CODEWITHSUNDEEP
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Kyle
Kyle

Reputation: 735

Yii2 using data from array in query

I have this bit of code which finds the ID of the logged in user, finds all of the customers that are related to that user and stores the 'custref'(id) into an array: 

 $chan = Yii::$app->user->identity->typedIdentity->getId();
 $cust = Customer::find()->where(['channelref' => $chan])->all();
 foreach($cust as $row){
        $ref[] = $row->custref;

}

I want to use the values in $ref inside of a query, something like this: 

$query = $model::findNongeo()->where(['custref'=>$ref->custref])->all();

But I have no idea how to do it, I have tried all sorts. Any help would be massively appreciated.

Upvotes: 0

Views: 2205

Answers (4)

Insane Skull
Insane Skull

Reputation: 9358

$ref = Customer::find()
    ->select('custref')
    ->where(['channelref' => Yii::$app->user->identity->typedIdentity->getId()])
    ->column();

$query = $model::findNongeo()
    ->where(['custref' => $ref])
    ->all();

Upvotes: 1

Naim Malek
Naim Malek

Reputation: 1184

Here is the shortest way.

$chanId = Yii::$app->user->identity->typedIdentity->getId();
$customerArray = Customer::find()->where(['channelref' => $chanId])->all();
$ref = yii\helpers\ArrayHelper::map($customerArray,'custref','custref');

You can use in other query like

$query = $model::findNongeo()->where(['custref'=>$ref])->all();

OR Like 

$query = $model::findNongeo()->where(['in', 'custref', $ref])->all();

Upvotes: 3

Radhe9254
Radhe9254

Reputation: 198

You are going right. Just change the code in loop like

$i = 0;
foreach($cust as $row){
        $ref[$i++] = $row->custref;

}

Upvotes: 1

mrateb
mrateb

Reputation: 2499

You can use the values in the array by using In

$model::find()->where(['in', 'custref', $ref])->all();

Upvotes: 1

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