How to convert sql query to codeigniter query - mysql

Question

Can somebody help me convert this Sql Query
SELECT *
FROM customer c
LEFT JOIN customer_order co
ON c.customer_number = co.customer_number
AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid'
AND c.order_status = 'unserve'
AND co.cus_ord_no IS null
into Codeigniter query just like the image below for example

Answer 1

When query statements do not have clauses that need to change conditionally then using $this->db-query() is the way to go.
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co
ON c.customer_number=co.customer_number AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid' AND c.order_status='unserve' AND co.cus_ord_no IS null";
$query = $this->db->query($sql)->result();
echo json_encode($query);
It might be wise to include a check on the return from query() though because if it fails (returns false) then the call to result() will throw an exception. One way that can be handled is like this.
$query = $this->db->query($sql);
if($query !== FALSE)
{
echo json_encode($query->result());
return;
}
echo json_encode([]); // respond with an empty array
Query Builder (QB) is a nice tool, but it is often overkill. It adds a lot of overhead to create a string that literally is passed to $db->query(). If you know the string and it doesn't need to be restructured for some reason you don't need QB.
QB is most useful when you want to make changes to your query statement conditionally. Sorting might be one possible case.
if($order === 'desc'){
$this->db->order_by('somefield','DESC');
} else {
$this->db->order_by('somefield','ASC');
}
$results = $this->db
->where('other_field', "Foo")
->get('some_table')
->result();
So if the value of $order is 'desc' the query statement would be
SELECT * FROM some_table WHERE other_field = 'Foo' ORDER BY somefield 'DESC'
But if you insist on using Query Builder I believe this your answer
$query = $this->db
->join('customer_order co', "c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared')", 'left')
->where('c.customer_status','unpaid')
->where('c.order_status','unserve')
->where('co.cus_ord_no IS NULL')
->get('customer c');
//another variation on how to check that the query worked
$result = $query ? $query->result() : [];
echo json_encode($result);

Answer 2

You can do
public function view_customers()
{
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co ON c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared') WHERE c.customer_status='unpaid' AND c.order_status = 'unserve' AND co.cus_ord_no IS null";
return $this->db->query($sql)->result();
}
You can use row() for one output to object, or row_array() if one output but array. result() is multiple objects and result_array() is multiple arrays.
My way do usually is like this:
Controller:
public function view()
{
$this->load->model('My_Model');
$data = new stdclass;
$data->user_lists = $this->my_model->view_users(array('nationality'=>'AMERICAN'));
}
Model:
public function view_users($param = null) //no value passed
{
$condition = '1';
if (!empty($param)) { //Having this will trap if you input an array or not
foreach ($param as $key=>$val) {
$condition .= " AND {$key}='{$val}'"; //Use double quote so the data $key and $val will be read.
}
}
$sql = "SELECT * FROM users WHERE {$condition}"; //Use double quote so the data $condition will be read.
// Final out is this "SELECT * FROM users WHERE 1 AND nationality='AMERICAN'";
return $this->db->query($sql)->result();
}