CodeIgniter: Function call result_array() gives error

Question

I'm getting this error for like a hour, what's this error on codeigniter

Here is my model:

i described property for fields too

class news_model extends CI_Model
{
    public function __construct()
    {
        parent::__construct();  
        $this->load->database();
    }

    public function get_news($id = false)
    {
        if ($id === false) {
            $query = $this->db->query("SELECT tbl_news.id,
                                            tbl_news.fa_name,
                                            tbl_news.en_name,
                                            tbl_news.fa_shrt_name,
                                            tbl_news.en_shrt_name,
                                            tbl_news.fa_text,
                                            tbl_news.en_text,
                                            tbl_news.image,
                                            tbl_news.grp_id,
                                            tbl_news_grp.fa_name FROM tbl_news JOIN tbl_news_grp ON tbl_news_grp.id = tbl_news.id ");

            return $query->result_array();
        }

        $query = $this->db->get_where('tbl_news',array('id' => $id));
        return $query->result_array();
    }
}

I get this error :

Fatal error: Call to a member function result_array() on a non-object in 
C:\xampp\htdocs\ipkoroosh\application\models\news_model.php on line 19

could you please give a bit more of code, looks like there is some issue with object making and useage — Arihant
– Arihant, Commented Apr 16, 2014 at 5:21
Did you loaded the database in constructor ??? $this->load->database(); — Pankaj Garg
– Pankaj Garg, Commented Apr 16, 2014 at 5:25
These fields are not selected with their related table names fa_shrt_name, en_shrt_name, fa_text, en_text, grp_id, image. They may be ambigous — Pankaj Garg
– Pankaj Garg, Commented Apr 16, 2014 at 5:30

Rakesh Sharma · Accepted Answer · 2014-04-16 05:47:00Z

1

How your where condition work you have no query above this and both condition will not run at a time so need to use else and refine query according you

public function get_news($id = false)
    {
        if ($id === false) {
            $query = $this->db->query("SELECT tblnews.*,tblnewsgrp.* FROM tbl_news tblnews JOIN tbl_news_grp tblnewsgrp ON tblnewsgrp.id = tblnews.id");

            return $query->result_array();
        }
        else {
           $query = $this->db->query("SELECT * FROM tbl_news WHERE id = '$id'");
            return $query->result_array();
        }
    }

edited Apr 16, 2014 at 5:47

answered Apr 16, 2014 at 5:24

Rakesh Sharma

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Community · Accepted Answer · 2017-05-23 10:28:02Z

0

Result has an optional $type parameter which decides what type of result is returned. By default ($type = "object"), it returns an object (result_object()). It can be set to "array", then it will return an array of result, that being equivalent of caling result_array(). The third version accepts a custom class to use as a result object.

/**
* Query result. Acts as a wrapper function for the following functions.
*
* @param string $type 'object', 'array' or a custom class name
* @return array
*/
public function result($type = 'object')
{
if ($type === 'array')
{
    return $this->result_array();
}
elseif ($type === 'object')
{
    return $this->result_object();
}
else
{
    return $this->custom_result_object($type);
}
}

Reference: Stackoverflow Answer

edited May 23, 2017 at 10:28

CommunityBot

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answered Apr 16, 2014 at 5:25

Arihant

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Comments

Pratik Butani · Accepted Answer · 2014-04-16 05:30:16Z

0

Change:

return $query->get()->result_array();

instead

return $query->result_array();

answered Apr 16, 2014 at 5:30

Pratik Butani

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1 Comment

hamid reza Over a year ago

i used $this->db->query() so i don't think that's gonna work here's the result by the way : Call to a member function get() on a non-object

Pankaj Garg · Accepted Answer · 2014-04-16 05:31:32Z

0

Try to print your SQL query by using $this->db->last_query(); and run the query on mysql console. There may be errors in SQL query.

Please check

answered Apr 16, 2014 at 5:31

Pankaj Garg

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Comments

Parimal · Accepted Answer · 2014-04-16 05:46:42Z

0

Check your query properly i think it is not giving result so this error is there.

Check it with this. $query->result();

edited Apr 16, 2014 at 5:46

answered Apr 16, 2014 at 5:27

Parimal

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