count(): Parameter must be an array or an object that implements Countable in php 7.2.1

Question

count(): Parameter must be an array or an object that implements Countable in php 7.2.1 using Yii2

How to resolve this issue?

 public static function findAdminByUsername($username)
{
    $adminUser = static::find()->where(['username' => $username,'userType'=>'ADMIN','user_status'=>self::STATUS_ACTIVE])->one();
    if(count($adminUser)>0){
        return $adminUser;
    }else{
        return null;
   }
}

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. — Yupik
– Yupik, Commented Feb 14, 2018 at 6:58
if my function returns object, I need to convert it to array or viceversa — Dev
– Dev, Commented Feb 14, 2018 at 7:02
Still it's unclear what you're asking, add some relevant code and describe problem — Yupik
– Yupik, Commented Feb 14, 2018 at 7:06
where is your code in which you are retrieving the object and then trying to count , add the code relevant to your problem — Muhammad Omer Aslam
– Muhammad Omer Aslam, Commented Feb 14, 2018 at 7:21

Muhammad Omer Aslam · Accepted Answer · 2019-12-23 22:47:40Z

4

The thing is you are checking for count > 1 with using ->one() which looks odd, looking at your code you want to return NULL if there is no record found and if you look into documentation the function one() already returns NULL if there are no records found so you are adding extra code and it could easily be reduced to

public static function findAdminByUsername($username)
{
    return static::find()->where(
        [
            'username' => $username,
            'userType' => 'ADMIN',
            'user_status' => self::STATUS_ACTIVE,
        ]
    )->one();
}

edited Dec 23, 2019 at 22:47

answered Feb 14, 2018 at 8:29

Muhammad Omer Aslam

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1 Comment

Dev Over a year ago

Appropriate one and very well explained. Thank you!

ScaisEdge · Accepted Answer · 2018-02-14 13:50:43Z

2

You are using find()......->one() so your query should return just an object .. without iteration capabilities.

if you want check if the find() return a value or not then you could check with isset. find()->one() return null if the query fail.

  public static function findAdminByUsername($username)
  {
      $adminUser = static::find()->where(['username' => $username,'userType'=>'ADMIN','user_status'=>self::STATUS_ACTIVE])->one();
      if( $adminUser !== null ){
          return $adminUser;
      }else{
          return null;
     }
  }

if you don't need others that return the result for find()->..one() you could simply return

return static::find()->
  where(['username' => $username,'userType'=>'ADMIN','user_status'=>self::STATUS_ACTIVE])
     ->one();

edited Feb 14, 2018 at 13:50

answered Feb 14, 2018 at 8:13

ScaisEdge

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2 Comments

NikiC Over a year ago

Never use isset() on plain variables. Use $adminUser !== null instead. At which point it also becomes obvious that the check is entirely useless in this case, as you just end up returning the same value anyway.

ScaisEdge Over a year ago

@NikiC the find()-->one() in yii2 return an object when a row is found but return null if nothings is found so isset(..) should be correct ..

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