Error while using Error of error in PHP

Question

I want to do the same actions thru some variables. So I created the variables of variables. But it is throwing me error - "Invalid argument supplied for foreach()" when I am looping thru $$a. I have check the type of the variable. It is array. Then what is the error ?

    $edu_data = Json::decode($model->education);
    $exp_data = Json::decode($model->experience);
    $avail_data = Json::decode($model->availability);
    $docs_data = Json::decode($model->documents);

    $model_edu = new \admin\models\ApplicantEducation();
    $model_exp = new \admin\models\ApplicantExperience();
    $model_avail = new \admin\models\Availability();
    $model_cre = new \admin\models\Credential();

    $all = array('edu_data' => 'model_edu', 'exp_data' => 'model_exp', 'avail_data' => 'model_avail', 'docs_data' => 'model_cre');
    foreach ($all as $a => $s)
    {
        $arr = $$a;
        foreach ($arr as $v)
        {
            $$s->applicant_id = $applicant_id;
            foreach ($arr[1] as $k1 => $v1)
            {
                $$s->$k1 = $v[$k1];
            }
            $$s->save();
        }
    }

Imanuel · Accepted Answer · 2017-04-22 11:21:14Z

3

Your array does not contain your variables (e.g. $model_edu), but only their respective names as string values ('model_edu'). Edit: My bad, I didn't notice this is intentional.

I suggest using a function:

function process_data($model, $data, $applicant_id) {
    foreach ($data as $v) {
        $model->applicant_id = $applicant_id;
        foreach ($data[1] as $k1 => $v1)
        {
            $model->$k1 = $v[$k1];
        }
        $model->save();
    }
}

process_data($model_edu, $edu_data);
process_data($model_exp, $exp_data);
process_data($model_avail, $avail_data);
process_data($model_docs, $docs_data);

Your code will be more easily comprehendable.

Apart from that, you can debug your code like this to find out exactly where and when the error happens:

foreach ($all as $a => $s)
{
    $arr = $$a;

    var_dump($arr);

    foreach ($arr as $v)
    {
        $$s->applicant_id = $applicant_id;

        var_dump($arr[1]);

        foreach ($arr[1] as $k1 => $v1)
        {
            $$s->$k1 = $v[$k1];
        }
        $$s->save();
    }
}

See if this is the expected value and proceed from there on.
Find out if the reason is an unexpected value in one of your variables or if it is an error in the code logic.

edited Apr 22, 2017 at 11:21

answered Apr 22, 2017 at 10:36

Imanuel

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3 Comments

Ish Over a year ago

same error. I think you didnt understand my question. I want to loop thru $edu_data using variable of variable functionality of PHP

Ish Over a year ago

yes, I get all the expected value and the type is also array. but I get that error. I tired t debug this way already.

Ish Over a year ago

I know the way you told me to make the function and move ahead. But I want to understand the flaw in my approach.

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