Message: Cannot use object of type stdClass as array in CodeIgnitor

Question

I am using CodeIgniter, I am returning more than one value in the model so I did like

return array('true' =>1,'customer_id' => $result->customer_id);

and In the controller, I am displaying like

$true=$result['true'];
$custId=$result['customer_id'];

So there is no issue in this.

Now let's talk about in details

In the model, I have this logic

//sql query
if ($result) {
    if ($result->password_change == 0) {
         return array('true' =>1,'customer_id' => $result->customer_id);//multiple returns
    }
    else{
       return $result;
    }
}
else{return false;}

In the controller

 $result=$this->Member_model->check_data(parameters);
 $true=$result['true'];
 $custId=$result['customer_id'];

 if ($result) {
     if($true == 1){
         //some code here
     }
     else{
         //some code here
     }
 }
 else{
 }

This is my second code.

if ($result) {
  if ($result - > password_change == 0) {
    if ($result - > id == $id) {
      return array('true' => 1, 'customer_id' => $result - > customer_id); //return more then 2
    } else {
      return false;
    }
  } else {
    if ()) // some condition
  {
    return (array) $result;
  } else {
    return false;
  }
}
} else {
  return false;
}

I am getting the error that

"Message: Cannot use object of type stdClass as array"

Because when returning if condition(from the model) then it's working but when it returns else (I mean $result from the model) then I am getting the error because it is not getting the $result['true'].

Hope you can understand my issue. Would you help me out on this issue?

Well you need to accommodate the different return permutations in your controller. You are currently just assuming the return value is the array you're creating. — Mitya
– Mitya, Commented Dec 20, 2018 at 14:27
@Utkanos, Can you explain me with the help of an example? Please — user9437856
– user9437856, Commented Dec 20, 2018 at 14:34
It's like this. Your model is returning different types of data, but your controller is not checking for this - it's assuming only one of those types. So either harmonise your model to always return a given type of data, or tell your controller to first CHECK the type of returned data before proceeding. See @Richard Parnaby-King's answer for more. — Mitya
– Mitya, Commented Dec 20, 2018 at 15:03

Richard Parnaby-King · Accepted Answer · 2018-12-20 14:48:34Z

1

You have two options. Either: ALWAYS return an array, or return an ArrayObject:

Alway return an Array

In your else, cast $result as an array:

//sql query
if ($result) {
    if ($result->password_change == 0) {
         return array('true' =>1,'customer_id' => $result->customer_id);//multiple returns
    }
    return (array) $result;
}
return false;

Always return an ArrayObject

Not knowing if this model is returning the result anywhere else in you application, I have no idea if other code expects $result to be an object. Casting as an array might break your code elsewhere. Let's convert $result from stdClass to an ArrayObject:

//sql query
if ($result) {
    if ($result->password_change == 0) {
         return array('true' =>1,'customer_id' => $result->customer_id);//multiple returns
    }
    return new \ArrayObject($result, \ArrayObject::ARRAY_AS_PROPS);
}
return false;

This approach will allow you to call $result['test'] AND $result->test.

answered Dec 20, 2018 at 14:48

Richard Parnaby-King

15k11 gold badges75 silver badges131 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

TimBrownlaw Over a year ago

One little issue in the case where it returns false which is boolean so the controller needs to test $result before trying to extract either arrays or objects...

user9437856 Over a year ago

@Richard, Give me some time to check.

user9437856 Over a year ago

I just want to know why you removed the else condition? because I have 2 more conditions in the else. I am still getting the error Message: Undefined index: true

Richard Parnaby-King Over a year ago

Sorry, that's my personal ocd. There's nothing wrong leaving them in :)

user9437856 Over a year ago

@RichardParnaby-King, Wait a min, I am updating my code in the question for more clarity.

|

Kamiyabali Sunasara kamu · Accepted Answer · 2018-12-21 14:08:34Z

1

Yeah, ok this is solution for multiple conditions : 

    $this->db->select('*'); // Here your column name
    $this->db->from('members'); // Table name
    $this->db->where($login_access); // Here your condition
    $query = $this->db->get();

    $result = $query->result_array();
    $query = $this->db->get('customer'); // Your code above if condition
    $result = $query->result_array(); // Get data in array Format using **result_array()**
        if ($result) {
          if ($query->num_rows() > 0) { // You have to change condition from **$result['password_change ']** To **$query->num_rows() > 0**
            if ($result - > id == $id) { // Also change here condition from **$result - > id** To $result['id']
              return array('true' => 1, 'customer_id' => $result['customer_id']); //Also change here from **$result - > customer_id** To $result['customer_id']
            } else {
              return false;
            }
          } else {
            if () // some condition
          {
            return $result; // Also Change here from **return (array) $result** To $result
          } else {
            return false;
          }
        }
        } else {
          return false;
        }

edited Dec 21, 2018 at 14:08

answered Dec 21, 2018 at 8:57

Kamiyabali Sunasara kamu

405 bronze badges

5 Comments

user9437856 Over a year ago

I don't want to remove the else condition because I have more condition inside it.

Kamiyabali Sunasara kamu Over a year ago

@user9437856 by mistake I posted another answer. You can find it in this updated answer.

user9437856 Over a year ago

I am getting the error in this like if ($result['password_change'] == 0) { //Message: Undefined index: password_change

Kamiyabali Sunasara kamu Over a year ago

ook let me check i will give you solution in as fast as possible.

user9437856 Over a year ago

Thanks for answer again. You used if ($query->num_rows() > 0) { May I know why you used that? I am fetching all the records and there is one column which is "password_change" and I am checking that column only.

Collectives™ on Stack Overflow

Message: Cannot use object of type stdClass as array in CodeIgnitor

2 Answers 2

Alway return an Array

Always return an ArrayObject

6 Comments

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Alway return an Array

Always return an ArrayObject

6 Comments

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related