'price'=>10],
],
'2016-06-16'=>[
['store'=>1,'price'=>10],['store'=>3,'price'=>10],
],
'2016-06-17'=>[
['store'=>3,'price'=>10],['store'=>4,'price'=>10],
],
];
//go through all dates. Notice we pass $stores as reference
//using "&" This allows us to modify it in the forEach
foreach ($source as $date => &$stores):
foreach($required_stores as $lookfor):
//$lookfor is the store number we want to add if it's missing
//will hold the store we look for, or be empty if it's not there
$found_store = array_filter(
$stores,
function($v) use ($lookfor){return $v['store']===$lookfor;}
);
//add the store to $stores if it was not found by array_filter
if(empty($found_store)) $stores[] = ['store'=>$lookfor,'price'=>null];
endforeach;
endforeach;
// here, $source is padded with all required stores
答案 1 :(得分：2)
正如Rizier123建议的那样，你可以使用array_column（）。 Yous可以编写一个简单的函数来接受存储数，通过引用和日期接受历史数组：
var outerObj = {
nestedObj: {
nestedObjMethod: function(key) {
var URLS = {
'url1': 'http://foo.com',
'url2': 'http://bar.com',
'url3': 'http://yay.com'
};
return URLS[key];
}
}
};
console.log(outerObj.nestedObj.nestedObjMethod('url1'));答案 2 :(得分：0)
<?php
$history = array(); // Assuming that's array's identifier.
$history['2016-05-28'] = array (
array('store' => 1, 'price' => 12),
array('store' => 7, 'price' => 18),
array('store' => 9, 'price' => 20)
);
// variables for the if condition
$day = '2016-05-28';
$store = 8;
$match_found = FALSE;
foreach($history[$day] as $element) {
if ($element['store'] == $store) {
$match_found = TRUE;
}
else {
continue;
}
}
if ($match_found == TRUE) {
// I included a break statement here. break works only in iterations, not conditionals.
} else {
array_push($history[$day], array('store' => $store, 'price' => null));
// I was pushing to $history[$date] instead of $history[$day] since the variable I created was $day, NOT $date
}
我重写了PHP代码段，因为键值声明给出了一些错误。例如，2016-05-28键元素应该是字符串或整数，根据PHP规范（http://php.net/manual/en/language.types.array.php）。所以你的代码片段就像上面的代码一样。
我编辑了代码，将数据附加到主日期元素而不是根
答案 3 :(得分：0)
一些嵌套循环应该这样做，我认为你可能想要创建一个新的日期元素，如果它在历史中也不存在。
代码被评论：
<?php
$history = Array
(
'2016-05-28' => Array
(
0 => Array
(
'store' => 1,
'price' => 12
),
1 => Array
(
'store' => 7,
'price' => 18
),
2 => Array
(
'store' => 9,
'price' => null
)
)
);
print_r($history);
$day = '2016-05-28';
$store = 8;
// loop through dates
foreach ($history as $key=>&$date){
// scan for date
$found_date = false;
if ($key != $day) continue;
$found_date = true;
// scan for store
foreach ($date as $item){
$found_store = false;
if ($item['store'] != $store) continue;
$found_store = true;
// stop looping if store found
break;
}
// create null element
if (!$found_store) {
$date []= array(
"store" => $store
, "price" => null
);
}
// stop looping if date found
break;
}
// if date not found, create all elements
if (!$found_date) {
$history[$day]= array(
0 => array(
"store" => $store
, "price" => null
)
);
}
print_r($history);
在：
Array
(
[2016-05-28] => Array
(
[0] => Array
(
[store] => 1
[price] => 12
)
[1] => Array
(
[store] => 7
[price] => 18
)
[2] => Array
(
[store] => 9
[price] =>
)
)
)
后：
Array
(
[2016-05-28] => Array
(
[0] => Array
(
[store] => 1
[price] => 12
)
[1] => Array
(
[store] => 7
[price] => 18
)
[2] => Array
(
[store] => 9
[price] =>
)
[3] => Array
(
[store] => 8
[price] =>
)
)
)
答案 4 :(得分：0)
感谢@ trincot评论的帮助，我设法得到了我想要做的事情，使用商店ID作为数组中的键，下面是工作代码。
$setPriceHistoryData = $daoObj->getSetPriceHistoryData($set['id']);
$chartDays = date('Y-m-d', strtotime('-30 days'));
$endDay = date('Y-m-d');
$priceHistoryData = array();
while ($chartDays <= $endDay) {
for ($i = 0; $i < count($setPriceData["price_history_store_data"]); $i++) {
$store = $setPriceData["price_history_store_data"][$i]["id"];
for ($j = 0; $j < count($setPriceHistoryData); $j++) {
if ($store == $setPriceHistoryData[$j]["vph_store"]
&& $chartDays == $setPriceHistoryData[$j]["vph_date"]
&& !isset($priceHistoryData[$chartDays][$store])) {
$priceHistoryData[$chartDays][$store] = $setPriceHistoryData[$j]["vph_price"];
} else {
if (!isset($priceHistoryData[$chartDays][$store])) {
$priceHistoryData[$chartDays][$store] = null;
}
}
}
}
// Increment day
$chartDays = date('Y-m-d', strtotime("+1 day", strtotime($chartDays)));
}
答案 5 :(得分：0)
由于您愿意更改数据结构，因此只需几行简单的代码即可实现另一种非常酷的方法。首先将数据源更改为新的形状，其中商店ID是键，价格是值
$history = [
'2016-06-15'=>[
1=>10, 2=>10, //On June 15, store 1 had price 10
],
'2016-06-16'=>[
1=>10, 3=>10,
],
'2016-06-17'=>[
3=>10, 4=>10,
],
];
现在你所要做的就是在你的源代码中循环，并在数组上使用+运算符添加任何缺少的存储（它会添加缺少的键）
$required_stores = [1,2,3,4]; // stores you wish to add if missing
//will be [1=>null, 2=>null, 3=>null, 4=>null]
$required_keys = array_combine(
$required_stores,
array_fill(0,count($required_stores),null)
);
//go through each day and add required keys if they're missing
foreach ($history as &$stores):
$stores += $required_keys
endforeach;
array_combined returns an array使用第一个参数作为键，第二个参数作为值。