what is going wrong in this query?

Question

public function insrt_into_aprlog($id)
{
    $query = 'insert into log_indent_apval (select id,indent_req_id,item_id,req_qty,approval_qty,approval_status,prod_categry,prod_type from tra_inent_rq_itm_dt where id ='.$id.')';
    $result =   $this->db->query($query);           
    $resultarr = $result->result_array();

}

my controller line:

$this->outpatient_model->insrt_into_aprlog($indent_id);

error : Call to a member function result_array() on a non-object

even i tried using in codeigniter model insert_batch(); still no use.

why are you using `result_array()` to insert data ?? – Drudge Rajen Mar 26 '16 at 08:20 — Drudge Rajen, Mar 26 '16 at 08:20
need to insert multiple lines.. – satish kilari Mar 26 '16 at 08:21 — satish kilari, Mar 26 '16 at 08:21
are you sure about your query .. ?? – santosh Mar 26 '16 at 08:32 — santosh, Mar 26 '16 at 08:32

score 3 · Accepted Answer · answered Mar 26 '16 at 08:27

result_array() is used to generate query result . You are trying to insert data in your table i guess, so you shouldn't use that . Try below code.

    public function insrt_into_aprlog($id)
    {
        $query = 'insert into log_indent_apval (select id,indent_req_id,item_id,req_qty,approval_qty,approval_status,prod_categry,prod_type from tra_inent_rq_itm_dt where id ='.$id.')';
return  $this->db->query($query);           

    }

Please visit this

score 3 · Answer 2 · edited May 23 '17 at 12:09

3

Reference Look into your query.

INSERT INTO table1 ( column1, column2, someInt, someVarChar )
SELECT  table2.column1, table2.column2, 8, 'some string etc.'
FROM    table2
WHERE   table2.ID = 7;

edited May 23 '17 at 12:09

Community

1
1

answered Mar 26 '16 at 08:34

Nabin Kunwar

1,965
14
29

score 0 · Answer 3 · answered Mar 26 '16 at 08:22

0

$query is being initialized to a string and then is accessed for an object. This is inconsistent for sure.

answered Mar 26 '16 at 08:22

rpy

3,953
2
20
31

score 0 · Answer 4 · answered Mar 26 '16 at 08:22

0

$query is a string, I guess you should use $result->result_array().

answered Mar 26 '16 at 08:22

Mark van Herpen

347
3
12

Yeah...i was a mistake i did it still no luck. – satish kilari Mar 26 '16 at 08:26
After looking again, I see that you are using an insert query which doesn't give you a result like the ones a select-query does. – Mark van Herpen Mar 26 '16 at 08:36

what is going wrong in this query?

4 Answers4