Array to string conversion error in Codeigniter

Question

I'm trying to display an average result from the database to the view but I keep getting this error:

A PHP Error was encountered

Severity: Notice

Message: Array to string conversion

Filename: views/resultview.php

Line Number: 38

Here is the code from the Controller:

$average['avg'] = $this->quiz->getAverage($quizid);

$this->load->view('resultview',array('quiz' => $quiz,
                                     'score' => $score, 
                                     'average_score' => $average));

The function from the model is the following:

   function getAverage($quiz) 
    {
    //get percentage from the database 

    $this->db->select_avg('score');
    $this->db->where('id', $quiz);
    $res = $this->db->get('userScoreQuiz');

    if ($res->num_rows() != 1) {
        // there should only be one row - anything else is an error
        return false;
    }
    return $res->result_array();
}

and the code from the view it is:

<h4> Avg. score on all previous attempts: <?php echo $average_score['avg'] ?>   %</h4>

I can't seam to find out why it does this.

Thank you for your help guys.

your function is returning an array return $res->result_array(); And $average_score['avg'] is an array that why you are getting this error — urfusion
– urfusion, Commented Jan 7, 2016 at 13:28
instead of this return false; pass return array(); and check isset() if necessary — Sundar
– Sundar, Commented Jan 7, 2016 at 13:32

Ilanus · Accepted Answer · 2016-01-07 14:20:10Z

5

That's too much of coding you got going on, Here is an Elegant solution:

function getAverage($quiz)
{
    //get percentage from the database
    $query = $this->db->select('AVG(score) as average_score')->from('userScoreQuiz')->where('id', $quiz)->get();
    return $query->row()->average_score;
}

For your view

$data['quiz']          = //fill this area
$data['average_score'] = $this->quiz->getAverage($quizid);
$data['score']         = //fill this area

$this->load->view('resultview', $data);

And they will be accessable as $quiz, $average_score, $score

edited Jan 7, 2016 at 14:20

answered Jan 7, 2016 at 13:59

Ilanus

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6 Comments

Nynaeve Over a year ago

the only problem is that I need the data from quiz and score to be passed to the view

Nynaeve Over a year ago

I did like you said and there is no error anymore it's just the avg doesn't display even though the scores are in the database

Ilanus Over a year ago

@Nynaeve with my code you have only 3 variables $quiz, $average_score, $score

Nynaeve Over a year ago

I have in my table scores that the previous users received and I want to display the avg score of all of them - that's the plan and the variables you have worked (two of them and but the average_score it's empty for some reason )

Nynaeve Over a year ago

I figured out why I had a problem with the query, your code was good btw, thanks so much for the help

|

Zahur Sh · Accepted Answer · 2016-01-07 13:37:10Z

3

function getAverage($quiz) 
{
    //get percentage from the database 

    $this->db->select_avg('score');
    $this->db->where('id', $quiz);
    $res = $this->db->get('userScoreQuiz');

    if ($res->num_rows() != 1) {
        // there should only be one row - anything else is an error
        return false;
    }
    return $res->row()->score;
}

from docs:

$this->db->select_avg('age');
$query = $this->db->get('members'); // Produces: SELECT AVG(age) as age FROM members

answered Jan 7, 2016 at 13:37

Zahur Sh

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4 Comments

Nynaeve Over a year ago

I don't get the error anymore but no score is displayed in the view even though it is in the table

Zahur Sh Over a year ago

Is it possible to you to make queries from phpmyadmin or something?

Zahur Sh Over a year ago

select avg(score) from userScoreQuiz where id=**yourid**

Nynaeve Over a year ago

I did that and I noticed a mistake in the query and that's why no result was showing in the view, thanks for the help

msvairam · Accepted Answer · 2016-01-07 13:32:55Z

0

replace the

echo $average_score['avg'];

to

print_r($average_score['avg'];

because that not one variable, that is set of array

answered Jan 7, 2016 at 13:32

msvairam

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3 Comments

urfusion Over a year ago

it will print array not the desire result.

msvairam Over a year ago

But we don't know about what desire result, then only I give that solution. If user see the array set, then he get that solution.

urfusion Over a year ago

this is obvious the average score Avg. score on all previous attempts: <?php echo $average_score['avg'] ?>

Feroz · Accepted Answer · 2016-01-07 15:37:10Z

0

codeigniter result_array() returns data as array. And array index starts with 0 index.

"$average_score['avg']" this should be undefined index. Try to access result array as $average_score['avg'][0]['score'] if you don't want to change in your function.

if you want to avoid extra array index then return results using row() instead on result_array().

answered Jan 7, 2016 at 15:37

Feroz

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