PHP: Check if array element exists skipping one level

Question

I'm trying to check if an array element exists already, and if it doesn't, I need to create an array element with just one value populated and the second value set to null. The added complexity is that I need to ignore the second level while checking the array without having to loop through the array again, as it can be quite a big array.

My array looks like this:

Array
(
    [2016-05-28] => Array
    (
        [0] => Array
        (
            [store] => 1
            [price] => 12
        )
        [1] => Array
        (
            [store] => 7
            [price] => 18
        )
        [2] => Array
        (
            [store] => 9
            [price] => 
        )
    )
)

I'm trying to check if there is an existing element that has a store value x and if it doesn't exist I create a new element, if it does exist I ignore it and move on.

For this example I have hard coded the $day and $store variables but this would usually be populated in a for loop and then in turn the below snippet would be run inside the for loop.

My code:

$day = '2016-05-28';
$store = 8;

if (!$history[$day][][$store]) {
    $history[$day][] = array(
        "store" => $store
        , "price" => null
    );
}

The problem is on checking if the element exists if (!$history[$day][][$store]) { , is it possible to ignore the second level between the $day element and $store element so that it will check for the store element to see if it exists, can I use a wild card or will in_array work?

Here's the full piece of code I'm currently using.

$setPriceHistoryData = $daoObj->getSetPriceHistoryData($set['id']);
$chartDays = date('Y-m-d', strtotime('-30 days'));
$priceHistoryData = array();
$endDay = date('Y-m-d');

while ($chartDays <= $endDay) {
    for ($i = 0; $i < count($setPriceData["price_history_store_data"]); $i++) {
        for ($j = 0; $j < count($setPriceHistoryData); $j++) {
            if ($setPriceData["price_history_store_data"][$i]["id"] == $setPriceHistoryData[$j]["vph_store"]
                && $chartDays == $setPriceHistoryData[$j]["vph_date"]) {
                $priceHistoryData[$chartDays][] = array(
                    "store" => $setPriceHistoryData[$j]["vph_store"]
                    , "price" => $setPriceHistoryData[$j]["vph_price"]
                );
            } else {
                if (!$priceHistoryData[$chartDays][]["store"]) {
                    $priceHistoryData[$chartDays][] = array(
                        "store" => $setPriceHistoryData[$j]["vph_store"]
                        , "price" => null
                    );
                }
            }
        }
    }

    // Increment day
    $chartDays = date('Y-m-d', strtotime("+1 day", strtotime($chartDays)));
} 

Answer 1

I would loop through all the dates. For each day, loop through all the store numbers you expect to find. Use array_filter to find the required stores. If you don't find a required store, add it.

$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

Answer 2

As Rizier123 suggested, you could go with array_column(). Yous could write a simple function that would accept a store num, a history array by reference and the day:

$history = [
    '2016-05-28' => [
        ['store' => 1, 'price' => 23],
        ['store' => 2, 'price' => 23],
        ['store' => 3, 'price' => 23]
    ]
];
$store   = 8;
$day     = '2016-05-28';
function storeHistory($store, &$history, $day)
{
    if ( ! isset($history[$day])) {
        return false;
    }
    $presentStores = array_column($history[$day], 'store');
    if ( ! in_array($store, $presentStores)) {
        $history[$day][] = ['store' => $store, 'price' => null];
    }
}

storeHistory($store, $history, $day);
var_dump($history);

array (size=1)
  '2016-05-28' => 
    array (size=4)
      0 => 
        array (size=2)
          'store' => int 1
          'price' => int 23
      1 => 
        array (size=2)
          'store' => int 2
          'price' => int 23
      2 => 
        array (size=2)
          'store' => int 3
          'price' => int 23
      3 => 
        array (size=2)
          'store' => int 8
          'price' => null

Answer 3

<?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
}

I rewrote the PHP snippet simply because the key-value declaration was giving some errors. For example the 2016-05-28 key element supposed to be either a string or an integer, according to PHP specifications ( http://php.net/manual/en/language.types.array.php ).
So your code snippet would have appeared like the code above.

I edited the code to append the data into the main date element instead of the root

Answer 4

A couple nested loops should do it, and I thought you might want to create a new date element if it doesn't exist in history either.

code is commented:

<?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);

Before:

Array
(
    [2016-05-28] => Array
        (
            [0] => Array
                (
                    [store] => 1
                    [price] => 12
                )

            [1] => Array
                (
                    [store] => 7
                    [price] => 18
                )

            [2] => Array
                (
                    [store] => 9
                    [price] => 
                )

        )

)

After:

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] => 
                )

        )

)

Answer 5

Thanks to help from @trincot's comment I managed to get what I was trying to do using the store id as the key in the array, working code below.

$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)));
}

Answer 6

Since you're willing to change the data structure, a different, really cool approach could do the job with just a few simple lines of code. First change the data source to a new shape in which the store ID is the key and the price is the value

$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,
    ],
];

Now all you have to do is cycle through the days in your source and add any missing stores using the + operator on arrays (it adds missing keys)

$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 by using the first parameter as keys and the second parameter as values.