warning : creating default object from empty value in cakephp

Question

I am working on CakePHP 3.2

I have this function to find product types if exists

if ($product['product_type'] != null) {
   $getProductType = $this->ProductTypes->find()
     ->where(['title LIKE' => '%'.$product['product_type'].'%'])
     ->first();

   debug($product['product_type']);   // this is not empty
   debug($getProductType);            // this shows 'id' => 1 in array
   debug($getProductType->id);        // this shows 1

   if (!empty($getProductType)) {
      $p->product_type_id = $getProductType->id;           // line 11
      $p->subcategory_id = $getProductType->subcategory_id;
      $p->category_id = $getProductType->category_id;

      return $this->saveNewBulkProduct($product, $p->category_id, $p->subcategory_id, $p->product_type_id);
   }
}

But this is giving warning as

Creating default object from empty value on line 11

Edit 2 Output of print_r($getProductType)

App\Model\Entity\ProductType Object 
( 
   [id] => 3 
   [category_id] => 1
   ....
)

Possible duplicate of Creating default object from empty value in PHP? — Mittul Chauhan
– Mittul Chauhan, Commented Aug 24, 2016 at 6:31

Haresh Vidja · Accepted Answer · 2016-08-24 06:41:08Z

2

I think no need to declare $p and not set property and its value in $p, you can pass values of $getProductType object as mentioned below.

if ($product['product_type'] != null) {
   $getProductType = $this->ProductTypes->find()
     ->where(['title LIKE' => '%'.$product['product_type'].'%'])
     ->first();
   debug($product['product_type']);   // this is not empty
   debug($getProductType);            // this shows 'id' => 1 in array
   debug($getProductType->id);        // this shows 1
   if (!empty($getProductType)) {
      return $this->saveNewBulkProduct($product, $getProductType->category_id,  $getProductType->subcategory_id, $getProductType->id);
   }
}

answered Aug 24, 2016 at 6:41

Haresh Vidja

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Comments

Dhaval Purohit · Accepted Answer · 2016-08-24 06:33:40Z

1

you have to declare first $p = new stdClass(); on line 11

answered Aug 24, 2016 at 6:33

Dhaval Purohit

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