CODEWITHSUNDEEP
php
www.indiawebdesigns.in
www.indiawebdesigns.in

Reputation: 99

Php in_array with session

My PHP statement looks like this

 $select_product= "SELECT * FROM `products` WHERE pro_name = '$page_name' and status = 'Active'";
 $sql101=$dbconn->prepare($select_product);
 $sql101->execute();
 $wlvd101=$sql101->fetchAll(PDO::FETCH_OBJ);
 foreach($wlvd101 as $rows101);
 
 $product_id = $rows101->id;
 // Gives me result of a sample id 101

Again I am another statement where I have fetched the ids of products from my cart. The statement looks like this:

$pidArr = array();
if(!empty($_SESSION['cart'])){
    
    foreach($_SESSION['cart'] as $id=>$val)
    {
        $rate=$val['product_mrp'] * $val['qty'];
        $total=$total+$rate;
        $pidArr[] = $val['uid'];
        $qtyArr[] = $val['qty'];
        $webArr[] = $val['ppid'];
    }
    $all_cart_products = "'" . implode("','", $pidArr) . "'";  
    //echo $all_cart_products;

    //It gives me a list of ids like this '100', '101', '102' etc
}

Now while using in_array, my statement is not working. The code looks like this:

$my_ids = $all_cart_products;

if (in_array("$product_id", $my_ids))
{
  echo "Match Found";
}
else
{
  echo "Match not found";
}

How to solve this problem?

Upvotes: 1

Views: 133

Answers (1)

AntG
AntG

Reputation: 1294

This line $all_cart_products = "'" . implode("','", $pidArr) . "'" creates a string.

Which you then assign $my_ids = $all_cart_products; so $my_ids is also now a string.

Pass it $pidArr instead.

Upvotes: 4

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